id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-1100 | Inner Automorphism Group is Isomorphic to Quotient Group with Center | Let $G$ be a group.
Let $\Inn G$ be the inner automorphism group of $G$.
Let $\map Z G$ be the center of $G$.
Let $G / \map Z G$ be the quotient group of $G$ by $\map Z G$.
Then $G / \map Z G \cong \Inn G$. | Let $G$ be a group.
Let the mapping $\kappa: G \to \Inn G$ be defined as:
:$\map \kappa a = \kappa_a$
where $\kappa_a$ is the inner automorphism of $G$ given by $a$.
From Kernel of Inner Automorphism Group is Center:
:$\map \ker \kappa = \map Z G$
and also that:
:$\Img \kappa = \Inn G$
From the First Isomorphism Theore... | Let $G$ be a [[Definition:Group|group]].
Let $\Inn G$ be the [[Definition:Inner Automorphism Group|inner automorphism group]] of $G$.
Let $\map Z G$ be the [[Definition:Center of Group|center]] of $G$.
Let $G / \map Z G$ be the [[Definition:Quotient Group|quotient group]] of $G$ by $\map Z G$.
Then $G / \map Z G \... | Let $G$ be a [[Definition:Group|group]].
Let the [[Definition:Mapping|mapping]] $\kappa: G \to \Inn G$ be defined as:
:$\map \kappa a = \kappa_a$
where $\kappa_a$ is the [[Definition:Inner Automorphism|inner automorphism]] of $G$ given by $a$.
From [[Kernel of Inner Automorphism Group is Center]]:
:$\map \ker \kap... | Inner Automorphism Group is Isomorphic to Quotient Group with Center | https://proofwiki.org/wiki/Inner_Automorphism_Group_is_Isomorphic_to_Quotient_Group_with_Center | https://proofwiki.org/wiki/Inner_Automorphism_Group_is_Isomorphic_to_Quotient_Group_with_Center | [
"Quotient Groups",
"Inner Automorphisms",
"Group Isomorphisms"
] | [
"Definition:Group",
"Definition:Inner Automorphism Group",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Quotient Group"
] | [
"Definition:Group",
"Definition:Mapping",
"Definition:Inner Automorphism",
"Kernel of Inner Automorphism Group is Center",
"First Isomorphism Theorem/Groups"
] |
proofwiki-1101 | Conjugates of Elements in Centralizer | Let $G$ be a group.
Let $\map {C_G} a$ be the centralizer of $a$ in $G$.
Then $\forall g, h \in G: g a g^{-1} = h a h^{-1}$ {{iff}} $g$ and $h$ belong to the same left coset of $\map {C_G} a$. | The centralizer of $a$ in $G$ is defined as:
:$\map {C_G} a = \set {x \in G: x \circ a = a \circ x}$
Let $g, h \in G$.
Then:
{{begin-eqn}}
{{eqn | l = g a g^{-1}
| r = h a h^{-1}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = g^{-1} \paren {g a g^{-1} } h
| r = g^{-1} \paren {h a h^{-1} } h
|... | Let $G$ be a [[Definition:Group|group]].
Let $\map {C_G} a$ be the [[Definition:Centralizer of Group Element|centralizer]] of $a$ in $G$.
Then $\forall g, h \in G: g a g^{-1} = h a h^{-1}$ {{iff}} $g$ and $h$ belong to the same [[Definition:Left Coset|left coset]] of $\map {C_G} a$. | The [[Definition:Centralizer of Group Element|centralizer]] of $a$ in $G$ is defined as:
:$\map {C_G} a = \set {x \in G: x \circ a = a \circ x}$
Let $g, h \in G$.
Then:
{{begin-eqn}}
{{eqn | l = g a g^{-1}
| r = h a h^{-1}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = g^{-1} \paren {g a g^{-1} } h
... | Conjugates of Elements in Centralizer | https://proofwiki.org/wiki/Conjugates_of_Elements_in_Centralizer | https://proofwiki.org/wiki/Conjugates_of_Elements_in_Centralizer | [
"Conjugacy",
"Centralizers",
"Cosets"
] | [
"Definition:Group",
"Definition:Centralizer/Group Element",
"Definition:Coset/Left Coset"
] | [
"Definition:Centralizer/Group Element",
"Elements in Same Left Coset iff Product with Inverse in Subgroup",
"Definition:Coset/Left Coset"
] |
proofwiki-1102 | Number of Conjugates is Number of Cosets of Centralizer | Let $G$ be a group.
Let $\map {C_G} a$ be the centralizer of $a$ in $G$.
Then the number of different conjugates of $a$ in $G$ equals the number of different (left) cosets of $\map {C_G} a$:
:$\card {\conjclass a} = \index G {\map {C_G} a}$
where:
:$\conjclass a$ is the conjugacy class of $a$ in $G$
:$\index G {\map {C... | Let $x, y \in \conjclass a$.
By definition of $\conjclass a$:
:$x a x^{-1} = y a y^{-1}$
By Conjugates of Elements in Centralizer, this is the case {{iff}} $x$ and $y$ belong to the same left coset of $\map {C_G} a$.
Hence:
:$\card {\conjclass a} = \index G {\map {C_G} a}$
It follows from Lagrange's Theorem that:
:$\ca... | Let $G$ be a [[Definition:Group|group]].
Let $\map {C_G} a$ be the [[Definition:Centralizer of Group Element|centralizer]] of $a$ in $G$.
Then the number of different [[Definition:Conjugate of Group Element|conjugates]] of $a$ in $G$ equals the number of different [[Definition:Left Coset|(left) cosets]] of $\map {C_... | Let $x, y \in \conjclass a$.
By definition of $\conjclass a$:
:$x a x^{-1} = y a y^{-1}$
By [[Conjugates of Elements in Centralizer]], this is the case {{iff}} $x$ and $y$ belong to the same [[Definition:Left Coset|left coset]] of $\map {C_G} a$.
Hence:
:$\card {\conjclass a} = \index G {\map {C_G} a}$
It follows... | Number of Conjugates is Number of Cosets of Centralizer | https://proofwiki.org/wiki/Number_of_Conjugates_is_Number_of_Cosets_of_Centralizer | https://proofwiki.org/wiki/Number_of_Conjugates_is_Number_of_Cosets_of_Centralizer | [
"Conjugacy",
"Centralizers",
"Cosets"
] | [
"Definition:Group",
"Definition:Centralizer/Group Element",
"Definition:Conjugate (Group Theory)/Element",
"Definition:Coset/Left Coset",
"Definition:Conjugacy Class",
"Definition:Index of Subgroup"
] | [
"Conjugates of Elements in Centralizer",
"Definition:Coset/Left Coset",
"Lagrange's Theorem (Group Theory)"
] |
proofwiki-1103 | Size of Conjugacy Class is Index of Normalizer | Let $G$ be a group.
Let $x \in G$.
Let $\conjclass x$ be the conjugacy class of $x$ in $G$.
Let $\map {N_G} x$ be the normalizer of $x$ in $G$.
Let $\index G {\map {N_G} x}$ be the index of $\map {N_G} x$ in $G$.
The number of elements in $\conjclass x$ is $\index G {\map {N_G} x}$. | The number of elements in $\conjclass x$ is the number of conjugates of the set $\set x$.
From Number of Distinct Conjugate Subsets is Index of Normalizer, the number of distinct subsets of a $G$ which are conjugates of $S \subseteq G$ is $\index G {\map {N_G} S}$.
The result follows.
{{qed}} | Let $G$ be a [[Definition:Group|group]].
Let $x \in G$.
Let $\conjclass x$ be the [[Definition:Conjugacy Class|conjugacy class]] of $x$ in $G$.
Let $\map {N_G} x$ be the [[Definition:Normalizer|normalizer of $x$ in $G$]].
Let $\index G {\map {N_G} x}$ be the [[Definition:Index of Subgroup|index of $\map {N_G} x$ in... | The number of [[Definition:Element|elements]] in $\conjclass x$ is the number of [[Definition:Conjugate of Group Subset|conjugates]] of the set $\set x$.
From [[Number of Distinct Conjugate Subsets is Index of Normalizer]], the number of distinct [[Definition:Subset|subsets]] of a $G$ which are [[Definition:Conjugate ... | Size of Conjugacy Class is Index of Normalizer | https://proofwiki.org/wiki/Size_of_Conjugacy_Class_is_Index_of_Normalizer | https://proofwiki.org/wiki/Size_of_Conjugacy_Class_is_Index_of_Normalizer | [
"Conjugacy Classes",
"Normalizers"
] | [
"Definition:Group",
"Definition:Conjugacy Class",
"Definition:Normalizer",
"Definition:Index of Subgroup",
"Definition:Element"
] | [
"Definition:Element",
"Definition:Conjugate (Group Theory)/Subset",
"Number of Distinct Conjugate Subsets is Index of Normalizer",
"Definition:Subset",
"Definition:Conjugate (Group Theory)/Subset"
] |
proofwiki-1104 | Conjugacy Class of Element of Center is Singleton | Let $G$ be a group.
Let $\map Z G$ denote the center of $G$.
The elements of $\map Z G$ form singleton conjugacy classes, and the elements of $G \setminus \map Z G$ belong to multi-element conjugacy classes. | Let $\conjclass a$ be the conjugacy class of $a$ in $G$.
{{begin-eqn}}
{{eqn | l = a
| o = \in
| r = \map Z G
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall x \in G
| l = x a
| r = a x
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall x \in G
| l = x a x^{-1}... | Let $G$ be a [[Definition:Group|group]].
Let $\map Z G$ denote the [[Definition:Center of Group|center]] of $G$.
The [[Definition:Element|elements]] of $\map Z G$ form [[Definition:Singleton|singleton]] [[Definition:Conjugacy Class|conjugacy classes]], and the elements of $G \setminus \map Z G$ belong to multi-eleme... | Let $\conjclass a$ be the [[Definition:Conjugacy Class|conjugacy class]] of $a$ in $G$.
{{begin-eqn}}
{{eqn | l = a
| o = \in
| r = \map Z G
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall x \in G
| l = x a
| r = a x
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \fora... | Conjugacy Class of Element of Center is Singleton | https://proofwiki.org/wiki/Conjugacy_Class_of_Element_of_Center_is_Singleton | https://proofwiki.org/wiki/Conjugacy_Class_of_Element_of_Center_is_Singleton | [
"Conjugacy Classes",
"Centers of Groups",
"Singletons"
] | [
"Definition:Group",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Element",
"Definition:Singleton",
"Definition:Conjugacy Class",
"Definition:Conjugacy Class"
] | [
"Definition:Conjugacy Class"
] |
proofwiki-1105 | Conjugacy Class Equation | Let $G$ be a group.
Let $\order G$ denote the order of $G$.
Let $\map Z G$ denote the center of $G$.
Let $x \in G$.
Let $\map {N_G} x$ denote the normalizer of $x$ in $G$.
Let $\index G {\map {N_G} x}$ denote the index of $\map {N_G} x$ in $G$.
Let $m$ be the number of non-singleton conjugacy classes of $G$.
Let $x_j: ... | From Conjugacy Class of Element of Center is Singleton, all elements of $\map Z G$ form their own singleton conjugacy classes.
=== Abelian Group ===
Suppose $G$ is abelian.
Then by definition of abelian group:
:$\map Z G = G$
So there are as many conjugacy classes as there are elements in $\map Z G$ and hence in $G$.
S... | Let $G$ be a [[Definition:Group|group]].
Let $\order G$ denote the [[Definition:Order of Group|order]] of $G$.
Let $\map Z G$ denote the [[Definition:Center of Group|center]] of $G$.
Let $x \in G$.
Let $\map {N_G} x$ denote the [[Definition:Normalizer|normalizer of $x$ in $G$]].
Let $\index G {\map {N_G} x}$ denot... | From [[Conjugacy Class of Element of Center is Singleton]], all [[Definition:Element|elements]] of $\map Z G$ form their own [[Definition:Singleton|singleton]] [[Definition:Conjugacy Class|conjugacy classes]].
=== Abelian Group ===
Suppose $G$ is [[Definition:Abelian Group|abelian]].
Then by definition of [[Definit... | Conjugacy Class Equation/Proof 1 | https://proofwiki.org/wiki/Conjugacy_Class_Equation | https://proofwiki.org/wiki/Conjugacy_Class_Equation/Proof_1 | [
"Conjugacy Class Equation",
"Conjugacy",
"Normal Subgroups",
"Named Theorems",
"Normalizers",
"Conjugacy Classes"
] | [
"Definition:Group",
"Definition:Order of Structure",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Normalizer",
"Definition:Index of Subgroup",
"Definition:Singleton",
"Definition:Conjugacy Class",
"Definition:Element",
"Definition:Conjugacy Class"
] | [
"Conjugacy Class of Element of Center is Singleton",
"Definition:Element",
"Definition:Singleton",
"Definition:Conjugacy Class",
"Definition:Abelian Group",
"Definition:Abelian Group",
"Definition:Conjugacy Class",
"Definition:Element",
"Definition:Abelian Group",
"Conjugacy Class of Element of Ce... |
proofwiki-1106 | Conjugacy Class Equation | Let $G$ be a group.
Let $\order G$ denote the order of $G$.
Let $\map Z G$ denote the center of $G$.
Let $x \in G$.
Let $\map {N_G} x$ denote the normalizer of $x$ in $G$.
Let $\index G {\map {N_G} x}$ denote the index of $\map {N_G} x$ in $G$.
Let $m$ be the number of non-singleton conjugacy classes of $G$.
Let $x_j: ... | Let the distinct orbits of $G$ under the conjugacy action be:
:$\Orb {x_1}, \Orb {x_2}, \ldots, \Orb {x_s}$
Then from the Partition Equation:
:$\order G = \order {\Orb {x_1} } + \order {\Orb {x_2} } + \cdots + \order {\Orb {x_s} }$
From the Orbit-Stabilizer Theorem:
:$\order {\Orb {x_i} } \divides \order G, i = 1, \ldo... | Let $G$ be a [[Definition:Group|group]].
Let $\order G$ denote the [[Definition:Order of Group|order]] of $G$.
Let $\map Z G$ denote the [[Definition:Center of Group|center]] of $G$.
Let $x \in G$.
Let $\map {N_G} x$ denote the [[Definition:Normalizer|normalizer of $x$ in $G$]].
Let $\index G {\map {N_G} x}$ denot... | Let the distinct [[Definition:Orbit (Group Theory)|orbits]] of $G$ under the [[Definition:Conjugacy Action|conjugacy action]] be:
:$\Orb {x_1}, \Orb {x_2}, \ldots, \Orb {x_s}$
Then from the [[Partition Equation]]:
:$\order G = \order {\Orb {x_1} } + \order {\Orb {x_2} } + \cdots + \order {\Orb {x_s} }$
From the [[Or... | Conjugacy Class Equation/Proof 2 | https://proofwiki.org/wiki/Conjugacy_Class_Equation | https://proofwiki.org/wiki/Conjugacy_Class_Equation/Proof_2 | [
"Conjugacy Class Equation",
"Conjugacy",
"Normal Subgroups",
"Named Theorems",
"Normalizers",
"Conjugacy Classes"
] | [
"Definition:Group",
"Definition:Order of Structure",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Normalizer",
"Definition:Index of Subgroup",
"Definition:Singleton",
"Definition:Conjugacy Class",
"Definition:Element",
"Definition:Conjugacy Class"
] | [
"Definition:Orbit (Group Theory)",
"Definition:Conjugacy Action",
"Partition Equation",
"Orbit-Stabilizer Theorem",
"Definition:Conjugacy Action"
] |
proofwiki-1107 | Group of Order Prime Squared is Abelian | A group whose order is the square of a prime is abelian. | Let $G$ be a group of order $p^2$, where $p$ is prime.
Let $\map Z G$ be the center of $G$.
By Center of Group is Subgroup, $\map Z G$ is a subgroup of $G$.
By Lagrange's Theorem:
:$\order {\map Z G} \divides \order G$
It follows that $\order {\map Z G} = 1, p$ or $p^2$.
By Center of Group of Prime Power Order is Non-T... | A [[Definition:Group|group]] whose [[Definition:Order of Structure|order]] is the [[Definition:Square Number|square]] of a [[Definition:Prime Number|prime]] is [[Definition:Abelian Group|abelian]]. | Let $G$ be a [[Definition:Group|group]] of order $p^2$, where $p$ is [[Definition:Prime Number|prime]].
Let $\map Z G$ be the [[Definition:Center of Group|center]] of $G$.
By [[Center of Group is Subgroup]], $\map Z G$ is a [[Definition:Subgroup|subgroup]] of $G$.
By [[Lagrange's Theorem (Group Theory)|Lagrange's Th... | Group of Order Prime Squared is Abelian | https://proofwiki.org/wiki/Group_of_Order_Prime_Squared_is_Abelian | https://proofwiki.org/wiki/Group_of_Order_Prime_Squared_is_Abelian | [
"Order of Groups",
"Abelian Groups",
"P-Groups",
"Groups of Order p^2"
] | [
"Definition:Group",
"Definition:Order of Structure",
"Definition:Square Number",
"Definition:Prime Number",
"Definition:Abelian Group"
] | [
"Definition:Group",
"Definition:Prime Number",
"Definition:Center (Abstract Algebra)/Group",
"Center of Group is Subgroup",
"Definition:Subgroup",
"Lagrange's Theorem (Group Theory)",
"Center of Group of Prime Power Order is Non-Trivial",
"Lagrange's Theorem (Group Theory)",
"Definition:Non-Trivial ... |
proofwiki-1108 | Center of Group of Prime Power Order is Non-Trivial | Let $G$ be a group whose order is the power of a prime.
Then the center of $G$ is non-trivial:
:$\forall G: \order G = p^r: p \in \mathbb P, r \in \N_{>0}: \map Z G \ne \set e$ | Suppose $G$ is abelian.
By definition of abelian group:
:$\map Z G = G$
and the result is seen to be true as $G$ is itself non-trivial.
From Prime Group is Cyclic and Cyclic Group is Abelian, this will always be the case for $r = 1$.
So, suppose $G$ is non-abelian.
Thus $\map Z G \ne G$ and therefore $G \setminus \map ... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Order of Structure|order]] is the [[Definition:Prime Power|power of a prime]].
Then the [[Definition:Center of Group|center]] of $G$ is [[Definition:Trivial Group|non-trivial]]:
:$\forall G: \order G = p^r: p \in \mathbb P, r \in \N_{>0}: \map Z G \ne \set e... | Suppose $G$ is [[Definition:Abelian Group|abelian]].
By definition of [[Definition:Abelian Group|abelian group]]:
:$\map Z G = G$
and the result is seen to be true as $G$ is itself [[Definition:Non-Trivial Group|non-trivial]].
From [[Prime Group is Cyclic]] and [[Cyclic Group is Abelian]], this will always be the cas... | Center of Group of Prime Power Order is Non-Trivial | https://proofwiki.org/wiki/Center_of_Group_of_Prime_Power_Order_is_Non-Trivial | https://proofwiki.org/wiki/Center_of_Group_of_Prime_Power_Order_is_Non-Trivial | [
"Group Theory",
"P-Groups",
"Centers of Groups"
] | [
"Definition:Group",
"Definition:Order of Structure",
"Definition:Prime Power",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Trivial Group"
] | [
"Definition:Abelian Group",
"Definition:Abelian Group",
"Definition:Non-Trivial Group",
"Prime Group is Cyclic",
"Cyclic Group is Abelian",
"Definition:Abelian Group",
"Definition:Conjugacy Class",
"Definition:Set Partition",
"Conjugacy Class of Element of Center is Singleton",
"Conjugacy Class Eq... |
proofwiki-1109 | Center of Group of Order Prime Cubed | Let $G$ be a group of order $p^3$, where $p$ is a prime.
Let $\map Z G$ be the center of $G$.
Then $\order {\map Z G} \ne p^2$. | {{AimForCont}} $\order {\map Z G} = p^2$.
Then $\order {G / \map Z G} = p$.
From Prime Group is Cyclic it follows that $G / \map Z G$ is cyclic.
From Quotient of Group by Center Cyclic implies Abelian it follows that $G$ is abelian.
by definition of abelian group it follows that $\order {\map Z G} = p^3$.
The result fo... | Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Structure|order]] $p^3$, where $p$ is a [[Definition:Prime Number|prime]].
Let $\map Z G$ be the [[Definition:Center of Group|center]] of $G$.
Then $\order {\map Z G} \ne p^2$. | {{AimForCont}} $\order {\map Z G} = p^2$.
Then $\order {G / \map Z G} = p$.
From [[Prime Group is Cyclic]] it follows that $G / \map Z G$ is [[Definition:Cyclic Group|cyclic]].
From [[Quotient of Group by Center Cyclic implies Abelian]] it follows that $G$ is [[Definition:Abelian Group|abelian]].
by definition of [... | Center of Group of Order Prime Cubed | https://proofwiki.org/wiki/Center_of_Group_of_Order_Prime_Cubed | https://proofwiki.org/wiki/Center_of_Group_of_Order_Prime_Cubed | [
"Normal Subgroups",
"Centers of Groups",
"P-Groups"
] | [
"Definition:Group",
"Definition:Order of Structure",
"Definition:Prime Number",
"Definition:Center (Abstract Algebra)/Group"
] | [
"Prime Group is Cyclic",
"Definition:Cyclic Group",
"Quotient of Group by Center Cyclic implies Abelian",
"Definition:Abelian Group",
"Definition:Abelian Group",
"Proof by Contradiction",
"Category:Normal Subgroups",
"Category:Centers of Groups",
"Category:P-Groups"
] |
proofwiki-1110 | Prime Power Group has Non-Trivial Proper Normal Subgroup | Let $G$ be a group, whose identity is $e$, such that $\order G = p^n: n > 1, p \in \mathbb P$.
Then $G$ has a proper normal subgroup which is non-trivial. | From Center of Group is Normal Subgroup, $\map Z G \lhd G$.
By Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is non-trivial.
If $\map Z G$ is a proper subgroup, the proof is finished.
Otherwise, $\map Z G = G$.
Then $G$ is abelian by definition.
However, then any $a \in G: a \ne e$ generates a non-tri... | Let $G$ be a [[Definition:Group|group]], whose [[Definition:Identity Element|identity]] is $e$, such that $\order G = p^n: n > 1, p \in \mathbb P$.
Then $G$ has a [[Definition:Proper Subgroup|proper]] [[Definition:Normal Subgroup|normal subgroup]] which is [[Definition:Non-Trivial Group|non-trivial]]. | From [[Center of Group is Normal Subgroup]], $\map Z G \lhd G$.
By [[Center of Group of Prime Power Order is Non-Trivial]], $\map Z G$ is [[Definition:Non-Trivial Group|non-trivial]].
If $\map Z G$ is a [[Definition:Proper Subgroup|proper subgroup]], the proof is finished.
Otherwise, $\map Z G = G$.
Then $G$ is [[... | Prime Power Group has Non-Trivial Proper Normal Subgroup | https://proofwiki.org/wiki/Prime_Power_Group_has_Non-Trivial_Proper_Normal_Subgroup | https://proofwiki.org/wiki/Prime_Power_Group_has_Non-Trivial_Proper_Normal_Subgroup | [
"Normal Subgroups",
"P-Groups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Proper Subgroup",
"Definition:Normal Subgroup",
"Definition:Non-Trivial Group"
] | [
"Center of Group is Normal Subgroup",
"Center of Group of Prime Power Order is Non-Trivial",
"Definition:Non-Trivial Group",
"Definition:Proper Subgroup",
"Definition:Abelian Group",
"Definition:Non-Trivial Group",
"Definition:Normal Subgroup",
"Subgroup of Abelian Group is Normal",
"Subgroup of Fin... |
proofwiki-1111 | Composition Series of Group of Prime Power Order | Let $G$ be a group whose identity is $e$, and whose order is a prime power:
:$\order G = p^n, p \in \mathbb P, n \ge 1$
Then $G$ has a composition series:
:$\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$
such that:
:$\order {G_k} = p^k$
where:
:$G_k \lhd G_{k + 1}$
:$G_{k + 1} / G_k$ is cyclic and of order $p... | To be proved by induction on $n$.
Let $P_n$ be the proposition for $\order G = p^n$. | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$, and whose [[Definition:Order of Structure|order]] is a [[Definition:Prime Power|prime power]]:
:$\order G = p^n, p \in \mathbb P, n \ge 1$
Then $G$ has a [[Definition:Composition Series|composition series]]:
:$\set e = G_0 ... | To be proved by [[Second Principle of Mathematical Induction|induction]] on $n$.
Let $P_n$ be the [[Definition:Proposition|proposition]] for $\order G = p^n$. | Composition Series of Group of Prime Power Order | https://proofwiki.org/wiki/Composition_Series_of_Group_of_Prime_Power_Order | https://proofwiki.org/wiki/Composition_Series_of_Group_of_Prime_Power_Order | [
"Cyclic Groups",
"Composition Series",
"P-Groups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Structure",
"Definition:Prime Power",
"Definition:Composition Series",
"Definition:Cyclic Group",
"Definition:Order of Structure"
] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition"
] |
proofwiki-1112 | Product Rule for Counting | Let it be possible to choose an element $\alpha$ from a given set $S$ in $m$ different ways.
Let it be possible to choose an element $\beta$ from a given set $T$ in $n$ different ways.
Then the ordered pair $\tuple {\alpha, \beta}$ can be chosen from the cartesian product $S \times T$ in $m n$ different ways. | {{handwaving}}
The validity of this rule follows directly from the definition of multiplication of integers.
The product $a b$ (for $a, b \in \N_{>0}$) is the number of sequences $\sequence {A, B}$, where $A$ can be any one of $a$ items and $B$ can be any one of $b$ items.
{{qed}} | Let it be possible to choose an [[Definition:Element|element]] $\alpha$ from a given [[Definition:Set|set]] $S$ in $m$ different ways.
Let it be possible to choose an [[Definition:Element|element]] $\beta$ from a given [[Definition:Set|set]] $T$ in $n$ different ways.
Then the [[Definition:Ordered Pair|ordered pair]... | {{handwaving}}
The validity of this rule follows directly from the definition of [[Definition:Integer Multiplication|multiplication of integers]].
The product $a b$ (for $a, b \in \N_{>0}$) is the number of [[Definition:Sequence|sequences]] $\sequence {A, B}$, where $A$ can be any one of $a$ items and $B$ can be any ... | Product Rule for Counting | https://proofwiki.org/wiki/Product_Rule_for_Counting | https://proofwiki.org/wiki/Product_Rule_for_Counting | [
"Combinatorics",
"Counting Arguments",
"Product Rule for Counting"
] | [
"Definition:Element",
"Definition:Set",
"Definition:Element",
"Definition:Set",
"Definition:Ordered Pair",
"Definition:Cartesian Product"
] | [
"Definition:Multiplication/Integers",
"Definition:Sequence"
] |
proofwiki-1113 | Commutativity of Group Direct Product | Let $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$ be groups.
Let $\struct {G \times H, \circ}$ be the group direct product of $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$, where the operation $\circ$ is defined as:
:$\tuple {g_1, h_1} \circ \tuple {g_2, h_2} = \tuple {g_1 \circ_g g_2, h_1 \circ_h h_2}$
Let $\s... | The mapping $\theta: G \times H \to H \times G$ defined as:
:$\forall g \in G, h \in H: \map \theta {g, h} = \tuple {h, g}$
is to be shown to be a group homomorphism, and that $\theta$ is bijective, as follows: | Let $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$ be [[Definition:Group|groups]].
Let $\struct {G \times H, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$, where the operation $\circ$ is defined as:
:$\tuple {g_1, h_1} \circ \tuple {g_... | The [[Definition:Mapping|mapping]] $\theta: G \times H \to H \times G$ defined as:
:$\forall g \in G, h \in H: \map \theta {g, h} = \tuple {h, g}$
is to be shown to be a [[Definition:Group Homomorphism|group homomorphism]], and that $\theta$ is [[Definition:Bijection|bijective]], as follows: | Commutativity of Group Direct Product | https://proofwiki.org/wiki/Commutativity_of_Group_Direct_Product | https://proofwiki.org/wiki/Commutativity_of_Group_Direct_Product | [
"Group Direct Products"
] | [
"Definition:Group",
"Definition:Group Direct Product",
"Definition:Group Direct Product",
"Definition:Group Direct Product",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Mapping",
"Definition:Group Homomorphism",
"Definition:Bijection",
"Definition:Group Homomorphism"
] |
proofwiki-1114 | External Direct Product of Abelian Groups is Abelian Group | Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups.
Then the group direct product $\struct {G \times H, \circ}$ is abelian {{iff}} both $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are abelian. | Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups whose identities are $e_G$ and $e_H$ respectively.
From External Direct Product of Groups is Group, $\struct {G \times H, \circ}$ is indeed a group whose identity is $\tuple {e_G, e_H}$.
Suppose $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are both ab... | Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be [[Definition:Group|groups]].
Then the [[Definition:Group Direct Product|group direct product]] $\struct {G \times H, \circ}$ is [[Definition:Abelian Group|abelian]] {{iff}} both $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are [[Definition:Abelian Group|ab... | Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$ respectively.
From [[External Direct Product of Groups is Group]], $\struct {G \times H, \circ}$ is indeed a [[Definition:Group|group]] whose [[Definition:Identity E... | External Direct Product of Abelian Groups is Abelian Group | https://proofwiki.org/wiki/External_Direct_Product_of_Abelian_Groups_is_Abelian_Group | https://proofwiki.org/wiki/External_Direct_Product_of_Abelian_Groups_is_Abelian_Group | [
"Group Direct Products",
"Abelian Groups"
] | [
"Definition:Group",
"Definition:Group Direct Product",
"Definition:Abelian Group",
"Definition:Abelian Group"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"External Direct Product of Groups is Group",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Abelian Group",
"External Direct Product Commutativity",
"Definition:Abelian Group... |
proofwiki-1115 | Associativity of Group Direct Product | The group direct product $G \times \paren {H \times K}$ is (group) isomorphic to $\paren {G \times H} \times K$. | Let $G, H, K$ be groups.
The mapping $\theta: G \times \paren {H \times K} \to \paren {G \times H} \times K$ defined as:
:$\forall g \in G, h \in H, k \in K: \map \theta {\tuple {g, \tuple {h, k} } } = \tuple {\tuple {g, h}, k}$
is shown to be a group isomorphism, as follows: | The [[Definition:Group Direct Product|group direct product]] $G \times \paren {H \times K}$ is [[Definition:Group Isomorphism|(group) isomorphic]] to $\paren {G \times H} \times K$. | Let $G, H, K$ be [[Definition:Group|groups]].
The [[Definition:Mapping|mapping]] $\theta: G \times \paren {H \times K} \to \paren {G \times H} \times K$ defined as:
:$\forall g \in G, h \in H, k \in K: \map \theta {\tuple {g, \tuple {h, k} } } = \tuple {\tuple {g, h}, k}$
is shown to be a [[Definition:Group Isomorp... | Associativity of Group Direct Product | https://proofwiki.org/wiki/Associativity_of_Group_Direct_Product | https://proofwiki.org/wiki/Associativity_of_Group_Direct_Product | [
"Group Direct Products",
"Associativity"
] | [
"Definition:Group Direct Product",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Group",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] |
proofwiki-1116 | Group Direct Product of Cyclic Groups | Let $G$ and $H$ both be finite cyclic groups with orders $n = \order G$ and $m = \order H$ respectively.
Then:
:The group direct product $G \times H$ is cyclic
{{iff}}:
:$n$ and $m$ are coprime, that is:
::$n \perp m$ | Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively. | Let $G$ and $H$ both be [[Definition:Finite Group|finite]] [[Definition:Cyclic Group|cyclic groups]] with [[Definition:Order of Group|orders]] $n = \order G$ and $m = \order H$ respectively.
Then:
:The [[Definition:Group Direct Product|group direct product]] $G \times H$ is [[Definition:Cyclic Group|cyclic]]
{{iff}}:
... | Let $G$ and $H$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$ respectively. | Group Direct Product of Cyclic Groups | https://proofwiki.org/wiki/Group_Direct_Product_of_Cyclic_Groups | https://proofwiki.org/wiki/Group_Direct_Product_of_Cyclic_Groups | [
"Group Direct Products",
"Cyclic Groups",
"Group Direct Product of Cyclic Groups"
] | [
"Definition:Finite Group",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Group Direct Product",
"Definition:Cyclic Group",
"Definition:Coprime/Integers"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-1117 | Group Direct Product of Infinite Cyclic Groups | The group direct product of two infinite cyclic groups is not cyclic. | Let $G_1 = \struct {G_1, \circ_1}$ and $G_2 = \struct {G_2, \circ_2}$ be infinite cyclic groups.
Let $G = \struct {G, \circ} = G_1 \times G_2$.
Let $G_1 = \gen {g_1}, G_2 = \gen {g_2}$.
From Generators of Infinite Cyclic Group:
:$g_1$ and $g_1^{-1}$ are the only generators of $G_1$
:$g_2$ and $g_2^{-1}$ are the only ge... | The [[Definition:Group Direct Product|group direct product]] of two [[Definition:Infinite Cyclic Group|infinite cyclic groups]] is not [[Definition:Cyclic Group|cyclic]]. | Let $G_1 = \struct {G_1, \circ_1}$ and $G_2 = \struct {G_2, \circ_2}$ be [[Definition:Infinite Cyclic Group|infinite cyclic groups]].
Let $G = \struct {G, \circ} = G_1 \times G_2$.
Let $G_1 = \gen {g_1}, G_2 = \gen {g_2}$.
From [[Generators of Infinite Cyclic Group]]:
:$g_1$ and $g_1^{-1}$ are the only [[Definition:... | Group Direct Product of Infinite Cyclic Groups | https://proofwiki.org/wiki/Group_Direct_Product_of_Infinite_Cyclic_Groups | https://proofwiki.org/wiki/Group_Direct_Product_of_Infinite_Cyclic_Groups | [
"Group Direct Products",
"Infinite Cyclic Group"
] | [
"Definition:Group Direct Product",
"Definition:Infinite Cyclic Group",
"Definition:Cyclic Group"
] | [
"Definition:Infinite Cyclic Group",
"Generators of Infinite Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Cyclic Group/Generator",
"Definition:Generator of Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Infinite Cyclic Group",
"Definition:Order of Gr... |
proofwiki-1118 | Order of Group Element in Group Direct Product | Let $G$ and $H$ be finite groups.
Let $g \in G: \order g = m, h \in H: \order h = n$.
Then $\order {\tuple {g, h} }$ in $G \times H$ is $\lcm \set {m, n}$. | Let $G$ and $H$ be a groups whose identities are $e_G$ and $e_H$.
Let $l = \lcm \set {m, n}$.
From the definition of lowest common multiple:
:$\exists x, y \in \Z: l = m x = n y$
From the definition of order of an element:
:$g^m = e_G, h^n = e_H$
Thus:
{{begin-eqn}}
{{eqn | l = \tuple {g, h}^l
| r = \tuple {g^l, ... | Let $G$ and $H$ be [[Definition:Finite Group|finite groups]].
Let $g \in G: \order g = m, h \in H: \order h = n$.
Then $\order {\tuple {g, h} }$ in $G \times H$ is $\lcm \set {m, n}$. | Let $G$ and $H$ be a [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$.
Let $l = \lcm \set {m, n}$.
From the definition of [[Definition:Lowest Common Multiple of Integers|lowest common multiple]]:
:$\exists x, y \in \Z: l = m x = n y$
From the definition of [[Definitio... | Order of Group Element in Group Direct Product | https://proofwiki.org/wiki/Order_of_Group_Element_in_Group_Direct_Product | https://proofwiki.org/wiki/Order_of_Group_Element_in_Group_Direct_Product | [
"Group Direct Products"
] | [
"Definition:Finite Group"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Lowest Common Multiple/Integers",
"Definition:Order of Group Element",
"Absolute Value of Integer is not less than Divisors",
"Definition:Positive/Integer",
"Definition:Common Multiple",
"Definition:Lowest Com... |
proofwiki-1119 | Subgroup Product is Internal Group Direct Product iff Surjective | Let $G$ be a group.
Let $\sequence {H_n}$ be a sequence of subgroups of $G$.
Let $\ds \phi: \prod_{k \mathop = 1}^n H_k \to G$ be a mapping defined by:
:$\ds \map \phi {h_1, h_2, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$
Then $\phi$ is surjective {{iff}}:
:$\ds G = \prod_{k \mathop = 1}^n H_k$
That is, {{iff}} $G$ is... | === Necessary Condition ===
Let $\phi$ be a surjection.
Then $\Img \phi$ consists of all the products $\ds \prod_{k \mathop = 1}^n h_k$ such that:
:$\forall k \in \closedint 1 n: h_k \in H_k$
Thus, as $\phi$ is surjective, every element of $G$ must be representable in this form.
Using the notation:
:$\ds \prod_{k \math... | Let $G$ be a [[Definition:Group|group]].
Let $\sequence {H_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Subgroup|subgroups]] of $G$.
Let $\ds \phi: \prod_{k \mathop = 1}^n H_k \to G$ be a [[Definition:Mapping|mapping]] defined by:
:$\ds \map \phi {h_1, h_2, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$
T... | === Necessary Condition ===
Let $\phi$ be a [[Definition:Surjection|surjection]].
Then $\Img \phi$ consists of all the products $\ds \prod_{k \mathop = 1}^n h_k$ such that:
:$\forall k \in \closedint 1 n: h_k \in H_k$
Thus, as $\phi$ is [[Definition:Surjection|surjective]], every [[Definition:Element|element]] of $G... | Subgroup Product is Internal Group Direct Product iff Surjective | https://proofwiki.org/wiki/Subgroup_Product_is_Internal_Group_Direct_Product_iff_Surjective | https://proofwiki.org/wiki/Subgroup_Product_is_Internal_Group_Direct_Product_iff_Surjective | [
"Internal Group Direct Products"
] | [
"Definition:Group",
"Definition:Sequence",
"Definition:Subgroup",
"Definition:Mapping",
"Definition:Surjection",
"Definition:Internal Group Direct Product"
] | [
"Definition:Surjection",
"Definition:Surjection",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Surjection"
] |
proofwiki-1120 | Internal Group Direct Product is Injective | Let $G$ be a group whose identity is $e$.
Let $H_1, H_2$ be subgroups of $G$.
Let $\phi: H_1 \times H_2 \to G$ be a mapping defined by:
:$\map \phi {h_1, h_2} = h_1 h_2$
Then $\phi$ is injective {{iff}}:
:$H_1 \cap H_2 = \set e$ | === Necessary Condition ===
Let $\phi$ be an injection.
Let $\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$.
As $\phi$ is injective, this means that:
:$\tuple {h_1, h_2} = \tuple {k_1, k_2}$
and thus:
:$h_1 = k_1, h_2 = k_2$
From the definition of $\phi$, this means:
:$h_1 h_2 = k_1 k_2$
Thus, each element of $G$ that ca... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H_1, H_2$ be [[Definition:Subgroup|subgroups]] of $G$.
Let $\phi: H_1 \times H_2 \to G$ be a [[Definition:Mapping|mapping]] defined by:
:$\map \phi {h_1, h_2} = h_1 h_2$
Then $\phi$ is [[Definition:Injection|injectiv... | === Necessary Condition ===
Let $\phi$ be an [[Definition:Injection|injection]].
Let $\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$.
As $\phi$ is [[Definition:Injection|injective]], this means that:
:$\tuple {h_1, h_2} = \tuple {k_1, k_2}$
and thus:
:$h_1 = k_1, h_2 = k_2$
From the definition of $\phi$, this means:
... | Internal Group Direct Product is Injective | https://proofwiki.org/wiki/Internal_Group_Direct_Product_is_Injective | https://proofwiki.org/wiki/Internal_Group_Direct_Product_is_Injective | [
"Internal Group Direct Products",
"Injections"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Mapping",
"Definition:Injection"
] | [
"Definition:Injection",
"Definition:Injection",
"Definition:Injection"
] |
proofwiki-1121 | Internal Group Direct Product Isomorphism | Let $G$ be a group.
Let $H_1, H_2$ be subgroups of $G$.
Let $\phi: H_1 \times H_2 \to G$ be the mapping defined by $\map \phi {h_1, h_2} := h_1 h_2$.
If $\phi$ is a (group) isomorphism, then both $H_1$ and $H_2$ are normal subgroups of $G$. | $\phi$ is an isomorphism, so in particular a (group) homomorphism.
Thus by Induced Group Product is Homomorphism iff Commutative, every element of $H_1$ commutes with every element of $H_2$.
Now suppose $a \in G$.
As $\phi$ is an isomorphism, it follows that $\phi$ is surjective.
Thus by Subgroup Product is Internal Gr... | Let $G$ be a [[Definition:Group|group]].
Let $H_1, H_2$ be [[Definition:Subgroup|subgroups]] of $G$.
Let $\phi: H_1 \times H_2 \to G$ be the [[Definition:Mapping|mapping]] defined by $\map \phi {h_1, h_2} := h_1 h_2$.
If $\phi$ is a [[Definition:Group Isomorphism|(group) isomorphism]], then both $H_1$ and $H_2$ are... | $\phi$ is an [[Definition:Group Isomorphism|isomorphism]], so in particular a [[Definition:Group Homomorphism|(group) homomorphism]].
Thus by [[Induced Group Product is Homomorphism iff Commutative]], every [[Definition:Element|element]] of $H_1$ [[Definition:Commuting Elements|commutes]] with every [[Definition:Eleme... | Internal Group Direct Product Isomorphism | https://proofwiki.org/wiki/Internal_Group_Direct_Product_Isomorphism | https://proofwiki.org/wiki/Internal_Group_Direct_Product_Isomorphism | [
"Internal Group Direct Products",
"Group Isomorphisms",
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Normal Subgroup"
] | [
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group Homomorphism",
"Induced Group Product is Homomorphism iff Commutative",
"Definition:Element",
"Definition:Commutative/Elements",
"Definition:Element",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Defini... |
proofwiki-1122 | Internal Group Direct Product of Normal Subgroups | Let $G$ be a group whose identity is $e$.
Let $H_1$ and $H_2$ be normal subgroups of $G$ such that $H_1 \cap H_2 = \set e$.
Let $\phi: H_1 \times H_2 \to G$ be a mapping defined by $\map \phi {h_1, h_2} = h_1 h_2$.
Then $\phi$ is a (group) homomorphism. | Let $H_1$ and $H_2$ be normal subgroups of $G$.
Let $h_1 \in H_1, h_2 \in H_2$.
Consider $x \in G: x = h_1 h_2 h_1^{-1} h_2^{-1}$.
{{begin-eqn}}
{{eqn | l = x
| r = h_1 h_2 h_1^{-1} h_2^{-1}
| c =
}}
{{eqn | r = \paren {h_1 h_2 h_1^{-1} } h_2^{-1}
| c =
}}
{{end-eqn}}
As $H_2$ is normal, we have $h_... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H_1$ and $H_2$ be [[Definition:Normal Subgroup|normal subgroups]] of $G$ such that $H_1 \cap H_2 = \set e$.
Let $\phi: H_1 \times H_2 \to G$ be a [[Definition:Mapping|mapping]] defined by $\map \phi {h_1, h_2} = h_1 h_... | Let $H_1$ and $H_2$ be [[Definition:Normal Subgroup|normal subgroups]] of $G$.
Let $h_1 \in H_1, h_2 \in H_2$.
Consider $x \in G: x = h_1 h_2 h_1^{-1} h_2^{-1}$.
{{begin-eqn}}
{{eqn | l = x
| r = h_1 h_2 h_1^{-1} h_2^{-1}
| c =
}}
{{eqn | r = \paren {h_1 h_2 h_1^{-1} } h_2^{-1}
| c =
}}
{{end-eqn... | Internal Group Direct Product of Normal Subgroups | https://proofwiki.org/wiki/Internal_Group_Direct_Product_of_Normal_Subgroups | https://proofwiki.org/wiki/Internal_Group_Direct_Product_of_Normal_Subgroups | [
"Internal Group Direct Products",
"Normal Subgroups",
"Group Homomorphisms"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Normal Subgroup",
"Definition:Mapping",
"Definition:Group Homomorphism"
] | [
"Definition:Normal Subgroup",
"Definition:Normal Subgroup",
"Product of Commuting Elements with Inverses",
"Definition:Commutative/Elements",
"Definition:Element",
"Definition:Element",
"Definition:Commutative/Elements",
"Definition:Element",
"Induced Group Product is Homomorphism iff Commutative",
... |
proofwiki-1123 | Inclusion Mapping is Surjection iff Identity | Let $T$ be a set.
Let $S\subseteq T$ be a subset.
Let $i_S: S \to T$ be the inclusion mapping.
Then:
:$i_S: S \to T$ is surjective {{iff}} $i_S: S \to T = I_S: S \to S$
where $I_S: S \to S$ denotes the identity mapping on $S$.
Alternatively, this theorem can be worded as:
:$i_S: S \to S = I_S: S \to S$
It follows direc... | It is apparent from the definitions of both the inclusion mapping and the identity mapping that:
:$(1): \quad \Dom {i_S} = S = \Dom {I_S}$
:$(2): \quad \forall s \in S: \map {i_S} s = s = \map {I_S} s$ | Let $T$ be a [[Definition:Set|set]].
Let $S\subseteq T$ be a [[Definition:Subset|subset]].
Let $i_S: S \to T$ be the [[Definition:Inclusion Mapping|inclusion mapping]].
Then:
:$i_S: S \to T$ is [[Definition:Surjection|surjective]] {{iff}} $i_S: S \to T = I_S: S \to S$
where $I_S: S \to S$ denotes the [[Definition:I... | It is apparent from the definitions of both the [[Definition:Inclusion Mapping|inclusion mapping]] and the [[Definition:Identity Mapping|identity mapping]] that:
:$(1): \quad \Dom {i_S} = S = \Dom {I_S}$
:$(2): \quad \forall s \in S: \map {i_S} s = s = \map {I_S} s$ | Inclusion Mapping is Surjection iff Identity | https://proofwiki.org/wiki/Inclusion_Mapping_is_Surjection_iff_Identity | https://proofwiki.org/wiki/Inclusion_Mapping_is_Surjection_iff_Identity | [
"Surjections",
"Inclusion Mappings",
"Identity Mappings"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Inclusion Mapping",
"Definition:Surjection",
"Definition:Identity Mapping",
"Restriction of Mapping to Image is Surjection",
"Definition:Surjective Restriction",
"Definition:Identity Mapping"
] | [
"Definition:Inclusion Mapping",
"Definition:Identity Mapping"
] |
proofwiki-1124 | Cyclic Group of Order 6 | Let $C_n$ be the cyclic group of order $n$.
Then:
: $C_2 \times C_3 \cong C_6$
: $C_6$ is the internal group direct product of $C_2$ and $C_3$. | From Group Direct Product of Cyclic Groups noting that $2 \perp 3$:
:$C_2 \times C_3 \cong C_6$
* $C_6$ is the internal group direct product of $C_2$ and $C_3$:
Let $\left({C_6, \circ}\right) = \left \langle {x} \right \rangle$, let $\left({C_2, \circ}\right) = \left \langle {x^3} \right \rangle$, and let $\left({C_3, ... | Let $C_n$ be the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $n$.
Then:
: $C_2 \times C_3 \cong C_6$
: $C_6$ is the [[Definition:Internal Group Direct Product|internal group direct product]] of $C_2$ and $C_3$. | From [[Group Direct Product of Cyclic Groups]] noting that $2 \perp 3$:
:$C_2 \times C_3 \cong C_6$
* $C_6$ is the [[Definition:Internal Group Direct Product|internal group direct product]] of $C_2$ and $C_3$:
Let $\left({C_6, \circ}\right) = \left \langle {x} \right \rangle$, let $\left({C_2, \circ}\right) = \left ... | Cyclic Group of Order 6 | https://proofwiki.org/wiki/Cyclic_Group_of_Order_6 | https://proofwiki.org/wiki/Cyclic_Group_of_Order_6 | [
"Examples of Cyclic Groups",
"Groups of Order 6"
] | [
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Internal Group Direct Product"
] | [
"Group Direct Product of Cyclic Groups",
"Definition:Internal Group Direct Product",
"Subgroup of Abelian Group is Normal",
"Category:Examples of Cyclic Groups",
"Category:Groups of Order 6"
] |
proofwiki-1125 | Internal Direct Product Generated by Subgroups | Let $G$ be a group whose identity is $e$.
Let $\sequence {H_n}$ be a sequence of subgroups of $G$.
Then:
:the subgroup generated by $\ds \bigcup_{k \mathop = 1}^n H_k$ is the internal group direct product of $\sequence {H_n}$
{{iff}}:
:$\sequence {H_n}$ is an independent sequence of subgroups such that every element of... | In the following, the notation $\closedint m n$ is to be understood to mean a (closed) integer interval:
:$\closedint m n := \set {x \in \Z: m \le x \le n}$
for $m, n \in \Z$.
For each $k \in \closedint 1 n$, let $\ds L_k = \prod_{j \mathop = 1}^k H_j$ be the cartesian product of the subgroups $H_1, H_2, \ldots, H_k$ o... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\sequence {H_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Subgroup|subgroups]] of $G$.
Then:
:the [[Definition:Generator of Subgroup|subgroup generated by $\ds \bigcup_{k \mathop = 1}^n H_k$]] is the [[De... | In the following, the notation $\closedint m n$ is to be understood to mean a [[Definition:Integer Interval|(closed) integer interval]]:
:$\closedint m n := \set {x \in \Z: m \le x \le n}$
for $m, n \in \Z$.
For each $k \in \closedint 1 n$, let $\ds L_k = \prod_{j \mathop = 1}^k H_j$ be the [[Definition:Finite Cartes... | Internal Direct Product Generated by Subgroups | https://proofwiki.org/wiki/Internal_Direct_Product_Generated_by_Subgroups | https://proofwiki.org/wiki/Internal_Direct_Product_Generated_by_Subgroups | [
"Internal Group Direct Products"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Sequence",
"Definition:Subgroup",
"Definition:Generator of Subgroup",
"Definition:Internal Group Direct Product",
"Definition:Independent Subgroups",
"Definition:Element",
"Definition:Commutative/Elements",
... | [
"Definition:Closed Interval/Integer Interval",
"Definition:Cartesian Product/Finite",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Subgroup"
] |
proofwiki-1126 | Internal Group Direct Product Commutativity | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
Let $\struct {G, \circ}$ be the internal group direct product of $H$ and $K$.
Then:
:$\forall h \in H, k \in K: h \circ k = k \circ h$ | Let $G$ be the internal group direct product of $H$ and $K$.
Then by definition the mapping:
:$\phi: H \times K \to G: \map \phi {h, k} = h \circ k$
is a (group) isomorphism from the (external) direct product $\struct {H, \circ \restriction_H} \times \struct {K, \circ \restriction_K}$ onto $\struct {G, \circ}$.
Let the... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
Let $\struct {G, \circ}$ be the [[Definition:Internal Group Direct Product|internal group direct product]] of $H$ and $K$.
Then:
:$\forall h \i... | Let $G$ be the [[Definition:Internal Group Direct Product|internal group direct product]] of $H$ and $K$.
Then by definition the [[Definition:Mapping|mapping]]:
:$\phi: H \times K \to G: \map \phi {h, k} = h \circ k$
is a [[Definition:Group Isomorphism|(group) isomorphism]] from the [[Definition:External Direct Produc... | Internal Group Direct Product Commutativity/Proof 1 | https://proofwiki.org/wiki/Internal_Group_Direct_Product_Commutativity | https://proofwiki.org/wiki/Internal_Group_Direct_Product_Commutativity/Proof_1 | [
"Internal Group Direct Products",
"Examples of Commutative Operations",
"Internal Group Direct Product Commutativity"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Internal Group Direct Product"
] | [
"Definition:Internal Group Direct Product",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:External Direct Product",
"Definition:Operation Induced by Direct Product",
"Definition:Homomorphism (Abstract Algebra)"
] |
proofwiki-1127 | Internal Group Direct Product Commutativity | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
Let $\struct {G, \circ}$ be the internal group direct product of $H$ and $K$.
Then:
:$\forall h \in H, k \in K: h \circ k = k \circ h$ | Let $\sqbrk {x, y}$ denote the commutator of $x, y \in G$:
:$\sqbrk {x, y} := x^{-1} y^{-1} x y$
We have that:
{{begin-eqn}}
{{eqn | n = 1
| l = y x \sqbrk {x, y}
| r = y x x^{-1} y^{-1} x y
| c = {{Defof|Commutator of Group Elements}}
}}
{{eqn | r = y y^{-1} x y
| c = {{Group-axiom|3}}
}}
{{eqn... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
Let $\struct {G, \circ}$ be the [[Definition:Internal Group Direct Product|internal group direct product]] of $H$ and $K$.
Then:
:$\forall h \i... | Let $\sqbrk {x, y}$ denote the [[Definition:Commutator of Group Elements|commutator]] of $x, y \in G$:
:$\sqbrk {x, y} := x^{-1} y^{-1} x y$
We have that:
{{begin-eqn}}
{{eqn | n = 1
| l = y x \sqbrk {x, y}
| r = y x x^{-1} y^{-1} x y
| c = {{Defof|Commutator of Group Elements}}
}}
{{eqn | r = y y^... | Internal Group Direct Product Commutativity/Proof 2 | https://proofwiki.org/wiki/Internal_Group_Direct_Product_Commutativity | https://proofwiki.org/wiki/Internal_Group_Direct_Product_Commutativity/Proof_2 | [
"Internal Group Direct Products",
"Examples of Commutative Operations",
"Internal Group Direct Product Commutativity"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Internal Group Direct Product"
] | [
"Definition:Commutator/Group",
"Commutator is Identity iff Elements Commute"
] |
proofwiki-1128 | Internal and External Group Direct Products are Isomorphic | Let $G$ be a group whose identity is $e$.
Then $G$ is the (external) group direct product of $G_1, G_2, \ldots, G_n$ {{iff}} $G$ is the internal group direct product of $N_1, N_2, \ldots, N_n$ such that:
:$\forall i \in \N_n: N_i \cong G_i$
where:
:$\cong$ denotes (group) isomorphism
:$\N_n$ denotes $\set {1, 2, \ldots... | === Necessary Condition ===
Let $G$ be the external direct product of groups $G_1, G_2, \ldots, G_n$.
For all $i \in \N_n$, let $N_i$ be defined as the set:
:$N_i = \set e \times \cdots \times \set e \times G_i \times \set e \times \cdots \times \set e$
of elements which have entry $e$ everywhere except possibly in the... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Then $G$ is the [[Definition:Group Direct Product/Finite Product|(external) group direct product]] of $G_1, G_2, \ldots, G_n$ {{iff}} $G$ is the [[Definition:Internal Group Direct Product/General Definition|internal group d... | === Necessary Condition ===
Let $G$ be the [[Definition:Group Direct Product/Finite Product|external direct product]] of [[Definition:Group|groups]] $G_1, G_2, \ldots, G_n$.
For all $i \in \N_n$, let $N_i$ be defined as the set:
:$N_i = \set e \times \cdots \times \set e \times G_i \times \set e \times \cdots \times... | Internal and External Group Direct Products are Isomorphic | https://proofwiki.org/wiki/Internal_and_External_Group_Direct_Products_are_Isomorphic | https://proofwiki.org/wiki/Internal_and_External_Group_Direct_Products_are_Isomorphic | [
"Group Direct Products",
"Internal Group Direct Products",
"Group Isomorphisms"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Direct Product/Finite Product",
"Definition:Internal Group Direct Product/General Definition",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Group Direct Product/Finite Product",
"Definition:Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Normal Subgroup",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Normal Subgroup"
] |
proofwiki-1129 | Direct Product of Central Subgroup with Inverse Isomorphism is Central Subgroup | Let $G$ and $H$ be groups.
Let $\map Z G$ denote the center of $G$.
Let $Z$ and $W$ be central subgroups of $G$ and $H$ respectively.
Let:
:$Z \cong W$
where $\cong$ denotes isomorphism.
Let such a group isomorphism be $\theta: Z \to W$.
Let $X$ be the set defined as:
:$X = \set {\tuple {x, \map \theta x^{-1} }: x \in ... | First note that by Group Homomorphism Preserves Inverses:
:$\forall x \in G: \map \phi {x^{-1} } = \paren {\map \phi x}^{-1} = \map \phi x^{-1}$
and so there is no amphiboly in the notation used.
It is established that $\tuple {x, \map \theta x^{-1} } \in G \times H$:
:$x \in G$
:$\map \theta x \in H$ and so $\map \the... | Let $G$ and $H$ be [[Definition:Group|groups]].
Let $\map Z G$ denote the [[Definition:Center of Group|center]] of $G$.
Let $Z$ and $W$ be [[Definition:Central Subgroup|central subgroups]] of $G$ and $H$ respectively.
Let:
:$Z \cong W$
where $\cong$ denotes [[Definition:Group Isomorphism|isomorphism]].
Let such a ... | First note that by [[Group Homomorphism Preserves Inverses]]:
:$\forall x \in G: \map \phi {x^{-1} } = \paren {\map \phi x}^{-1} = \map \phi x^{-1}$
and so there is no [[Definition:Amphiboly|amphiboly]] in the notation used.
It is established that $\tuple {x, \map \theta x^{-1} } \in G \times H$:
:$x \in G$
:$\map \... | Direct Product of Central Subgroup with Inverse Isomorphism is Central Subgroup | https://proofwiki.org/wiki/Direct_Product_of_Central_Subgroup_with_Inverse_Isomorphism_is_Central_Subgroup | https://proofwiki.org/wiki/Direct_Product_of_Central_Subgroup_with_Inverse_Isomorphism_is_Central_Subgroup | [
"Quotient Groups",
"Group Isomorphisms",
"Central Subgroups"
] | [
"Definition:Group",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Central Subgroup",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Set",
"Definition:Central Subgroup"
] | [
"Group Homomorphism Preserves Inverses",
"Definition:Amphiboly",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Group Homomorphism Preserves Identity",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:I... |
proofwiki-1130 | Pullback of Quotient Group Isomorphism is Subgroup | Let $\struct {G, \circ}$ be a group whose identity element is $e_G$.
Let $\struct {H, *}$ be a group whose identity element is $e_H$.
Let $N \lhd G, K \lhd H$ be normal subgroups of $G$ and $H$ respectively.
Let:
:$G / N \cong H / K$
where:
:$G / N$ denotes the quotient of $G$ by $N$
:$\cong$ denotes group isomorphism.... | This result is proved by an application of the Two-Step Subgroup Test: | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $e_G$.
Let $\struct {H, *}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $e_H$.
Let $N \lhd G, K \lhd H$ be [[Definition:Normal Subgroup|normal subgroups]] of... | This result is proved by an application of the [[Two-Step Subgroup Test]]: | Pullback of Quotient Group Isomorphism is Subgroup | https://proofwiki.org/wiki/Pullback_of_Quotient_Group_Isomorphism_is_Subgroup | https://proofwiki.org/wiki/Pullback_of_Quotient_Group_Isomorphism_is_Subgroup | [
"Pullbacks of Quotient Group Isomorphisms",
"Quotient Groups",
"Group Isomorphisms",
"Subgroups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Normal Subgroup",
"Definition:Quotient Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Isomor... | [
"Two-Step Subgroup Test",
"Two-Step Subgroup Test"
] |
proofwiki-1131 | Group/Examples/x+y over 1+xy | Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to \R$ be the binary operation defined as:
:$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
The algebraic structure $\struct {G, \circ}$ is a group. | Let $-1 < x, y, z < 1$.
We check the group axioms in turn: | Let $G = \set {x \in \R: -1 < x < 1}$ be the [[Definition:Set|set]] of all [[Definition:Real Number|real numbers]] whose [[Definition:Absolute Value|absolute value]] is less than $1$.
Let $\circ: G \times G \to \R$ be the [[Definition:Binary Operation|binary operation]] defined as:
:$\forall x, y \in G: x \circ y = \d... | Let $-1 < x, y, z < 1$.
We check the [[Axiom:Group Axioms|group axioms]] in turn: | Group/Examples/x+y over 1+xy | https://proofwiki.org/wiki/Group/Examples/x+y_over_1+xy | https://proofwiki.org/wiki/Group/Examples/x+y_over_1+xy | [
"Examples of Groups",
"Examples of Groups/x+y over 1+xy"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Absolute Value",
"Definition:Operation/Binary Operation",
"Definition:Algebraic Structure/One Operation",
"Definition:Group"
] | [
"Axiom:Group Axioms"
] |
proofwiki-1132 | Group/Examples/x+y over 1+xy | Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to \R$ be the binary operation defined as:
:$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
The algebraic structure $\struct {G, \circ}$ is a group. | To prove $G$ is isomorphic to $\struct {\R, +}$, we need to find a bijective homorphism $\phi: \openint {-1} 1 \to \R$:
:$\forall x, y \in G: \map \phi {x \circ y} = \map \phi x + \map \phi y$
From Group Examples: $\dfrac {x + y} {1 + x y}$:
:the identity element of $G$ is $0$
:the inverse of $x$ in $G$ is $-x$.
This a... | Let $G = \set {x \in \R: -1 < x < 1}$ be the [[Definition:Set|set]] of all [[Definition:Real Number|real numbers]] whose [[Definition:Absolute Value|absolute value]] is less than $1$.
Let $\circ: G \times G \to \R$ be the [[Definition:Binary Operation|binary operation]] defined as:
:$\forall x, y \in G: x \circ y = \d... | To prove $G$ is [[Definition:Isomorphism|isomorphic]] to $\struct {\R, +}$, we need to find a [[Definition:Bijection|bijective]] [[Definition:Group Homomorphism|homorphism]] $\phi: \openint {-1} 1 \to \R$:
:$\forall x, y \in G: \map \phi {x \circ y} = \map \phi x + \map \phi y$
From [[Group/Examples/x+y over 1+xy|Gro... | Group/Examples/x+y over 1+xy/Isomorphic to Real Numbers/Proof 1 | https://proofwiki.org/wiki/Group/Examples/x+y_over_1+xy | https://proofwiki.org/wiki/Group/Examples/x+y_over_1+xy/Isomorphic_to_Real_Numbers/Proof_1 | [
"Examples of Groups",
"Examples of Groups/x+y over 1+xy"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Absolute Value",
"Definition:Operation/Binary Operation",
"Definition:Algebraic Structure/One Operation",
"Definition:Group"
] | [
"Definition:Isomorphism",
"Definition:Bijection",
"Definition:Group Homomorphism",
"Group/Examples/x+y over 1+xy",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Odd Function",
"Definition:Real Interval/Open",
"Definition:Inv... |
proofwiki-1133 | Group/Examples/x+y over 1+xy | Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to \R$ be the binary operation defined as:
:$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
The algebraic structure $\struct {G, \circ}$ is a group. | To prove $G$ is isomorphic to $\struct {\R, +}$, it is sufficient to find a bijective homorphism $\phi: \to \R \to G$:
:$\forall x, y \in G: \map \phi {x + y} = \map \phi x \circ \map \phi y$
From Group Examples: $\dfrac {x + y} {1 + x y}$:
:the identity element of $G$ is $0$
:the inverse of $x$ in $G$ is $-x$.
This al... | Let $G = \set {x \in \R: -1 < x < 1}$ be the [[Definition:Set|set]] of all [[Definition:Real Number|real numbers]] whose [[Definition:Absolute Value|absolute value]] is less than $1$.
Let $\circ: G \times G \to \R$ be the [[Definition:Binary Operation|binary operation]] defined as:
:$\forall x, y \in G: x \circ y = \d... | To prove $G$ is [[Definition:Isomorphism|isomorphic]] to $\struct {\R, +}$, it is sufficient to find a [[Definition:Bijection|bijective]] [[Definition:Group Homomorphism|homorphism]] $\phi: \to \R \to G$:
:$\forall x, y \in G: \map \phi {x + y} = \map \phi x \circ \map \phi y$
From [[Group/Examples/x+y over 1+xy|Grou... | Group/Examples/x+y over 1+xy/Isomorphic to Real Numbers/Proof 2 | https://proofwiki.org/wiki/Group/Examples/x+y_over_1+xy | https://proofwiki.org/wiki/Group/Examples/x+y_over_1+xy/Isomorphic_to_Real_Numbers/Proof_2 | [
"Examples of Groups",
"Examples of Groups/x+y over 1+xy"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Absolute Value",
"Definition:Operation/Binary Operation",
"Definition:Algebraic Structure/One Operation",
"Definition:Group"
] | [
"Definition:Isomorphism",
"Definition:Bijection",
"Definition:Group Homomorphism",
"Group/Examples/x+y over 1+xy",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Odd Function",
"Definition:Image (Set Theory)/Mapping/Mapping",
... |
proofwiki-1134 | Group/Examples/inv x = 1 - x | Let $S = \set {x \in \R: 0 < x < 1}$.
Then an operation $\circ$ can be found such that $\struct {S, \circ}$ is a group such that the inverse of $x \in S$ is $1 - x$. | Define $f: \openint 0 1 \to \R$ by:
:$\map f x := \map \ln {\dfrac {1 - x} x}$
Let us show that $f$ is a bijection by constructing an inverse mapping $g: \R \to \openint 0 1$:
:$\map g z := \dfrac 1 {1 + \exp z}$ | Let $S = \set {x \in \R: 0 < x < 1}$.
Then an operation $\circ$ can be found such that $\struct {S, \circ}$ is a [[Definition:Group|group]] such that the [[Definition:Inverse Element|inverse]] of $x \in S$ is $1 - x$. | Define $f: \openint 0 1 \to \R$ by:
:$\map f x := \map \ln {\dfrac {1 - x} x}$
Let us show that $f$ is a [[Definition:Bijection|bijection]] by constructing an [[Definition:Inverse Mapping|inverse mapping]] $g: \R \to \openint 0 1$:
:$\map g z := \dfrac 1 {1 + \exp z}$ | Group/Examples/inv x = 1 - x | https://proofwiki.org/wiki/Group/Examples/inv_x_=_1_-_x | https://proofwiki.org/wiki/Group/Examples/inv_x_=_1_-_x | [
"Examples of Groups",
"Group/Examples/inv x = 1 - x"
] | [
"Definition:Group",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Bijection",
"Definition:Inverse Mapping",
"Definition:Bijection"
] |
proofwiki-1135 | Group/Examples/Self-Inverse and Cancellable Elements | Let $S$ be a set with an operation which assigns to each $\tuple {a, b} \in S \times S$ an element $a \ast b \in S$ such that:
:$(1): \quad \exists e \in S: a \ast b = e \iff a = b$
:$(2): \quad \forall a, b, c \in S: \paren {a \ast c} \ast \paren {b \ast c} = a \ast b$
Then $\struct {S, \circ}$ is a group, where $\cir... | We verify the group axioms, in the following order (for convenience): | Let $S$ be a [[Definition:Set|set]] with an [[Definition:Binary Operation|operation]] which assigns to each $\tuple {a, b} \in S \times S$ an element $a \ast b \in S$ such that:
:$(1): \quad \exists e \in S: a \ast b = e \iff a = b$
:$(2): \quad \forall a, b, c \in S: \paren {a \ast c} \ast \paren {b \ast c} = a \ast ... | We verify the [[Axiom:Group Axioms|group axioms]], in the following order (for convenience): | Group/Examples/Self-Inverse and Cancellable Elements | https://proofwiki.org/wiki/Group/Examples/Self-Inverse_and_Cancellable_Elements | https://proofwiki.org/wiki/Group/Examples/Self-Inverse_and_Cancellable_Elements | [
"Examples of Groups"
] | [
"Definition:Set",
"Definition:Operation/Binary Operation",
"Definition:Group"
] | [
"Axiom:Group Axioms"
] |
proofwiki-1136 | Complex Numbers under Addition form Infinite Abelian Group | Let $\C$ be the set of complex numbers.
The structure $\struct {\C, +}$ is an infinite abelian group. | From Complex Numbers under Addition form Group, $\struct {\C, +}$ is a group.
Then:
:Complex Addition is Commutative
and:
:Complex Numbers are Uncountable.
{{qed}} | Let $\C$ be the set of [[Definition:Complex Number|complex numbers]].
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\C, +}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]]. | From [[Complex Numbers under Addition form Group]], $\struct {\C, +}$ is a [[Definition:Group|group]].
Then:
:[[Complex Addition is Commutative]]
and:
:[[Complex Numbers are Uncountable]].
{{qed}} | Complex Numbers under Addition form Infinite Abelian Group | https://proofwiki.org/wiki/Complex_Numbers_under_Addition_form_Infinite_Abelian_Group | https://proofwiki.org/wiki/Complex_Numbers_under_Addition_form_Infinite_Abelian_Group | [
"Additive Group of Complex Numbers",
"Complex Addition",
"Examples of Abelian Groups",
"Examples of Infinite Groups"
] | [
"Definition:Complex Number",
"Definition:Algebraic Structure/One Operation",
"Definition:Infinite Group",
"Definition:Abelian Group"
] | [
"Complex Numbers under Addition form Group",
"Definition:Group",
"Complex Addition is Commutative",
"Complex Numbers are Uncountable"
] |
proofwiki-1137 | Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group | Let $\C_{\ne 0}$ be the set of complex numbers without zero, that is:
:$\C_{\ne 0} = \C \setminus \set 0$
The structure $\struct {\C_{\ne 0}, \times}$ is an infinite abelian group. | From Non-Zero Complex Numbers under Multiplication form Group, $\struct {\C_{\ne 0}, \times}$ is a group.
Then we have:
:Complex Multiplication is Commutative
and:
:Complex Numbers are Uncountable.
{{qed}} | Let $\C_{\ne 0}$ be the set of [[Definition:Complex Number|complex numbers]] without [[Definition:Zero (Number)|zero]], that is:
:$\C_{\ne 0} = \C \setminus \set 0$
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\C_{\ne 0}, \times}$ is an [[Definition:Infinite Group|infinite]] [[Definiti... | From [[Non-Zero Complex Numbers under Multiplication form Group]], $\struct {\C_{\ne 0}, \times}$ is a [[Definition:Group|group]].
Then we have:
:[[Complex Multiplication is Commutative]]
and:
:[[Complex Numbers are Uncountable]].
{{qed}} | Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group | https://proofwiki.org/wiki/Non-Zero_Complex_Numbers_under_Multiplication_form_Infinite_Abelian_Group | https://proofwiki.org/wiki/Non-Zero_Complex_Numbers_under_Multiplication_form_Infinite_Abelian_Group | [
"Complex Multiplication",
"Examples of Abelian Groups",
"Examples of Infinite Groups"
] | [
"Definition:Complex Number",
"Definition:Zero (Number)",
"Definition:Algebraic Structure/One Operation",
"Definition:Infinite Group",
"Definition:Abelian Group"
] | [
"Non-Zero Complex Numbers under Multiplication form Group",
"Definition:Group",
"Complex Multiplication is Commutative",
"Complex Numbers are Uncountable"
] |
proofwiki-1138 | Real Numbers under Addition form Infinite Abelian Group | Let $\R$ be the set of real numbers.
The structure $\struct {\R, +}$ is an infinite abelian group. | From Real Numbers under Addition form Group, $\struct {\R, +}$ is a group.
Then:
:Real Addition is Commutative
and:
:Real Numbers are Uncountably Infinite.
{{qed}} | Let $\R$ be the set of [[Definition:Real Number|real numbers]].
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\R, +}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]]. | From [[Real Numbers under Addition form Group]], $\struct {\R, +}$ is a [[Definition:Group|group]].
Then:
:[[Real Addition is Commutative]]
and:
:[[Real Numbers are Uncountably Infinite]].
{{qed}} | Real Numbers under Addition form Infinite Abelian Group | https://proofwiki.org/wiki/Real_Numbers_under_Addition_form_Infinite_Abelian_Group | https://proofwiki.org/wiki/Real_Numbers_under_Addition_form_Infinite_Abelian_Group | [
"Additive Group of Real Numbers"
] | [
"Definition:Real Number",
"Definition:Algebraic Structure/One Operation",
"Definition:Infinite Group",
"Definition:Abelian Group"
] | [
"Real Numbers under Addition form Group",
"Definition:Group",
"Real Addition is Commutative",
"Real Numbers are Uncountably Infinite"
] |
proofwiki-1139 | Non-Zero Real Numbers under Multiplication form Abelian Group | Let $\R_{\ne 0}$ be the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
The structure $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group. | Taking the group axioms in turn:
=== {{Group-axiom|0|nolink}} ===
From Non-Zero Real Numbers Closed under Multiplication: Proof 2, $\R_{\ne 0}$ is closed under multiplication.
Note that proof 2 needs to be used specifically here, as proof 1 rests on this result.
{{qed|lemma}}
=== {{Group-axiom|1|nolink}} ===
Real Multi... | Let $\R_{\ne 0}$ be the set of [[Definition:Real Number|real numbers]] without [[Definition:Zero (Number)|zero]]:
:$\R_{\ne 0} = \R \setminus \set 0$
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\R_{\ne 0}, \times}$ is an [[Definition:Uncountable Group|uncountable]] [[Definition:Abelia... | Taking the [[Axiom:Group Axioms|group axioms]] in turn:
=== {{Group-axiom|0|nolink}} ===
From [[Non-Zero Real Numbers Closed under Multiplication/Proof 2|Non-Zero Real Numbers Closed under Multiplication: Proof 2]], $\R_{\ne 0}$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Multiplicati... | Non-Zero Real Numbers under Multiplication form Abelian Group/Proof 1 | https://proofwiki.org/wiki/Non-Zero_Real_Numbers_under_Multiplication_form_Abelian_Group | https://proofwiki.org/wiki/Non-Zero_Real_Numbers_under_Multiplication_form_Abelian_Group/Proof_1 | [
"Real Multiplication",
"Examples of Abelian Groups",
"Examples of Infinite Groups",
"Non-Zero Real Numbers under Multiplication form Abelian Group"
] | [
"Definition:Real Number",
"Definition:Zero (Number)",
"Definition:Algebraic Structure/One Operation",
"Definition:Infinite Group/Uncountable",
"Definition:Abelian Group"
] | [
"Axiom:Group Axioms",
"Non-Zero Real Numbers Closed under Multiplication/Proof 2",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Real Numbers",
"Non-Zero Real Numbers Closed under Multiplication/Proof 2",
"Non-Zero Real Numbers Closed under Multiplication/Proof 1"... |
proofwiki-1140 | Non-Zero Real Numbers under Multiplication form Abelian Group | Let $\R_{\ne 0}$ be the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
The structure $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group. | We have Real Numbers under Multiplication form Monoid.
From Inverse for Real Multiplication, the non-zero numbers are exactly the invertible elements of real multiplication.
Thus from Invertible Elements of Monoid form Subgroup of Cancellable Elements, the non-zero real numbers under multiplication form a group.
From:
... | Let $\R_{\ne 0}$ be the set of [[Definition:Real Number|real numbers]] without [[Definition:Zero (Number)|zero]]:
:$\R_{\ne 0} = \R \setminus \set 0$
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\R_{\ne 0}, \times}$ is an [[Definition:Uncountable Group|uncountable]] [[Definition:Abelia... | We have [[Real Numbers under Multiplication form Monoid]].
From [[Inverse for Real Multiplication]], the non-[[Definition:Zero (Number)|zero]] [[Definition:Number|numbers]] are exactly the [[Definition:Invertible Element|invertible elements]] of [[Definition:Real Multiplication|real multiplication]].
Thus from [[Inve... | Non-Zero Real Numbers under Multiplication form Abelian Group/Proof 2 | https://proofwiki.org/wiki/Non-Zero_Real_Numbers_under_Multiplication_form_Abelian_Group | https://proofwiki.org/wiki/Non-Zero_Real_Numbers_under_Multiplication_form_Abelian_Group/Proof_2 | [
"Real Multiplication",
"Examples of Abelian Groups",
"Examples of Infinite Groups",
"Non-Zero Real Numbers under Multiplication form Abelian Group"
] | [
"Definition:Real Number",
"Definition:Zero (Number)",
"Definition:Algebraic Structure/One Operation",
"Definition:Infinite Group/Uncountable",
"Definition:Abelian Group"
] | [
"Real Numbers under Multiplication form Monoid",
"Inverse for Real Multiplication",
"Definition:Zero (Number)",
"Definition:Number",
"Definition:Invertible Element",
"Definition:Multiplication/Real Numbers",
"Invertible Elements of Monoid form Subgroup of Cancellable Elements",
"Definition:Zero (Numbe... |
proofwiki-1141 | Non-Zero Real Numbers under Multiplication form Abelian Group | Let $\R_{\ne 0}$ be the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
The structure $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group. | From Non-Zero Real Numbers under Multiplication form Group, $\struct {\R_{\ne 0}, \times}$ forms a group.
{{qed|lemma}}
From Real Multiplication is Commutative it follows that $\struct {\R_{\ne 0}, \times}$ is abelian.
{{qed|lemma}}
From Real Numbers are Uncountably Infinite it follows that $\struct {\R_{\ne 0}, \times... | Let $\R_{\ne 0}$ be the set of [[Definition:Real Number|real numbers]] without [[Definition:Zero (Number)|zero]]:
:$\R_{\ne 0} = \R \setminus \set 0$
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\R_{\ne 0}, \times}$ is an [[Definition:Uncountable Group|uncountable]] [[Definition:Abelia... | From [[Non-Zero Real Numbers under Multiplication form Group]], $\struct {\R_{\ne 0}, \times}$ forms a [[Definition:Group|group]].
{{qed|lemma}}
From [[Real Multiplication is Commutative]] it follows that $\struct {\R_{\ne 0}, \times}$ is [[Definition:Abelian Group|abelian]].
{{qed|lemma}}
From [[Real Numbers are U... | Non-Zero Real Numbers under Multiplication form Abelian Group/Proof 3 | https://proofwiki.org/wiki/Non-Zero_Real_Numbers_under_Multiplication_form_Abelian_Group | https://proofwiki.org/wiki/Non-Zero_Real_Numbers_under_Multiplication_form_Abelian_Group/Proof_3 | [
"Real Multiplication",
"Examples of Abelian Groups",
"Examples of Infinite Groups",
"Non-Zero Real Numbers under Multiplication form Abelian Group"
] | [
"Definition:Real Number",
"Definition:Zero (Number)",
"Definition:Algebraic Structure/One Operation",
"Definition:Infinite Group/Uncountable",
"Definition:Abelian Group"
] | [
"Non-Zero Real Numbers under Multiplication form Group",
"Definition:Group",
"Real Multiplication is Commutative",
"Definition:Abelian Group",
"Real Numbers are Uncountably Infinite",
"Definition:Infinite Group/Uncountable",
"Definition:Abelian Group"
] |
proofwiki-1142 | Rational Numbers under Addition form Infinite Abelian Group | Let $\Q$ be the set of rational numbers.
The structure $\struct {\Q, +}$ is a countably infinite abelian group. | The rational numbers are, by definition, the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
Hence by definition, $\struct {\Q, +, \times}$ is a field.
The fact that $\struct {\Q, +}$ forms an abelian group follows directly from the definition of a field.
From Rational Numbers are Count... | Let $\Q$ be the set of [[Definition:Rational Number|rational numbers]].
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\Q, +}$ is a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Abelian Group|abelian group]]. | The [[Definition:Rational Number|rational numbers]] are, by definition, the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Integral Domain|integral domain]] $\struct {\Z, +, \times}$ of [[Definition:Integer|integers]].
Hence by definition, $\struct {\Q, +, \times}$ is a [[Definition:Field (Ab... | Rational Numbers under Addition form Infinite Abelian Group | https://proofwiki.org/wiki/Rational_Numbers_under_Addition_form_Infinite_Abelian_Group | https://proofwiki.org/wiki/Rational_Numbers_under_Addition_form_Infinite_Abelian_Group | [
"Rational Addition",
"Examples of Abelian Groups",
"Examples of Infinite Groups"
] | [
"Definition:Rational Number",
"Definition:Algebraic Structure/One Operation",
"Definition:Countably Infinite/Set",
"Definition:Abelian Group"
] | [
"Definition:Rational Number",
"Definition:Field of Quotients",
"Definition:Integral Domain",
"Definition:Integer",
"Definition:Field (Abstract Algebra)",
"Definition:Abelian Group",
"Definition:Field (Abstract Algebra)",
"Rational Numbers are Countably Infinite",
"Definition:Countably Infinite/Set"
... |
proofwiki-1143 | Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group | Let $\Q_{\ne 0}$ be the set of non-zero rational numbers:
:$\Q_{\ne 0} = \Q \setminus \set 0$
The structure $\struct {\Q_{\ne 0}, \times}$ is a countably infinite abelian group. | From the definition of rational numbers, the structure $\struct {\Q, + \times}$ is constructed as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
Hence from Multiplicative Group of Field is Abelian Group, $\struct {\Q_{\ne 0}, \times}$ is an abelian group.
From Rational Numbers are ... | Let $\Q_{\ne 0}$ be the [[Definition:Set|set]] of non-[[Definition:Zero (Number)|zero]] [[Definition:Rational Number|rational numbers]]:
:$\Q_{\ne 0} = \Q \setminus \set 0$
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\Q_{\ne 0}, \times}$ is a [[Definition:Countably Infinite Group|coun... | From the [[Definition:Rational Number|definition of rational numbers]], the [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\Q, + \times}$ is constructed as the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Integral Domain|integral domain]] $\struct {\Z, +, \times}$ ... | Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group | https://proofwiki.org/wiki/Non-Zero_Rational_Numbers_under_Multiplication_form_Infinite_Abelian_Group | https://proofwiki.org/wiki/Non-Zero_Rational_Numbers_under_Multiplication_form_Infinite_Abelian_Group | [
"Rational Multiplication",
"Examples of Abelian Groups",
"Examples of Infinite Groups"
] | [
"Definition:Set",
"Definition:Zero (Number)",
"Definition:Rational Number",
"Definition:Algebraic Structure/One Operation",
"Definition:Infinite Group/Countable",
"Definition:Abelian Group"
] | [
"Definition:Rational Number",
"Definition:Algebraic Structure/One Operation",
"Definition:Field of Quotients",
"Definition:Integral Domain",
"Definition:Integer",
"Multiplicative Group of Field is Abelian Group",
"Definition:Abelian Group",
"Rational Numbers are Countably Infinite",
"Definition:Infi... |
proofwiki-1144 | Integers under Multiplication form Countably Infinite Commutative Monoid | The set of integers under multiplication $\struct {\Z, \times}$ is a countably infinite commutative monoid. | First we note that Integers under Multiplication form Monoid.
{{qed|lemma}}
Then we have: | The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]] $\struct {\Z, \times}$ is a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Commutative Monoid|commutative monoid]]. | First we note that [[Integers under Multiplication form Monoid]].
{{qed|lemma}}
Then we have: | Integers under Multiplication form Countably Infinite Commutative Monoid | https://proofwiki.org/wiki/Integers_under_Multiplication_form_Countably_Infinite_Commutative_Monoid | https://proofwiki.org/wiki/Integers_under_Multiplication_form_Countably_Infinite_Commutative_Monoid | [
"Integer Multiplication",
"Examples of Commutative Monoids"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Countably Infinite/Set",
"Definition:Commutative Monoid"
] | [
"Integers under Multiplication form Monoid"
] |
proofwiki-1145 | Additive Group of Rationals is Normal Subgroup of Reals | Let $\struct {\Q, +}$ be the additive group of rational numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Then $\struct {\Q, +}$ is a normal subgroup of $\struct {\R, +}$. | From the definition of real numbers, $\Q$ is a subset of $\R$.
As $\struct {\R, +}$ is a group, and $\struct {\Q, +}$ is a group, it follows from the definition of subgroup that $\struct {\Q, +}$ is a subgroup of $\struct {\R, +}$.
As $\struct {\R, +}$ is abelian, it follows from Subgroup of Abelian Group is Normal tha... | Let $\struct {\Q, +}$ be the [[Definition:Additive Group of Rational Numbers|additive group of rational numbers]].
Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
Then $\struct {\Q, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\R, +}$... | From the definition of [[Definition:Real Number|real numbers]], $\Q$ is a [[Definition:Subset|subset]] of $\R$.
As $\struct {\R, +}$ is a [[Definition:Group|group]], and $\struct {\Q, +}$ is a [[Definition:Group|group]], it follows from the definition of [[Definition:Subgroup|subgroup]] that $\struct {\Q, +}$ is a sub... | Additive Group of Rationals is Normal Subgroup of Reals | https://proofwiki.org/wiki/Additive_Group_of_Rationals_is_Normal_Subgroup_of_Reals | https://proofwiki.org/wiki/Additive_Group_of_Rationals_is_Normal_Subgroup_of_Reals | [
"Rational Addition",
"Additive Group of Real Numbers",
"Additive Group of Rational Numbers",
"Examples of Normal Subgroups"
] | [
"Definition:Additive Group of Rational Numbers",
"Definition:Additive Group of Real Numbers",
"Definition:Normal Subgroup"
] | [
"Definition:Real Number",
"Definition:Subset",
"Definition:Group",
"Definition:Group",
"Definition:Subgroup",
"Definition:Abelian Group",
"Subgroup of Abelian Group is Normal",
"Definition:Normal Subgroup"
] |
proofwiki-1146 | Additive Group of Integers is Subgroup of Rationals | Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\Q, +}$ be the additive group of rational numbers.
Then $\struct {\Z, +}$ is a subgroup of $\struct {\Q, +}$. | Recall that Integers form Integral Domain.
The set $\Q$ of rational numbers is defined as the field of quotients of the integers.
The fact that the integers are a subgroup of the rationals follows from the work done in proving the Existence of Field of Quotients from an integral domain.
{{qed}} | Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
Let $\struct {\Q, +}$ be the [[Definition:Additive Group of Rational Numbers|additive group of rational numbers]].
Then $\struct {\Z, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Q, +}$. | Recall that [[Integers form Integral Domain]].
The set $\Q$ of [[Definition:Rational Number|rational numbers]] is defined as the [[Definition:Rational Number|field of quotients of the integers]].
The fact that the [[Definition:Integer|integers]] are a [[Definition:Subgroup|subgroup]] of the [[Definition:Rational Numb... | Additive Group of Integers is Subgroup of Rationals | https://proofwiki.org/wiki/Additive_Group_of_Integers_is_Subgroup_of_Rationals | https://proofwiki.org/wiki/Additive_Group_of_Integers_is_Subgroup_of_Rationals | [
"Additive Group of Integers",
"Additive Group of Rational Numbers",
"Examples of Subgroups"
] | [
"Definition:Additive Group of Integers",
"Definition:Additive Group of Rational Numbers",
"Definition:Subgroup"
] | [
"Integers form Integral Domain",
"Definition:Rational Number",
"Definition:Rational Number",
"Definition:Integer",
"Definition:Subgroup",
"Definition:Rational Number",
"Existence of Field of Quotients",
"Definition:Integral Domain"
] |
proofwiki-1147 | Additive Group of Reals is Normal Subgroup of Complex | Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {\C, +}$ be the additive group of complex numbers.
Then $\struct {\R, +}$ is a normal subgroup of $\struct {\C, +}$. | From Additive Group of Reals is Subgroup of Complex, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.
Then from Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is abelian.
The result follows from Subgroup of Abelian Group is Normal.
{{Qed}}
Category:Additive Group of Real Numbers
Cate... | Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
Let $\struct {\C, +}$ be the [[Definition:Additive Group of Complex Numbers|additive group of complex numbers]].
Then $\struct {\R, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\C, +}$. | From [[Additive Group of Reals is Subgroup of Complex]], $\struct {\R, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\C, +}$.
Then from [[Complex Numbers under Addition form Infinite Abelian Group]], $\struct {\C, +}$ is [[Definition:Abelian Group|abelian]].
The result follows from [[Subgroup of Abelian Grou... | Additive Group of Reals is Normal Subgroup of Complex | https://proofwiki.org/wiki/Additive_Group_of_Reals_is_Normal_Subgroup_of_Complex | https://proofwiki.org/wiki/Additive_Group_of_Reals_is_Normal_Subgroup_of_Complex | [
"Additive Group of Real Numbers",
"Additive Group of Complex Numbers",
"Examples of Normal Subgroups"
] | [
"Definition:Additive Group of Real Numbers",
"Definition:Additive Group of Complex Numbers",
"Definition:Normal Subgroup"
] | [
"Additive Group of Reals is Subgroup of Complex",
"Definition:Subgroup",
"Complex Numbers under Addition form Infinite Abelian Group",
"Definition:Abelian Group",
"Subgroup of Abelian Group is Normal",
"Category:Additive Group of Real Numbers",
"Category:Additive Group of Complex Numbers",
"Category:E... |
proofwiki-1148 | Multiplicative Group of Reals is Normal Subgroup of Complex | Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Then $\struct {\R_{\ne 0}, \times}$ is a normal subgroup of $\struct {\C_{\ne 0}, \times}$. | Let $x, y \in \C_{\ne 0}$ such that $x = x_1 + 0 i, y = y_1 + 0 i$.
As $x$ and $y$ are wholly real, we have that $x, y \in \R_{\ne 0}$.
Then $x + y = x_1 y_1 - 0 \cdot 0 + \left({0 \cdot y_1 + x_1 \cdot 0}\right) i = x_1 y_1 + 0i$ which is also wholly real.
Also, the inverse of $x$ is $\dfrac 1 x = \dfrac 1 {x_1} + 0 i... | Let $\struct {\R_{\ne 0}, \times}$ be the [[Definition:Multiplicative Group of Real Numbers|multiplicative group of real numbers]].
Let $\struct {\C_{\ne 0}, \times}$ be the [[Definition:Multiplicative Group of Complex Numbers|multiplicative group of complex numbers]].
Then $\struct {\R_{\ne 0}, \times}$ is a [[Defi... | Let $x, y \in \C_{\ne 0}$ such that $x = x_1 + 0 i, y = y_1 + 0 i$.
As $x$ and $y$ are [[Definition:Wholly Real|wholly real]], we have that $x, y \in \R_{\ne 0}$.
Then $x + y = x_1 y_1 - 0 \cdot 0 + \left({0 \cdot y_1 + x_1 \cdot 0}\right) i = x_1 y_1 + 0i$ which is also [[Definition:Wholly Real|wholly real]].
Also,... | Multiplicative Group of Reals is Normal Subgroup of Complex | https://proofwiki.org/wiki/Multiplicative_Group_of_Reals_is_Normal_Subgroup_of_Complex | https://proofwiki.org/wiki/Multiplicative_Group_of_Reals_is_Normal_Subgroup_of_Complex | [
"Multiplicative Group of Real Numbers",
"Multiplicative Group of Complex Numbers",
"Examples of Normal Subgroups"
] | [
"Definition:Multiplicative Group of Real Numbers",
"Definition:Multiplicative Group of Complex Numbers",
"Definition:Normal Subgroup"
] | [
"Definition:Complex Number/Wholly Real",
"Definition:Complex Number/Wholly Real",
"Definition:Complex Number/Wholly Real",
"Two-Step Subgroup Test",
"Definition:Subgroup",
"Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group",
"Definition:Abelian Group",
"Subgroup of Abelian Grou... |
proofwiki-1149 | Multiplicative Group of Rationals is Normal Subgroup of Reals | Let $\struct {\Q_{\ne 0}, \times}$ be the multiplicative group of rational numbers.
Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.
Then $\struct {\Q_{\ne 0}, \times}$ is a normal subgroup of $\left({\R_{\ne 0}, \times}\right)$. | From the definition of real numbers, it follows that $\Q_{\ne 0}$ is a subset of $\R_{\ne 0}$.
As $\struct {\R_{\ne 0}, \times}$ is a group, and $\struct {\Q_{\ne 0}, \times}$ is a group, it follows from the definition of subgroup that $\struct {\Q_{\ne 0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$.
As $... | Let $\struct {\Q_{\ne 0}, \times}$ be the [[Definition:Multiplicative Group of Rational Numbers|multiplicative group of rational numbers]].
Let $\struct {\R_{\ne 0}, \times}$ be the [[Definition:Multiplicative Group of Real Numbers|multiplicative group of real numbers]].
Then $\struct {\Q_{\ne 0}, \times}$ is a [[De... | From the definition of [[Definition:Real Number|real numbers]], it follows that $\Q_{\ne 0}$ is a [[Definition:Subset|subset]] of $\R_{\ne 0}$.
As $\struct {\R_{\ne 0}, \times}$ is a [[Definition:Group|group]], and $\struct {\Q_{\ne 0}, \times}$ is a [[Definition:Group|group]], it follows from the definition of [[Defi... | Multiplicative Group of Rationals is Normal Subgroup of Reals | https://proofwiki.org/wiki/Multiplicative_Group_of_Rationals_is_Normal_Subgroup_of_Reals | https://proofwiki.org/wiki/Multiplicative_Group_of_Rationals_is_Normal_Subgroup_of_Reals | [
"Multiplicative Group of Rational Numbers",
"Multiplicative Group of Real Numbers",
"Examples of Normal Subgroups"
] | [
"Definition:Multiplicative Group of Rational Numbers",
"Definition:Multiplicative Group of Real Numbers",
"Definition:Normal Subgroup"
] | [
"Definition:Real Number",
"Definition:Subset",
"Definition:Group",
"Definition:Group",
"Definition:Subgroup",
"Definition:Abelian Group",
"Subgroup of Abelian Group is Normal",
"Definition:Normal Subgroup"
] |
proofwiki-1150 | Circle Group is Infinite Abelian Group | The circle group $\struct {K, \times}$ is an uncountably infinite abelian group. | From Circle Group is Group, we have that $\struct {K, \times}$ is a group.
As $K \subseteq \C$, it follows by definition that $\struct {K, \times}$ is a subgroup of $\struct {\C, \times}$.
From Complex Multiplication is Commutative, $\times$ is commutative on $\C$.
From Restriction of Commutative Operation is Commutati... | The [[Definition:Circle Group|circle group]] $\struct {K, \times}$ is an [[Definition:Uncountable Set|uncountably infinite]] [[Definition:Abelian Group|abelian group]]. | From [[Circle Group is Group]], we have that $\struct {K, \times}$ is a [[Definition:Group|group]].
As $K \subseteq \C$, it follows by definition that $\struct {K, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\C, \times}$.
From [[Complex Multiplication is Commutative]], $\times$ is [[Definition:Commuta... | Circle Group is Infinite Abelian Group | https://proofwiki.org/wiki/Circle_Group_is_Infinite_Abelian_Group | https://proofwiki.org/wiki/Circle_Group_is_Infinite_Abelian_Group | [
"Circle Group",
"Complex Numbers",
"Circle Group is Infinite Abelian Group"
] | [
"Definition:Circle Group",
"Definition:Uncountable/Set",
"Definition:Abelian Group"
] | [
"Circle Group is Group",
"Definition:Group",
"Definition:Subgroup",
"Complex Multiplication is Commutative",
"Definition:Commutative/Operation",
"Restriction of Commutative Operation is Commutative",
"Definition:Commutative/Operation",
"Definition:Abelian Group",
"Circle Group is Uncountably Infinit... |
proofwiki-1151 | Homomorphism from Reals to Circle Group | Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {K, \times}$ be the circle group.
Let $\phi: \struct {\R, +} \to \struct {K, \times}$ be the mapping defined as:
:$\forall x \in \R: \map \phi x = e^{i x}$
Then $\phi$ is a (group) homomorphism. | Let $x, y \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi x \times \map \phi y
| r = e^{i x} e^{i y}
| c =
}}
{{eqn | r = e^{i \paren {x + y} }
| c =
}}
{{eqn | r = \map \phi {x + y}
| c =
}}
{{end-eqn}}
{{qed}} | Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
Let $\struct {K, \times}$ be the [[Definition:Circle Group|circle group]].
Let $\phi: \struct {\R, +} \to \struct {K, \times}$ be the [[Definition:Mapping|mapping]] defined as:
:$\forall x \in \R: \map \phi x = ... | Let $x, y \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi x \times \map \phi y
| r = e^{i x} e^{i y}
| c =
}}
{{eqn | r = e^{i \paren {x + y} }
| c =
}}
{{eqn | r = \map \phi {x + y}
| c =
}}
{{end-eqn}}
{{qed}} | Homomorphism from Reals to Circle Group | https://proofwiki.org/wiki/Homomorphism_from_Reals_to_Circle_Group | https://proofwiki.org/wiki/Homomorphism_from_Reals_to_Circle_Group | [
"Circle Group",
"Examples of Group Homomorphisms"
] | [
"Definition:Additive Group of Real Numbers",
"Definition:Circle Group",
"Definition:Mapping",
"Definition:Group Homomorphism"
] | [] |
proofwiki-1152 | Integers under Addition form Infinite Cyclic Group | The additive group of integers $\struct {\Z, +}$ is an infinite cyclic group which is generated by the element $1 \in \Z$. | By Epimorphism from Integers to Cyclic Group and integer multiplication:
:$\forall n \in \Z: n = \map {+^n} 1 \in \gen 1$
Thus:
:$\struct {\Z, +} = \gen 1$
and thus, by the definition of a cyclic group, is cyclic.
{{qed}} | The [[Definition:Additive Group of Integers|additive group of integers]] $\struct {\Z, +}$ is an [[Definition:Infinite Cyclic Group|infinite cyclic group]] which is [[Definition:Generator of Cyclic Group|generated by]] the element $1 \in \Z$. | By [[Epimorphism from Integers to Cyclic Group]] and [[Definition:Integer Multiplication|integer multiplication]]:
:$\forall n \in \Z: n = \map {+^n} 1 \in \gen 1$
Thus:
:$\struct {\Z, +} = \gen 1$
and thus, by the definition of a [[Definition:Cyclic Group|cyclic group]], is [[Definition:Cyclic Group|cyclic]].
{{qed... | Integers under Addition form Infinite Cyclic Group | https://proofwiki.org/wiki/Integers_under_Addition_form_Infinite_Cyclic_Group | https://proofwiki.org/wiki/Integers_under_Addition_form_Infinite_Cyclic_Group | [
"Infinite Cyclic Group",
"Integer Addition",
"Additive Group of Integers"
] | [
"Definition:Additive Group of Integers",
"Definition:Infinite Cyclic Group",
"Definition:Cyclic Group/Generator"
] | [
"Epimorphism from Integers to Cyclic Group",
"Definition:Multiplication/Integers",
"Definition:Cyclic Group",
"Definition:Cyclic Group"
] |
proofwiki-1153 | Generators of Additive Group of Integers | The only generators of the additive group of integers $\struct {\Z, +}$ are $1$ and $-1$. | From Integers under Addition form Infinite Cyclic Group, $\struct {\Z, +}$ is an infinite cyclic group generated by $1$.
From Generators of Infinite Cyclic Group, there is only one other generator of such a group, and that is the inverse of that generator.
The result follows.
{{qed}} | The only [[Definition:Generator of Cyclic Group|generators]] of the [[Definition:Additive Group of Integers|additive group of integers]] $\struct {\Z, +}$ are $1$ and $-1$. | From [[Integers under Addition form Infinite Cyclic Group]], $\struct {\Z, +}$ is an [[Definition:Infinite Cyclic Group|infinite cyclic group]] generated by $1$.
From [[Generators of Infinite Cyclic Group]], there is only one other [[Definition:Generator of Cyclic Group|generator]] of such a group, and that is the inv... | Generators of Additive Group of Integers | https://proofwiki.org/wiki/Generators_of_Additive_Group_of_Integers | https://proofwiki.org/wiki/Generators_of_Additive_Group_of_Integers | [
"Integer Addition",
"Infinite Cyclic Group",
"Additive Group of Integers"
] | [
"Definition:Cyclic Group/Generator",
"Definition:Additive Group of Integers"
] | [
"Integers under Addition form Infinite Cyclic Group",
"Definition:Infinite Cyclic Group",
"Generators of Infinite Cyclic Group",
"Definition:Cyclic Group/Generator"
] |
proofwiki-1154 | Inverse of Generator of Cyclic Group is Generator | Let $\gen g = G$ be a cyclic group.
Then:
:$G = \gen {g^{-1} }$
where $g^{-1}$ denotes the inverse of $g$.
Thus, in general, a generator of a cyclic group is not unique. | Let $\gen g = G$.
Then from Set of Words Generates Group:
:$\map W {\set {g, g^{-1} } } = G$
But:
:$\gen {g^{-1} } = \map W {\set {g, g^{-1} } }$
and the result follows.
{{qed}} | Let $\gen g = G$ be a [[Definition:Cyclic Group|cyclic group]].
Then:
:$G = \gen {g^{-1} }$
where $g^{-1}$ denotes the [[Definition:Inverse Element|inverse]] of $g$.
Thus, in general, a [[Definition:Generator of Cyclic Group|generator]] of a [[Definition:Cyclic Group|cyclic group]] is not [[Definition:Unique|unique... | Let $\gen g = G$.
Then from [[Set of Words Generates Group]]:
:$\map W {\set {g, g^{-1} } } = G$
But:
:$\gen {g^{-1} } = \map W {\set {g, g^{-1} } }$
and the result follows.
{{qed}} | Inverse of Generator of Cyclic Group is Generator/Proof 1 | https://proofwiki.org/wiki/Inverse_of_Generator_of_Cyclic_Group_is_Generator | https://proofwiki.org/wiki/Inverse_of_Generator_of_Cyclic_Group_is_Generator/Proof_1 | [
"Inverse of Generator of Cyclic Group is Generator",
"Cyclic Groups"
] | [
"Definition:Cyclic Group",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Cyclic Group/Generator",
"Definition:Cyclic Group",
"Definition:Unique"
] | [
"Set of Words Generates Group"
] |
proofwiki-1155 | Inverse of Generator of Cyclic Group is Generator | Let $\gen g = G$ be a cyclic group.
Then:
:$G = \gen {g^{-1} }$
where $g^{-1}$ denotes the inverse of $g$.
Thus, in general, a generator of a cyclic group is not unique. | Let $C_n = \gen g$ be the cyclic group of order $n$.
By definition, $g^n = e$.
We have that $n - 1$ is coprime to $n$.
So it follows from that Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order that:
: $C_n = \gen {g^{n - 1} }$
But from Inverse Element is Power of Order Less 1:
:$g^{n - 1} ... | Let $\gen g = G$ be a [[Definition:Cyclic Group|cyclic group]].
Then:
:$G = \gen {g^{-1} }$
where $g^{-1}$ denotes the [[Definition:Inverse Element|inverse]] of $g$.
Thus, in general, a [[Definition:Generator of Cyclic Group|generator]] of a [[Definition:Cyclic Group|cyclic group]] is not [[Definition:Unique|unique... | Let $C_n = \gen g$ be the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order $n$]].
By definition, $g^n = e$.
We have that $n - 1$ is [[Definition:Coprime Integers|coprime]] to $n$.
So it follows from that [[Power of Generator of Cyclic Group is Generator iff Power is Coprime with Orde... | Inverse of Generator of Cyclic Group is Generator/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Generator_of_Cyclic_Group_is_Generator | https://proofwiki.org/wiki/Inverse_of_Generator_of_Cyclic_Group_is_Generator/Proof_2 | [
"Inverse of Generator of Cyclic Group is Generator",
"Cyclic Groups"
] | [
"Definition:Cyclic Group",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Cyclic Group/Generator",
"Definition:Cyclic Group",
"Definition:Unique"
] | [
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Coprime/Integers",
"Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order",
"Inverse Element is Power of Order Less 1"
] |
proofwiki-1156 | Generators of Infinite Cyclic Group | Let $\gen g = G$ be an infinite cyclic group.
Then the only other generator of $G$ is $g^{-1}$.
Thus an infinite cyclic group has exactly $2$ generators. | By definition, the infinite cyclic group with generator $g$ is:
:$\gen g = \set {\ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots}$
where $e$ denotes the identity $e = g^0$.
The fact that $g^{-1}$ generates $G$ is shown by Inverse of Generator of Cyclic Group is Generator.
Futhermore:
:$\gen e = \set e \ne G$
By definition of... | Let $\gen g = G$ be an [[Definition:Infinite Cyclic Group|infinite cyclic group]].
Then the only other [[Definition:Generator of Cyclic Group|generator]] of $G$ is $g^{-1}$.
Thus an [[Definition:Infinite Cyclic Group|infinite cyclic group]] has exactly $2$ [[Definition:Generator of Cyclic Group|generators]]. | By definition, the [[Definition:Infinite Cyclic Group|infinite cyclic group]] with [[Definition:Generator of Cyclic Group|generator]] $g$ is:
:$\gen g = \set {\ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots}$
where $e$ denotes the [[Definition:Identity Element|identity]] $e = g^0$.
The fact that $g^{-1}$ [[Definition:Gene... | Generators of Infinite Cyclic Group | https://proofwiki.org/wiki/Generators_of_Infinite_Cyclic_Group | https://proofwiki.org/wiki/Generators_of_Infinite_Cyclic_Group | [
"Infinite Cyclic Group",
"Examples of Generators of Groups"
] | [
"Definition:Infinite Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Infinite Cyclic Group",
"Definition:Cyclic Group/Generator"
] | [
"Definition:Infinite Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Cyclic Group/Generator",
"Inverse of Generator of Cyclic Group is Generator",
"Definition:Infinite Cyclic Group"
] |
proofwiki-1157 | Subgroup of Integers is Ideal | Let $\struct {\Z, +}$ be the additive group of integers.
Every subgroup of $\struct {\Z, +}$ is an ideal of the ring $\struct {\Z, +, \times}$. | Let $H$ be a subgroup of $\struct {\Z, +}$.
Let $n \in \Z, h \in H$.
Then from the definition of cyclic group and Negative Index Law for Monoids:
:$n h = n \cdot h \in \gen h \subseteq H$
The result follows.
{{Qed}} | Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
Every [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$ is an [[Definition:Ideal of Ring|ideal]] of the [[Definition:Ring (Abstract Algebra)|ring]] $\struct {\Z, +, \times}$. | Let $H$ be a subgroup of $\struct {\Z, +}$.
Let $n \in \Z, h \in H$.
Then from the definition of [[Definition:Cyclic Group|cyclic group]] and [[Index Laws for Monoids/Negative Index|Negative Index Law for Monoids]]:
:$n h = n \cdot h \in \gen h \subseteq H$
The result follows.
{{Qed}} | Subgroup of Integers is Ideal | https://proofwiki.org/wiki/Subgroup_of_Integers_is_Ideal | https://proofwiki.org/wiki/Subgroup_of_Integers_is_Ideal | [
"Subgroups",
"Integers",
"Ideal Theory"
] | [
"Definition:Additive Group of Integers",
"Definition:Subgroup",
"Definition:Ideal of Ring",
"Definition:Ring (Abstract Algebra)"
] | [
"Definition:Cyclic Group",
"Index Laws for Monoids/Negative Index"
] |
proofwiki-1158 | Additive Group of Integers is Subgroup of Reals | Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Then $\struct {\Z, +}$ is a subgroup of $\struct {\R, +}$. | From Additive Group of Integers is Subgroup of Rationals, $\struct {\Z, +}$ is a subgroup of $\struct {\Q, +}$.
From Additive Group of Rationals is Normal Subgroup of Reals, $\struct {\Q, +}$ is a subgroup of $\struct {\R, +}$.
Thus $\struct {\Z, +}$ is a subgroup of $\struct {\R, +}$.
{{qed}} | Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
Then $\struct {\Z, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\R, +}$. | From [[Additive Group of Integers is Subgroup of Rationals]], $\struct {\Z, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Q, +}$.
From [[Additive Group of Rationals is Normal Subgroup of Reals]], $\struct {\Q, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\R, +}$.
Thus $\struct {\Z, +}$ is a [[Defi... | Additive Group of Integers is Subgroup of Reals | https://proofwiki.org/wiki/Additive_Group_of_Integers_is_Subgroup_of_Reals | https://proofwiki.org/wiki/Additive_Group_of_Integers_is_Subgroup_of_Reals | [
"Additive Group of Integers",
"Additive Group of Real Numbers"
] | [
"Definition:Additive Group of Integers",
"Definition:Additive Group of Real Numbers",
"Definition:Subgroup"
] | [
"Additive Group of Integers is Subgroup of Rationals",
"Definition:Subgroup",
"Additive Group of Rationals is Normal Subgroup of Reals",
"Definition:Subgroup",
"Definition:Subgroup"
] |
proofwiki-1159 | Quotient Group of Reals by Integers is Circle Group | Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $K$ be the circle group.
Then the quotient group of $\struct {\R, +}$ by $\struct {\Z, +}$ is isomorphic to $K$. | Define $\phi: \R / \Z \to K$ by:
:$\map \phi {x + \Z} = \map \exp {2 \pi i x}$
Then $\phi$ is well-defined.
For, if $x + \Z = y + \Z$, then $y = x + n$ for some $n \in \Z$, and:
:$\map \exp {2 \pi i \paren {x + n} } = \map \exp {2 \pi i x}$
by Complex Exponential Function has Imaginary Period.
Moreover, by Exponential ... | Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
Let $K$ be the [[Definition:Circle Group|circle group]].
Then the [[Definition:Quotient Group|quotient group]]... | Define $\phi: \R / \Z \to K$ by:
:$\map \phi {x + \Z} = \map \exp {2 \pi i x}$
Then $\phi$ is [[Definition:Well-Defined Mapping|well-defined]].
For, if $x + \Z = y + \Z$, then $y = x + n$ for some $n \in \Z$, and:
:$\map \exp {2 \pi i \paren {x + n} } = \map \exp {2 \pi i x}$
by [[Complex Exponential Function has ... | Quotient Group of Reals by Integers is Circle Group | https://proofwiki.org/wiki/Quotient_Group_of_Reals_by_Integers_is_Circle_Group | https://proofwiki.org/wiki/Quotient_Group_of_Reals_by_Integers_is_Circle_Group | [
"Integers",
"Real Numbers",
"Complex Numbers",
"Circle Group",
"Examples of Quotient Groups"
] | [
"Definition:Additive Group of Integers",
"Definition:Additive Group of Real Numbers",
"Definition:Circle Group",
"Definition:Quotient Group",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Well-Defined/Mapping",
"Period of Complex Exponential Function",
"Exponential of Sum/Complex Numbers",
"Definition:Group Homomorphism",
"Euler's Formula",
"Sine and Cosine are Periodic on Reals",
"Kernel is Trivial iff Monomorphism/Group",
"Definition:Group Monomorphism",
"Definition:Com... |
proofwiki-1160 | Integers Modulo m under Addition form Cyclic Group | Let $\Z_m$ be the set of integers modulo $m$.
Let $+_m$ be the operation of addition modulo $m$.
Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$.
Then $\struct {\Z_m, +_m}$ is a cyclic group of order $m$, generated by the element $\eqclass 1 m \in \Z_m$. | From the definition of integers modulo $m$, we have:
:$\Z_m = \dfrac \Z {\RR_m} = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
It is established that Modulo Addition is Well-Defined:
:$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$
The group axioms are fulfilled:
:'''{{Group-axiom|0}}''': Additio... | Let $\Z_m$ be the set of [[Definition:Integers Modulo m|integers modulo $m$]].
Let $+_m$ be the [[Definition:Binary Operation|operation]] of [[Definition:Modulo Addition|addition modulo $m$]].
Let $\struct {\Z_m, +_m}$ denote the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $m$]... | From the definition of [[Definition:Integers Modulo m|integers modulo $m$]], we have:
:$\Z_m = \dfrac \Z {\RR_m} = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
It is established that [[Modulo Addition is Well-Defined]]:
:$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$
The [[Axiom:Group Axiom... | Integers Modulo m under Addition form Cyclic Group | https://proofwiki.org/wiki/Integers_Modulo_m_under_Addition_form_Cyclic_Group | https://proofwiki.org/wiki/Integers_Modulo_m_under_Addition_form_Cyclic_Group | [
"Additive Groups of Integers Modulo m",
"Modulo Addition"
] | [
"Definition:Integers Modulo m",
"Definition:Operation/Binary Operation",
"Definition:Modulo Addition",
"Definition:Additive Group of Integers Modulo m",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Cyclic Group/Generator"
] | [
"Definition:Integers Modulo m",
"Modulo Addition is Well-Defined",
"Axiom:Group Axioms",
"Modulo Addition is Closed",
"Modulo Addition is Associative",
"Modulo Addition has Identity",
"Modulo Addition has Inverses",
"Modulo Addition is Commutative",
"Integers under Addition form Infinite Cyclic Grou... |
proofwiki-1161 | Integers Modulo m under Multiplication form Commutative Monoid | Let $\struct {\Z_m, \times_m}$ denote the algebraic structure such that:
:$\Z_m$ is the set of integers modulo $m$
:$\times_m$ denotes the operation of multiplication modulo $m$.
Then $\struct {\Z_m, \times_m}$ is a commutative monoid. | Multiplication modulo $m$ is closed.
Multiplication modulo $m$ is associative.
Multiplication modulo $m$ has an identity:
:$\forall k \in \Z: \eqclass k m \times_m \eqclass 1 m = \eqclass k m = \eqclass 1 m \times_m \eqclass k m$
This identity is unique.
Multiplication modulo $m$ is commutative.
Thus all the conditions... | Let $\struct {\Z_m, \times_m}$ denote the [[Definition:Algebraic Structure with One Operation|algebraic structure]] such that:
:$\Z_m$ is the [[Definition:Set|set]] of [[Definition:Integers Modulo m|integers modulo $m$]]
:$\times_m$ denotes the operation of [[Definition:Modulo Multiplication|multiplication modulo $m$]]... | [[Modulo Multiplication is Closed|Multiplication modulo $m$ is closed]].
[[Modulo Multiplication is Associative|Multiplication modulo $m$ is associative]].
[[Modulo Multiplication has Identity|Multiplication modulo $m$ has an identity]]:
:$\forall k \in \Z: \eqclass k m \times_m \eqclass 1 m = \eqclass k m = \eqclas... | Integers Modulo m under Multiplication form Commutative Monoid | https://proofwiki.org/wiki/Integers_Modulo_m_under_Multiplication_form_Commutative_Monoid | https://proofwiki.org/wiki/Integers_Modulo_m_under_Multiplication_form_Commutative_Monoid | [
"Modulo Multiplication",
"Examples of Commutative Monoids"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Integers Modulo m",
"Definition:Modulo Multiplication",
"Definition:Commutative Monoid"
] | [
"Modulo Multiplication is Closed",
"Modulo Multiplication is Associative",
"Modulo Multiplication has Identity",
"Identity of Monoid is Unique",
"Modulo Multiplication is Commutative",
"Definition:Commutative Monoid"
] |
proofwiki-1162 | Multiplicative Inverse in Ring of Integers Modulo m | Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$.
Then $\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$ {{iff}} $k \perp m$. | First, suppose $k \perp m$.
That is:
:$\gcd \set {k, m} = 1$
Then, by Bézout's Identity:
:$\exists u, v \in \Z: u k + v m = 1$
Thus:
:$\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$
Thus $\eqclass u m$ is an inverse of $\eqclass k m$.
Suppose that:
:$\exists u \in \Z: \eqclass u m... | Let $\struct {\Z_m, +_m, \times_m}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
Then $\eqclass k m \in \Z_m$ has an [[Definition:Multiplicative Inverse|inverse]] in $\struct {\Z_m, \times_m}$ {{iff}} $k \perp m$. | First, suppose $k \perp m$.
That is:
:$\gcd \set {k, m} = 1$
Then, by [[Bézout's Identity]]:
:$\exists u, v \in \Z: u k + v m = 1$
Thus:
:$\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$
Thus $\eqclass u m$ is an [[Definition:Multiplicative Inverse|inverse]] of $\eqclass k m$.... | Multiplicative Inverse in Ring of Integers Modulo m/Proof 1 | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Ring_of_Integers_Modulo_m | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Ring_of_Integers_Modulo_m/Proof_1 | [
"Ring of Integers Modulo m",
"Multiplicative Inverse in Ring of Integers Modulo m",
"Multiplicative Inverses"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Multiplicative Inverse"
] | [
"Bézout's Identity",
"Definition:Multiplicative Inverse",
"Bézout's Identity"
] |
proofwiki-1163 | Multiplicative Inverse in Ring of Integers Modulo m | Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$.
Then $\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$ {{iff}} $k \perp m$. | From Ring of Integers Modulo m is Ring, $\left({\Z_m, +_m, \times_m}\right)$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity $\left[\!\left[{1}\right]\!\right]_m$]].
Thus by definition $\left({\Z_m, \times_m}\right)$ is a commutative monoid.
The result follows from Multiplicative Inverse in M... | Let $\struct {\Z_m, +_m, \times_m}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
Then $\eqclass k m \in \Z_m$ has an [[Definition:Multiplicative Inverse|inverse]] in $\struct {\Z_m, \times_m}$ {{iff}} $k \perp m$. | From [[Ring of Integers Modulo m is Ring]], $\left({\Z_m, +_m, \times_m}\right)$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity $\left[\!\left[{1}\right]\!\right]_m$]].
Thus by definition $\left({\Z_m, \times_m}\right)$ is a [[Definition:Commutative Monoid|commutative monoid]].
The result ... | Multiplicative Inverse in Ring of Integers Modulo m/Proof 2 | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Ring_of_Integers_Modulo_m | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Ring_of_Integers_Modulo_m/Proof_2 | [
"Ring of Integers Modulo m",
"Multiplicative Inverse in Ring of Integers Modulo m",
"Multiplicative Inverses"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Multiplicative Inverse"
] | [
"Ring of Integers Modulo m is Ring",
"Definition:Commutative Monoid",
"Multiplicative Inverse in Monoid of Integers Modulo m"
] |
proofwiki-1164 | Reduced Residue System is Subset of Set of All Residue Classes | Let $\Z_m$ be the set of set of residue classes modulo $m$.
Let $\Z'_m$ be the reduced residue system modulo $m$.
Then:
:$\forall m \in \Z_{> 1}: \O \subset \Z'_m \subset \Z_m$ | By definition of reduced residue system modulo $m$:
:$\Z'_m = \set {x \in \Z_m: x \perp m}$
From Subset of Set with Propositional Function:
:$\Z'_m \subseteq \Z_m$
We have that:
:$\gcd \set {m, 0} = m$
Thus it follows that:
:$m > 1 \implies \gcd \set {m, 0} \ne 1$
So:
:$\eqclass 0 m \notin \Z'_m$
However:
:$\eqclass 0 ... | Let $\Z_m$ be the set of [[Definition:Set of Residue Classes|set of residue classes modulo $m$]].
Let $\Z'_m$ be the [[Definition:Reduced Residue System|reduced residue system modulo $m$]].
Then:
:$\forall m \in \Z_{> 1}: \O \subset \Z'_m \subset \Z_m$ | By definition of [[Definition:Reduced Residue System|reduced residue system modulo $m$]]:
:$\Z'_m = \set {x \in \Z_m: x \perp m}$
From [[Subset of Set with Propositional Function]]:
:$\Z'_m \subseteq \Z_m$
We have that:
:$\gcd \set {m, 0} = m$
Thus it follows that:
:$m > 1 \implies \gcd \set {m, 0} \ne 1$
So:
:$\e... | Reduced Residue System is Subset of Set of All Residue Classes | https://proofwiki.org/wiki/Reduced_Residue_System_is_Subset_of_Set_of_All_Residue_Classes | https://proofwiki.org/wiki/Reduced_Residue_System_is_Subset_of_Set_of_All_Residue_Classes | [
"Residue Classes",
"Reduced Residue Systems"
] | [
"Definition:Set of Residue Classes",
"Definition:Reduced Residue System"
] | [
"Definition:Reduced Residue System",
"Subset of Set with Propositional Function",
"Category:Residue Classes",
"Category:Reduced Residue Systems"
] |
proofwiki-1165 | Reduced Residue System under Multiplication forms Abelian Group | Let $\Z_m$ be the set of set of residue classes modulo $m$.
Let $\struct {\Z'_m, \times}$ denote the multiplicative group of reduced residues modulo $m$.
Then $\struct {\Z'_m, \times}$ is an abelian group, precisely equal to the group of units of $\Z_m$. | From Ring of Integers Modulo m is Ring, $\struct {\Z_m, +, \times}$ forms a (commutative) ring with unity.
Then we have that the units of a ring with unity form a group.
By Multiplicative Inverse in Ring of Integers Modulo m we have that the elements of $\struct {\Z'_m, \times}$ are precisely those that have inverses, ... | Let $\Z_m$ be the set of [[Definition:Set of Residue Classes|set of residue classes modulo $m$]].
Let $\struct {\Z'_m, \times}$ denote the [[Definition:Multiplicative Group of Reduced Residues|multiplicative group of reduced residues modulo $m$]].
Then $\struct {\Z'_m, \times}$ is an [[Definition:Abelian Group|abel... | From [[Ring of Integers Modulo m is Ring]], $\struct {\Z_m, +, \times}$ forms a [[Definition:Commutative and Unitary Ring|(commutative) ring with unity]].
Then we have that the [[Group of Units is Group|units of a ring with unity form a group]].
By [[Multiplicative Inverse in Ring of Integers Modulo m]] we have that ... | Reduced Residue System under Multiplication forms Abelian Group/Proof 1 | https://proofwiki.org/wiki/Reduced_Residue_System_under_Multiplication_forms_Abelian_Group | https://proofwiki.org/wiki/Reduced_Residue_System_under_Multiplication_forms_Abelian_Group/Proof_1 | [
"Reduced Residue Systems",
"Examples of Abelian Groups",
"Multiplicative Groups of Reduced Residues",
"Reduced Residue System under Multiplication forms Abelian Group"
] | [
"Definition:Set of Residue Classes",
"Definition:Multiplicative Group of Reduced Residues",
"Definition:Abelian Group",
"Definition:Group of Units"
] | [
"Ring of Integers Modulo m is Ring",
"Definition:Commutative and Unitary Ring",
"Group of Units is Group",
"Multiplicative Inverse in Ring of Integers Modulo m",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Unit of Ring",
"Definition:Abelian Group",
"Restriction of Commutative Operatio... |
proofwiki-1166 | Reduced Residue System under Multiplication forms Abelian Group | Let $\Z_m$ be the set of set of residue classes modulo $m$.
Let $\struct {\Z'_m, \times}$ denote the multiplicative group of reduced residues modulo $m$.
Then $\struct {\Z'_m, \times}$ is an abelian group, precisely equal to the group of units of $\Z_m$. | Taking the group axioms in turn:
=== {{Group-axiom|0|nolink}} ===
From Modulo Multiplication on Reduced Residue System is Closed:
:$\struct {\Z'_m, \times}$ is closed.
{{qed|lemma}}
=== {{Group-axiom|1|nolink}} ===
We have that Modulo Multiplication is Associative.
{{qed|lemma}}
=== {{Group-axiom|2|nolink}} ===
From Mo... | Let $\Z_m$ be the set of [[Definition:Set of Residue Classes|set of residue classes modulo $m$]].
Let $\struct {\Z'_m, \times}$ denote the [[Definition:Multiplicative Group of Reduced Residues|multiplicative group of reduced residues modulo $m$]].
Then $\struct {\Z'_m, \times}$ is an [[Definition:Abelian Group|abel... | Taking the [[Axiom:Group Axioms|group axioms]] in turn:
=== {{Group-axiom|0|nolink}} ===
From [[Modulo Multiplication on Reduced Residue System is Closed]]:
:$\struct {\Z'_m, \times}$ is [[Definition:Closed Algebraic Structure|closed]].
{{qed|lemma}}
=== {{Group-axiom|1|nolink}} ===
We have that [[Modulo Multipl... | Reduced Residue System under Multiplication forms Abelian Group/Proof 2 | https://proofwiki.org/wiki/Reduced_Residue_System_under_Multiplication_forms_Abelian_Group | https://proofwiki.org/wiki/Reduced_Residue_System_under_Multiplication_forms_Abelian_Group/Proof_2 | [
"Reduced Residue Systems",
"Examples of Abelian Groups",
"Multiplicative Groups of Reduced Residues",
"Reduced Residue System under Multiplication forms Abelian Group"
] | [
"Definition:Set of Residue Classes",
"Definition:Multiplicative Group of Reduced Residues",
"Definition:Abelian Group",
"Definition:Group of Units"
] | [
"Axiom:Group Axioms",
"Modulo Multiplication on Reduced Residue System is Closed",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Modulo Multiplication is Associative",
"Modulo Multiplication has Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Multiplicative Inver... |
proofwiki-1167 | Reduced Residue System under Multiplication forms Abelian Group | Let $\Z_m$ be the set of set of residue classes modulo $m$.
Let $\struct {\Z'_m, \times}$ denote the multiplicative group of reduced residues modulo $m$.
Then $\struct {\Z'_m, \times}$ is an abelian group, precisely equal to the group of units of $\Z_m$. | Taking the finite group axioms in turn:
=== $\text {FG} 0$: Closure ===
From Modulo Multiplication on Reduced Residue System is Closed:
:$\struct {\Z'_m, \times}$ is closed.
{{qed|lemma}}
=== $\text {FG} 1$: Associativity ===
We have that Modulo Multiplication is Associative.
{{qed|lemma}}
=== $\text {FG} 2$: Finitenes... | Let $\Z_m$ be the set of [[Definition:Set of Residue Classes|set of residue classes modulo $m$]].
Let $\struct {\Z'_m, \times}$ denote the [[Definition:Multiplicative Group of Reduced Residues|multiplicative group of reduced residues modulo $m$]].
Then $\struct {\Z'_m, \times}$ is an [[Definition:Abelian Group|abel... | Taking the [[Axiom:Finite Group Axioms|finite group axioms]] in turn:
=== $\text {FG} 0$: Closure ===
From [[Modulo Multiplication on Reduced Residue System is Closed]]:
:$\struct {\Z'_m, \times}$ is [[Definition:Closed Algebraic Structure|closed]].
{{qed|lemma}}
=== $\text {FG} 1$: Associativity ===
We have tha... | Reduced Residue System under Multiplication forms Abelian Group/Proof 3 | https://proofwiki.org/wiki/Reduced_Residue_System_under_Multiplication_forms_Abelian_Group | https://proofwiki.org/wiki/Reduced_Residue_System_under_Multiplication_forms_Abelian_Group/Proof_3 | [
"Reduced Residue Systems",
"Examples of Abelian Groups",
"Multiplicative Groups of Reduced Residues",
"Reduced Residue System under Multiplication forms Abelian Group"
] | [
"Definition:Set of Residue Classes",
"Definition:Multiplicative Group of Reduced Residues",
"Definition:Abelian Group",
"Definition:Group of Units"
] | [
"Axiom:Finite Group/Axioms",
"Modulo Multiplication on Reduced Residue System is Closed",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Modulo Multiplication is Associative",
"Definition:Euler Phi Function",
"Definition:Order of Structure/Finite Structure",
"Modulo Multiplication on Redu... |
proofwiki-1168 | Ring of Integers Modulo Prime is Field | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then:
:$m$ is prime
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a field. | === Prime Modulus ===
$\struct {\Z_m, +, \times}$ is a commutative ring with unity by definition.
From Reduced Residue System under Multiplication forms Abelian Group, $\struct {\Z'_m, \times}$ is an abelian group.
$\Z'_m$ consists of all the elements of $\Z_m$ coprime to $m$.
Now when $m$ is prime, we have, from Reduc... | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
Then:
:$m$ is [[Definition:Prime Number|prime]]
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]]. | === Prime Modulus ===
$\struct {\Z_m, +, \times}$ is a [[Definition:Commutative Ring with Unity|commutative ring with unity]] by definition.
From [[Reduced Residue System under Multiplication forms Abelian Group]], $\struct {\Z'_m, \times}$ is an [[Definition:Abelian Group|abelian group]].
$\Z'_m$ consists of all th... | Ring of Integers Modulo Prime is Field/Proof 1 | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_Prime_is_Field | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_Prime_is_Field/Proof_1 | [
"Ring of Integers Modulo Prime is Field",
"Ring of Integers Modulo m",
"Galois Fields"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Prime Number",
"Definition:Field (Abstract Algebra)"
] | [
"Definition:Commutative and Unitary Ring",
"Reduced Residue System under Multiplication forms Abelian Group",
"Definition:Abelian Group",
"Definition:Coprime/Integers",
"Definition:Prime Number",
"Reduced Residue System Modulo Prime",
"Definition:Set Difference",
"Definition:Composite Number",
"Ring... |
proofwiki-1169 | Ring of Integers Modulo Prime is Field | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then:
:$m$ is prime
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a field. | Let $p$ be prime.
From Irreducible Elements of Ring of Integers, we have that $p$ is irreducible in the ring of integers $\struct {\Z, +, \times}$.
From Ring of Integers is Principal Ideal Domain, $\struct {\Z, +, \times}$ is a principal ideal domain.
Let $\ideal p$ denote the principal ideal of $\struct {\Z, +, \times... | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
Then:
:$m$ is [[Definition:Prime Number|prime]]
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]]. | Let $p$ be [[Definition:Prime Number|prime]].
From [[Irreducible Elements of Ring of Integers]], we have that $p$ is [[Definition:Irreducible Element of Ring|irreducible]] in the [[Definition:Ring of Integers|ring of integers]] $\struct {\Z, +, \times}$.
From [[Ring of Integers is Principal Ideal Domain]], $\struct {... | Ring of Integers Modulo Prime is Field/Proof 2 | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_Prime_is_Field | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_Prime_is_Field/Proof_2 | [
"Ring of Integers Modulo Prime is Field",
"Ring of Integers Modulo m",
"Galois Fields"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Prime Number",
"Definition:Field (Abstract Algebra)"
] | [
"Definition:Prime Number",
"Irreducible Elements of Ring of Integers",
"Definition:Irreducible Element of Ring",
"Definition:Ring of Integers",
"Ring of Integers is Principal Ideal Domain",
"Definition:Principal Ideal Domain",
"Definition:Principal Ideal of Ring",
"Principal Ideal of Principal Ideal D... |
proofwiki-1170 | Ring of Integers Modulo Prime is Field | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then:
:$m$ is prime
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a field. | Let $m$ be prime.
From Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain.
Let $\eqclass a m \ne \eqclass 0 m$ be a residue class modulo $m$.
We need to find a residue class modulo $m$ $\eqclass x m$ such that $\eqclass a m \eqclass x m = \eqclass 1 m$.
Because $m$ is pr... | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
Then:
:$m$ is [[Definition:Prime Number|prime]]
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]]. | Let $m$ be [[Definition:Prime Number|prime]].
From [[Ring of Integers Modulo Prime is Integral Domain]], $\struct {\Z_m, +, \times}$ is an [[Definition:Integral Domain|integral domain]].
Let $\eqclass a m \ne \eqclass 0 m$ be a [[Definition:Residue Class|residue class modulo $m$]].
We need to find a [[Definition:Res... | Ring of Integers Modulo Prime is Field/Proof 3 | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_Prime_is_Field | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_Prime_is_Field/Proof_3 | [
"Ring of Integers Modulo Prime is Field",
"Ring of Integers Modulo m",
"Galois Fields"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Prime Number",
"Definition:Field (Abstract Algebra)"
] | [
"Definition:Prime Number",
"Ring of Integers Modulo Prime is Integral Domain",
"Definition:Integral Domain",
"Definition:Residue Class",
"Definition:Residue Class",
"Definition:Prime Number",
"Bézout's Identity",
"Definition:Ring Zero",
"Definition:Residue Class",
"Definition:Product Inverse",
"... |
proofwiki-1171 | Ring of Integers Modulo Prime is Field | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then:
:$m$ is prime
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a field. | Let $m$ be prime.
From Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain.
From Finite Integral Domain is Galois Field, $\struct {\Z_m, +, \times}$ is a field.
{{qed|lemma}}
Now suppose $m \in \Z: m \ge 2$ is composite.
From Ring of Integers Modulo Composite is not Integ... | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
Then:
:$m$ is [[Definition:Prime Number|prime]]
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]]. | Let $m$ be [[Definition:Prime Number|prime]].
From [[Ring of Integers Modulo Prime is Integral Domain]], $\struct {\Z_m, +, \times}$ is an [[Definition:Integral Domain|integral domain]].
From [[Finite Integral Domain is Galois Field]], $\struct {\Z_m, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]].
{{... | Ring of Integers Modulo Prime is Field/Proof 4 | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_Prime_is_Field | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_Prime_is_Field/Proof_4 | [
"Ring of Integers Modulo Prime is Field",
"Ring of Integers Modulo m",
"Galois Fields"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Prime Number",
"Definition:Field (Abstract Algebra)"
] | [
"Definition:Prime Number",
"Ring of Integers Modulo Prime is Integral Domain",
"Definition:Integral Domain",
"Finite Integral Domain is Galois Field",
"Definition:Field (Abstract Algebra)",
"Definition:Composite Number",
"Ring of Integers Modulo Composite is not Integral Domain",
"Definition:Integral ... |
proofwiki-1172 | Subgroups of Additive Group of Integers | Let $\struct {\Z, +}$ be the additive group of integers.
Let $n \Z$ be the additive group of integer multiples of $n$.
Every non-trivial subgroup of $\struct {\Z, +}$ has the form $n \Z$. | First we note that, from Integer Multiples under Addition form Infinite Cyclic Group, $\struct {n \Z, +}$ is an infinite cyclic group.
From Cyclic Group is Abelian, it follows that $\struct {n \Z, +}$ is an infinite abelian group.
Let $H$ be a non-trivial subgroup of $\struct {\Z, +}$.
Because $H$ is non-trivial:
:$\ex... | Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
Let $n \Z$ be the [[Definition:Additive Group of Integer Multiples|additive group of integer multiples]] of $n$.
Every [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$ ... | First we note that, from [[Integer Multiples under Addition form Infinite Cyclic Group]], $\struct {n \Z, +}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Cyclic Group|cyclic group]].
From [[Cyclic Group is Abelian]], it follows that $\struct {n \Z, +}$ is an [[Definition:Infinite Group|infinite]] [[Defin... | Subgroups of Additive Group of Integers | https://proofwiki.org/wiki/Subgroups_of_Additive_Group_of_Integers | https://proofwiki.org/wiki/Subgroups_of_Additive_Group_of_Integers | [
"Subgroups",
"Additive Group of Integers",
"Additive Groups of Integer Multiples",
"Subgroups of Additive Group of Integers"
] | [
"Definition:Additive Group of Integers",
"Definition:Additive Group of Integer Multiples",
"Definition:Non-Trivial Group",
"Definition:Subgroup"
] | [
"Integer Multiples under Addition form Infinite Cyclic Group",
"Definition:Infinite Group",
"Definition:Cyclic Group",
"Cyclic Group is Abelian",
"Definition:Infinite Group",
"Definition:Abelian Group",
"Definition:Non-Trivial Group",
"Definition:Subgroup",
"Definition:Non-Trivial Group",
"Definit... |
proofwiki-1173 | Integer Multiples under Addition form Infinite Cyclic Group | Let $n \Z$ be the set of integer multiples of $n$.
Then $\struct {n \Z, +}$ is a countably infinite cyclic group.
It is generated by $n$ and $-n$:
:$n \Z = \gen n$
:$n \Z = \gen {-n}$
Hence $\struct {n \Z, +}$ can be justifiably referred to as the additive group of integer multiples. | From Integer Multiples under Addition form Subgroup of Integers, $\struct {n \Z, +}$ is a subgroup of the additive group of integers $\struct {\Z, +}$.
From Integers under Addition form Infinite Cyclic Group, $\struct {\Z, +}$ is a cyclic group.
So by Subgroup of Cyclic Group is Cyclic, $\struct {n \Z, +}$ is a cyclic ... | Let $n \Z$ be the [[Definition:Set of Integer Multiples|set of integer multiples]] of $n$.
Then $\struct {n \Z, +}$ is a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Cyclic Group|cyclic group]].
It is generated by $n$ and $-n$:
:$n \Z = \gen n$
:$n \Z = \gen {-n}$
Hence $\struct {n \Z, +}$ c... | From [[Integer Multiples under Addition form Subgroup of Integers]], $\struct {n \Z, +}$ is a [[Definition:Subgroup|subgroup]] of the [[Definition:Additive Group of Integers|additive group of integers]] $\struct {\Z, +}$.
From [[Integers under Addition form Infinite Cyclic Group]], $\struct {\Z, +}$ is a [[Definition:... | Integer Multiples under Addition form Infinite Cyclic Group | https://proofwiki.org/wiki/Integer_Multiples_under_Addition_form_Infinite_Cyclic_Group | https://proofwiki.org/wiki/Integer_Multiples_under_Addition_form_Infinite_Cyclic_Group | [
"Additive Groups of Integer Multiples",
"Infinite Cyclic Group",
"Sets of Integer Multiples"
] | [
"Definition:Set of Integer Multiples",
"Definition:Countably Infinite/Set",
"Definition:Cyclic Group",
"Definition:Additive Group of Integer Multiples"
] | [
"Integer Multiples under Addition form Subgroup of Integers",
"Definition:Subgroup",
"Definition:Additive Group of Integers",
"Integers under Addition form Infinite Cyclic Group",
"Definition:Cyclic Group",
"Subgroup of Cyclic Group is Cyclic",
"Definition:Cyclic Group",
"Subgroup of Infinite Cyclic G... |
proofwiki-1174 | Quotient Group of Integers by Multiples | Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {m \Z, +}$ be the additive group of integer multiples of $m$.
Let $\struct {\Z_m, +_m}$ be the additive group of integers modulo $m$.
Then the quotient group of $\struct {\Z, +}$ by $\struct {m \Z, +}$ is $\struct {\Z_m, +_m}$.
Thus:
:$\index \Z {m \... | From Subgroups of Additive Group of Integers, $\struct {m \Z, +}$ is a subgroup of $\struct {\Z, +}$.
From Subgroup of Abelian Group is Normal, $\struct {m \Z, +}$ is normal in $\struct {\Z, +}$.
Therefore the quotient group $\dfrac {\struct {\Z, +} } {\struct {m \Z, +} }$ is defined.
Now $\Z$ modulo $m \Z$ is Congruen... | Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
Let $\struct {m \Z, +}$ be the [[Definition:Additive Group of Integer Multiples|additive group of integer multiples]] of $m$.
Let $\struct {\Z_m, +_m}$ be the [[Definition:Additive Group of Integers Modulo m|additive gr... | From [[Subgroups of Additive Group of Integers]], $\struct {m \Z, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$.
From [[Subgroup of Abelian Group is Normal]], $\struct {m \Z, +}$ is [[Definition:Normal Subgroup|normal]] in $\struct {\Z, +}$.
Therefore the [[Definition:Quotient Group|quotient group]] ... | Quotient Group of Integers by Multiples | https://proofwiki.org/wiki/Quotient_Group_of_Integers_by_Multiples | https://proofwiki.org/wiki/Quotient_Group_of_Integers_by_Multiples | [
"Examples of Quotient Groups",
"Modulo Arithmetic",
"Additive Group of Integers",
"Additive Groups of Integer Multiples",
"Additive Groups of Integers Modulo m"
] | [
"Definition:Additive Group of Integers",
"Definition:Additive Group of Integer Multiples",
"Definition:Additive Group of Integers Modulo m",
"Definition:Quotient Group"
] | [
"Subgroups of Additive Group of Integers",
"Definition:Subgroup",
"Subgroup of Abelian Group is Normal",
"Definition:Normal Subgroup",
"Definition:Quotient Group",
"Definition:Congruence Modulo Subgroup",
"Definition:Congruence (Number Theory)/Integers",
"Definition:Quotient Set",
"Definition:Coset/... |
proofwiki-1175 | Euler's Theorem (Number Theory) | Let $a, m \in \Z$ be coprime integers: $a \perp m$.
Let $\map \phi m$ be the Euler $\phi$ function of $m$.
Then:
:$a^{\map \phi m} \equiv 1 \pmod m$ | Let $\eqclass a m$ denote the residue class modulo $m$ of $a$.
Since $a \perp m$, it follows by Reduced Residue System under Multiplication forms Abelian Group that $\eqclass a m$ belongs to the abelian group $\struct {\Z'_m, \times}$.
Let $k = \order {\eqclass a m}$ where $\order {\, \cdot \,}$ denotes the order of a ... | Let $a, m \in \Z$ be [[Definition:Coprime Integers|coprime integers]]: $a \perp m$.
Let $\map \phi m$ be the [[Definition:Euler Phi Function|Euler $\phi$ function]] of $m$.
Then:
:$a^{\map \phi m} \equiv 1 \pmod m$ | Let $\eqclass a m$ denote the [[Definition:Residue Class|residue class modulo $m$ of $a$]].
Since $a \perp m$, it follows by [[Reduced Residue System under Multiplication forms Abelian Group]] that $\eqclass a m$ belongs to the [[Definition:Abelian Group|abelian group]] $\struct {\Z'_m, \times}$.
Let $k = \order {\eq... | Euler's Theorem (Number Theory) | https://proofwiki.org/wiki/Euler's_Theorem_(Number_Theory) | https://proofwiki.org/wiki/Euler's_Theorem_(Number_Theory) | [
"Euler's Theorem (Number Theory)",
"Euler's Theorems",
"Number Theory"
] | [
"Definition:Coprime/Integers",
"Definition:Euler Phi Function"
] | [
"Definition:Residue Class",
"Reduced Residue System under Multiplication forms Abelian Group",
"Definition:Abelian Group",
"Definition:Order of Group Element",
"Order of Element Divides Order of Finite Group",
"Definition:Euler Phi Function",
"Congruence of Powers"
] |
proofwiki-1176 | Symmetry Group is Group | Let $P$ be a geometric figure.
Let $S_P$ be the set of all symmetries of $P$.
Let $\circ$ denote composition of mappings.
The symmetry group $\struct {S_P, \circ}$ is indeed a group. | By definition, a symmetry is a bijection, and hence a permutation.
From Symmetric Group is Group, the set of all permutations on $P$ form the symmetric group $\struct {\map \Gamma P, \circ}$ on $P$.
Thus $S_P$ is a subset of $\struct {\map \Gamma P, \circ}$.
Let $A$ and $B$ be symmetries on $P$.
From Composition of Sym... | Let $P$ be a [[Definition:Geometric Figure|geometric figure]].
Let $S_P$ be the [[Definition:Set|set]] of all [[Definition:Symmetry (Geometry)|symmetries]] of $P$.
Let $\circ$ denote [[Definition:Composition of Mappings|composition of mappings]].
The [[Definition:Symmetry Group|symmetry group]] $\struct {S_P, \circ... | By definition, a [[Definition:Symmetry (Geometry)|symmetry]] is a [[Definition:Bijection|bijection]], and hence a [[Definition:Permutation|permutation]].
From [[Symmetric Group is Group]], the [[Definition:Set|set]] of all [[Definition:Permutation|permutations]] on $P$ form the [[Definition:Symmetric Group|symmetric g... | Symmetry Group is Group | https://proofwiki.org/wiki/Symmetry_Group_is_Group | https://proofwiki.org/wiki/Symmetry_Group_is_Group | [
"Symmetry Groups"
] | [
"Definition:Geometric Figure",
"Definition:Set",
"Definition:Symmetry (Geometry)",
"Definition:Composition of Mappings",
"Definition:Symmetry Group",
"Definition:Group"
] | [
"Definition:Symmetry (Geometry)",
"Definition:Bijection",
"Definition:Permutation",
"Symmetric Group is Group",
"Definition:Set",
"Definition:Permutation",
"Definition:Symmetric Group",
"Definition:Subset",
"Definition:Symmetry (Geometry)",
"Composition of Symmetries is Symmetry",
"Definition:Sy... |
proofwiki-1177 | Internal Angles of Square | The internal angles of a square are right angles. | By definition, a square is a regular quadrilateral.
From Internal Angles of Regular Polygon, the internal angles of a square measure $\dfrac {180 \degrees \paren {4 - 2} } 4 = 90 \degrees$.
The result follows from Measurement of Right Angle.
{{qed}}
Category:Squares
Category:Internal Angles
e7un0prr5q4j459ggtc6jabni989... | The [[Definition:Internal Angle|internal angles]] of a [[Definition:Square (Geometry)|square]] are [[Definition:Right Angle|right angles]]. | By definition, a [[Definition:Square (Geometry)|square]] is a [[Definition:Regular Polygon|regular]] [[Definition:Quadrilateral|quadrilateral]].
From [[Internal Angles of Regular Polygon]], the [[Definition:Internal Angle|internal angles]] of a [[Definition:Square (Geometry)|square]] measure $\dfrac {180 \degrees \par... | Internal Angles of Square | https://proofwiki.org/wiki/Internal_Angles_of_Square | https://proofwiki.org/wiki/Internal_Angles_of_Square | [
"Squares",
"Internal Angles"
] | [
"Definition:Polygon/Internal Angle",
"Definition:Quadrilateral/Square",
"Definition:Right Angle"
] | [
"Definition:Quadrilateral/Square",
"Definition:Polygon/Regular",
"Definition:Quadrilateral",
"Internal Angles of Regular Polygon",
"Definition:Polygon/Internal Angle",
"Definition:Quadrilateral/Square",
"Measurements of Common Angles/Right Angle",
"Category:Squares",
"Category:Internal Angles"
] |
proofwiki-1178 | Area of Square | A square has an area of $L^2$ where $L$ is the length of a side of the square.
Thus we have that the area is a function of the length of the side:
:$\forall L \in \R_{\ge 0}: \map \Area L = L^2$
where it is noted that the domain of $L$ is the set of non-negative real numbers. | === Integer Side Length ===
In the case where $L = 1$, the statement follows from the definition of area.
If $L \in \N, L > 1$, then we can divide the square into smaller squares, each of side length one.
Since there will be $L$ squares of side length one on each side, it follows that there will be $L \cdot L = L^2$ sq... | A [[Definition:Square (Geometry)|square]] has an [[Definition:Area|area]] of $L^2$ where $L$ is the [[Definition:Length of Line|length]] of a [[Definition:Side of Polygon|side]] of the square.
Thus we have that the [[Definition:Area|area]] is a [[Definition:Real Function|function]] of the [[Definition:Length of Line|l... | === Integer Side Length ===
In the case where $L = 1$, the statement follows from the [[Definition:Area|definition of area]].
If $L \in \N, L > 1$, then we can divide the square into smaller squares, each of side length one.
Since there will be $L$ squares of side length one on each side, it follows that there will ... | Area of Square/Proof 1 | https://proofwiki.org/wiki/Area_of_Square | https://proofwiki.org/wiki/Area_of_Square/Proof_1 | [
"Areas of Quadrilaterals",
"Area of Square"
] | [
"Definition:Quadrilateral/Square",
"Definition:Area",
"Definition:Linear Measure/Length",
"Definition:Polygon/Side",
"Definition:Area",
"Definition:Real Function",
"Definition:Linear Measure/Length",
"Definition:Polygon/Side",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Positive/Real Num... | [
"Definition:Area",
"Definition:Area",
"Definition:Quadrilateral/Square",
"Definition:Rational Number",
"Definition:Quadrilateral/Square",
"Definition:Polygon/Side",
"Definition:Linear Measure/Length",
"Area of Square/Proof 1",
"Definition:Polygon/Side",
"Definition:Quadrilateral/Square",
"Defini... |
proofwiki-1179 | Area of Square | A square has an area of $L^2$ where $L$ is the length of a side of the square.
Thus we have that the area is a function of the length of the side:
:$\forall L \in \R_{\ge 0}: \map \Area L = L^2$
where it is noted that the domain of $L$ is the set of non-negative real numbers. | Let $\Box ABCD$ be a square whose side $AB$ is of length $L$.
Let $\Box EFGH$ be a square whose side $EF$ is of length $1$.
:300px
From the Axioms of Area, the area of $\Box EFGH$ is $1$.
By definition, $AB : EF = L : 1$.
From Similar Polygons are composed of Similar Triangles, the ratio of the areas of $\Box ABCD$ to ... | A [[Definition:Square (Geometry)|square]] has an [[Definition:Area|area]] of $L^2$ where $L$ is the [[Definition:Length of Line|length]] of a [[Definition:Side of Polygon|side]] of the square.
Thus we have that the [[Definition:Area|area]] is a [[Definition:Real Function|function]] of the [[Definition:Length of Line|l... | Let $\Box ABCD$ be a [[Definition:Square (Geometry)|square]] whose [[Definition:Side of Polygon|side]] $AB$ is of [[Definition:Linear Measure|length]] $L$.
Let $\Box EFGH$ be a [[Definition:Square (Geometry)|square]] whose [[Definition:Side of Polygon|side]] $EF$ is of [[Definition:Linear Measure|length]] $1$.
:[[Fil... | Area of Square/Proof 2 | https://proofwiki.org/wiki/Area_of_Square | https://proofwiki.org/wiki/Area_of_Square/Proof_2 | [
"Areas of Quadrilaterals",
"Area of Square"
] | [
"Definition:Quadrilateral/Square",
"Definition:Area",
"Definition:Linear Measure/Length",
"Definition:Polygon/Side",
"Definition:Area",
"Definition:Real Function",
"Definition:Linear Measure/Length",
"Definition:Polygon/Side",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Positive/Real Num... | [
"Definition:Quadrilateral/Square",
"Definition:Polygon/Side",
"Definition:Linear Measure",
"Definition:Quadrilateral/Square",
"Definition:Polygon/Side",
"Definition:Linear Measure",
"File:AreaOfSquare.png",
"Axiom:Area Axioms",
"Definition:Area",
"Similar Polygons are composed of Similar Triangles... |
proofwiki-1180 | Area of Square | A square has an area of $L^2$ where $L$ is the length of a side of the square.
Thus we have that the area is a function of the length of the side:
:$\forall L \in \R_{\ge 0}: \map \Area L = L^2$
where it is noted that the domain of $L$ is the set of non-negative real numbers. | Let a square have a side length $a \in \R$.
This square is equivalent to the area under the graph of $\map f x = a$ from $0$ to $a$.
Thus from the geometric interpretation of the definite integral, the area of the square will be the integral:
:$\ds A = \int_0^a a \rd l$
Thus:
{{begin-eqn}}
{{eqn | l = A
| r = \in... | A [[Definition:Square (Geometry)|square]] has an [[Definition:Area|area]] of $L^2$ where $L$ is the [[Definition:Length of Line|length]] of a [[Definition:Side of Polygon|side]] of the square.
Thus we have that the [[Definition:Area|area]] is a [[Definition:Real Function|function]] of the [[Definition:Length of Line|l... | Let a [[Definition:Square (Geometry)|square]] have a [[Definition:Side of Polygon|side length]] $a \in \R$.
This square is equivalent to the area under the graph of $\map f x = a$ from $0$ to $a$.
Thus from [[Definition:Geometric Interpretation of Definite Integral|the geometric interpretation of the definite integra... | Area of Square/Proof 3 | https://proofwiki.org/wiki/Area_of_Square | https://proofwiki.org/wiki/Area_of_Square/Proof_3 | [
"Areas of Quadrilaterals",
"Area of Square"
] | [
"Definition:Quadrilateral/Square",
"Definition:Area",
"Definition:Linear Measure/Length",
"Definition:Polygon/Side",
"Definition:Area",
"Definition:Real Function",
"Definition:Linear Measure/Length",
"Definition:Polygon/Side",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Positive/Real Num... | [
"Definition:Quadrilateral/Square",
"Definition:Polygon/Side",
"Definition:Darboux Integral/Geometric Interpretation",
"Integral of Constant",
"Definition:Definite Integral",
"Definition:Area",
"Area of Parallelogram/Rectangle",
"Definition:Linear Measure/Breadth",
"Definition:Linear Measure/Height",... |
proofwiki-1181 | Order of Symmetric Group | Let $S$ be a finite set of cardinality $n$.
Let $\struct {\map \Gamma S, \circ}$ be the symmetric group on $S$.
Then $\struct {\map \Gamma S, \circ}$ has $n!$ elements (see factorial). | A direct application of Cardinality of Set of Bijections.
{{qed}} | Let $S$ be a [[Definition:Finite Set|finite set]] of [[Definition:Cardinality|cardinality]] $n$.
Let $\struct {\map \Gamma S, \circ}$ be the [[Definition:Symmetric Group|symmetric group]] on $S$.
Then $\struct {\map \Gamma S, \circ}$ has $n!$ elements (see [[Definition:Factorial|factorial]]). | A direct application of [[Cardinality of Set of Bijections]].
{{qed}} | Order of Symmetric Group | https://proofwiki.org/wiki/Order_of_Symmetric_Group | https://proofwiki.org/wiki/Order_of_Symmetric_Group | [
"Symmetric Groups"
] | [
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Symmetric Group",
"Definition:Factorial"
] | [
"Cardinality of Set of Bijections"
] |
proofwiki-1182 | Powers of Permutation Element | Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$, and let $i \in \N^*_n$.
Let $k \in \Z: k > 0$ be the smallest such that:
:$\map {\pi^k} i \in \set {i, \map \pi i, \map {\pi^2} i, \ldots, \map {\pi^{k - 1} } i}$
Then $\map {\pi^k} i = i$. | {{AimForCont}} $\map {\pi^k} i = \map {\pi^r} i$ for some $r > 0$.
Then, since $\pi$ has an inverse, $\map {\pi^{k - r} } i = i$.
This contradicts the definition of $k$, so $r = 0$.
{{qed}}
Category:Symmetric Groups
07qq48opp9x93zw1utfe45zwng026lm | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $\pi \in S_n$, and let $i \in \N^*_n$.
Let $k \in \Z: k > 0$ be the smallest such that:
:$\map {\pi^k} i \in \set {i, \map \pi i, \map {\pi^2} i, \ldots, \map {\pi^{k - 1} } i}$
Then $\map {\pi^k} i = i$. | {{AimForCont}} $\map {\pi^k} i = \map {\pi^r} i$ for some $r > 0$.
Then, since $\pi$ has an [[Definition:Inverse Element|inverse]], $\map {\pi^{k - r} } i = i$.
This contradicts the definition of $k$, so $r = 0$.
{{qed}}
[[Category:Symmetric Groups]]
07qq48opp9x93zw1utfe45zwng026lm | Powers of Permutation Element | https://proofwiki.org/wiki/Powers_of_Permutation_Element | https://proofwiki.org/wiki/Powers_of_Permutation_Element | [
"Symmetric Groups"
] | [
"Definition:Symmetric Group/n Letters"
] | [
"Definition:Inverse (Abstract Algebra)/Inverse",
"Category:Symmetric Groups"
] |
proofwiki-1183 | Fixed Elements form 1-Cycles | Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$.
Let $\Fix \pi$ be the set of elements fixed by $\pi$.
For any $\pi \in S_n$, all the elements of $\Fix \pi$ form $1$-cycles. | Let $\pi$ be a permutation, and let $x \in \Fix \pi$.
From the definition of a fixed element:
:$\map \pi x = x$
From the definition of a $k$-cycle, we see that $1$ is the smallest $k \in \Z: k > 0$ such that:
:$\map {\pi^k} x = x$
The result follows.
{{qed}}
Category:Fixed Elements under Permutations
Category:Symmetric... | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $\pi \in S_n$.
Let $\Fix \pi$ be the [[Definition:Set of Fixed Elements|set of elements fixed by $\pi$]].
For any $\pi \in S_n$, all the elements of $\Fix \pi$ form [[Definition:Cyclic Permutation|$1$-cycles]]. | Let $\pi$ be a [[Definition:Permutation|permutation]], and let $x \in \Fix \pi$.
From the definition of a [[Definition:Fixed Element under Permutation|fixed element]]:
:$\map \pi x = x$
From the definition of a [[Definition:Cyclic Permutation|$k$-cycle]], we see that $1$ is the smallest $k \in \Z: k > 0$ such that:
:... | Fixed Elements form 1-Cycles | https://proofwiki.org/wiki/Fixed_Elements_form_1-Cycles | https://proofwiki.org/wiki/Fixed_Elements_form_1-Cycles | [
"Fixed Elements under Permutations",
"Symmetric Groups"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Fixed Element under Permutation/Set of Fixed Elements",
"Definition:Cyclic Permutation"
] | [
"Definition:Permutation",
"Definition:Fixed Element under Permutation",
"Definition:Cyclic Permutation",
"Category:Fixed Elements under Permutations",
"Category:Symmetric Groups"
] |
proofwiki-1184 | Equality of Cycles | Let $S_n$ denote the symmetric group on $n$ letters, realised as the permutations of $\set {1, \ldots, n}$.
Let:
:$\rho = \begin {bmatrix} a_0 & \cdots & a_{k - 1} \end {bmatrix} \in S_n$
:$\sigma = \begin {bmatrix} b_0 & \cdots & b_{k - 1} \end {bmatrix} \in S_n$
be $k$-cycles of $S_n$.
For $d \in \Z$, by Integer is C... | {{ProofWanted}}
Category:Symmetric Groups
1zc47z6t6vjz3fckdm4zldg8bqcv4kj | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]], realised as the [[Definition:Permutation on n Letters|permutations of $\set {1, \ldots, n}$]].
Let:
:$\rho = \begin {bmatrix} a_0 & \cdots & a_{k - 1} \end {bmatrix} \in S_n$
:$\sigma = \begin {bmatrix} b_0 & \cdots & b_{k... | {{ProofWanted}}
[[Category:Symmetric Groups]]
1zc47z6t6vjz3fckdm4zldg8bqcv4kj | Equality of Cycles | https://proofwiki.org/wiki/Equality_of_Cycles | https://proofwiki.org/wiki/Equality_of_Cycles | [
"Symmetric Groups"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Permutation on n Letters",
"Definition:Cyclic Permutation",
"Integer is Congruent to Integer less than Modulus",
"Definition:Integer"
] | [
"Category:Symmetric Groups"
] |
proofwiki-1185 | Identity Permutation is Disjoint from All | Let $S_n$ denote the symmetric group on $n$ letters.
Let $e \in S_n$ be the identity permutation on $S_n$.
Then $e$ is disjoint from every permutation $\pi$ on $S_n$ (including itself). | By definition of the identity permutation:
:$\forall i \in \N_{>0}: \map e i = i$
Thus $e$ fixes all elements of $S_n$.
Thus each element moved by a permutation $\pi$ is fixed by $e$.
The set of elements moved by $e$ is $\O$, so the converse is true vacuously.
{{qed}}
Category:Symmetric Groups
Category:Identity Mapping... | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $e \in S_n$ be the [[Definition:Identity Mapping|identity permutation]] on $S_n$.
Then $e$ is [[Definition:Disjoint Permutations|disjoint]] from every [[Definition:Permutation|permutation]] $\pi$ on $S_n$ (including ... | By definition of the [[Definition:Identity Mapping|identity permutation]]:
:$\forall i \in \N_{>0}: \map e i = i$
Thus $e$ [[Definition:Fixed Element under Permutation|fixes]] all elements of $S_n$.
Thus each element [[Definition:Moved Element under Permutation|moved]] by a permutation $\pi$ is [[Definition:Fixed Ele... | Identity Permutation is Disjoint from All | https://proofwiki.org/wiki/Identity_Permutation_is_Disjoint_from_All | https://proofwiki.org/wiki/Identity_Permutation_is_Disjoint_from_All | [
"Symmetric Groups",
"Identity Mappings"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Identity Mapping",
"Definition:Disjoint Permutations",
"Definition:Permutation"
] | [
"Definition:Identity Mapping",
"Definition:Fixed Element under Permutation",
"Definition:Fixed Element under Permutation/Moved",
"Definition:Fixed Element under Permutation",
"Category:Symmetric Groups",
"Category:Identity Mappings"
] |
proofwiki-1186 | Disjoint Permutations Commute | Let $S_n$ denote the symmetric group on $n$ letters.
Let $\rho, \sigma \in S_n$ such that $\rho$ and $\sigma$ are disjoint.
Then $\rho \sigma = \sigma \rho$. | Let $\rho$ and $\sigma$ be disjoint permutations.
Let $i \in \Fix \rho$.
Then:
:$\map {\sigma \rho} i = \map \sigma i$
whereas:
:$\map {\rho \sigma} i = \map \rho {\map \sigma i}$
{{AimForCont}} $\map \sigma i \notin \Fix \rho$.
Then because $\sigma$ and $\rho$ are disjoint it follows that:
{{begin-eqn}}
{{eqn | l = \m... | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $\rho, \sigma \in S_n$ such that $\rho$ and $\sigma$ are [[Definition:Disjoint Permutations|disjoint]].
Then $\rho \sigma = \sigma \rho$. | Let $\rho$ and $\sigma$ be [[Definition:Disjoint Permutations|disjoint permutations]].
Let $i \in \Fix \rho$.
Then:
:$\map {\sigma \rho} i = \map \sigma i$
whereas:
:$\map {\rho \sigma} i = \map \rho {\map \sigma i}$
{{AimForCont}} $\map \sigma i \notin \Fix \rho$.
Then because $\sigma$ and $\rho$ are [[Definitio... | Disjoint Permutations Commute | https://proofwiki.org/wiki/Disjoint_Permutations_Commute | https://proofwiki.org/wiki/Disjoint_Permutations_Commute | [
"Disjoint Permutations",
"Commutativity",
"Symmetric Groups"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Disjoint Permutations"
] | [
"Definition:Disjoint Permutations",
"Definition:Disjoint Permutations",
"Definition:Contradiction"
] |
proofwiki-1187 | Permutation Induces Equivalence Relation | Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$.
Let $\RR_\pi$ be the relation defined by:
:$i \mathrel {\RR_\pi} j \iff \exists k \in \Z: \map {\pi^k} i = j$
Then $\RR_\pi$ is an equivalence relation. | Let $\pi \in S_n$.
From Element of Finite Group is of Finite Order, every element of a finite group has finite order.
Thus $\pi$ has finite order.
So:
:$\exists r \in \Z: \pi^r = e$
Checking in turn each of the criteria for equivalence: | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $\pi \in S_n$.
Let $\RR_\pi$ be the [[Definition:Relation|relation]] defined by:
:$i \mathrel {\RR_\pi} j \iff \exists k \in \Z: \map {\pi^k} i = j$
Then $\RR_\pi$ is an [[Definition:Equivalence Relation|equivalenc... | Let $\pi \in S_n$.
From [[Element of Finite Group is of Finite Order]], every [[Definition:Element|element]] of a [[Definition:Finite Group|finite group]] has [[Definition:Finite Order Element|finite order]].
Thus $\pi$ has [[Definition:Finite Order Element|finite order]].
So:
:$\exists r \in \Z: \pi^r = e$
Checki... | Permutation Induces Equivalence Relation | https://proofwiki.org/wiki/Permutation_Induces_Equivalence_Relation | https://proofwiki.org/wiki/Permutation_Induces_Equivalence_Relation | [
"Symmetric Groups",
"Examples of Equivalence Relations"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Relation",
"Definition:Equivalence Relation"
] | [
"Element of Finite Group is of Finite Order",
"Definition:Element",
"Definition:Finite Group",
"Definition:Order of Group Element/Finite",
"Definition:Order of Group Element/Finite",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-1188 | Existence and Uniqueness of Cycle Decomposition | Let $S_n$ denote the symmetric group on $n$ letters.
Every element of $S_n$ may be uniquely expressed as a cycle decomposition, up to the order of factors. | By definition, a cycle decomposition of an element of $S_n$ is a product of disjoint cycles. | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Every [[Definition:Element|element]] of $S_n$ may be [[Definition:Unique|uniquely expressed]] as a [[Definition:Cycle Decomposition|cycle decomposition]], up to the order of factors. | By definition, a [[Definition:Cycle Decomposition|cycle decomposition]] of an [[Definition:Element|element]] of $S_n$ is a product of [[Definition:Disjoint Permutations|disjoint]] [[Definition:Cyclic Permutation|cycles]]. | Existence and Uniqueness of Cycle Decomposition | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Cycle_Decomposition | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Cycle_Decomposition | [
"Symmetric Groups",
"Cycle Decompositions"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Element",
"Definition:Unique",
"Definition:Cycle Decomposition"
] | [
"Definition:Cycle Decomposition",
"Definition:Element",
"Definition:Disjoint Permutations",
"Definition:Cyclic Permutation",
"Definition:Disjoint Permutations",
"Definition:Cyclic Permutation",
"Definition:Cycle Decomposition",
"Definition:Cycle Decomposition",
"Definition:Disjoint Permutations",
... |
proofwiki-1189 | Powers of Disjoint Permutations | Let $S_n$ denote the symmetric group on $n$ letters.
Let $\rho, \sigma$ be disjoint permutations.
Then:
: $\forall k \in \Z: \paren {\sigma \rho}^k = \sigma^k \rho^k$ | A direct application of Power of Product of Commutative Elements in Group, and the fact that Disjoint Permutations Commute.
{{qed}} | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $\rho, \sigma$ be [[Definition:Disjoint Permutations|disjoint permutations]].
Then:
: $\forall k \in \Z: \paren {\sigma \rho}^k = \sigma^k \rho^k$ | A direct application of [[Power of Product of Commutative Elements in Group]], and the fact that [[Disjoint Permutations Commute]].
{{qed}} | Powers of Disjoint Permutations | https://proofwiki.org/wiki/Powers_of_Disjoint_Permutations | https://proofwiki.org/wiki/Powers_of_Disjoint_Permutations | [
"Symmetric Groups"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Disjoint Permutations"
] | [
"Power of Product of Commutative Elements in Group",
"Disjoint Permutations Commute"
] |
proofwiki-1190 | Order of Product of Disjoint Permutations | Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi$ be a product of disjoint permutations of orders $k_1, k_2, \ldots, k_r$.
Then:
:$\order \pi = \lcm \set {k_1, k_2, \ldots, k_r}$
where:
:$\order \pi$ denotes the order of $\pi$ in $S_n$
:$\lcm$ denotes lowest common multiple. | Suppose $\pi$ is a cycle.
Then from Order of Cycle is Length of Cycle, $\order \pi$ is its length.
As the LCM of $n \in \Z$ is $n$ itself, the result follows.
Let $\pi = \rho_1 \rho_2 \cdots \rho_r$ where:
: each $\rho_s$ is of order $k_s$
: $\rho_1$ to $\rho_r$ are mutually disjoint permutations.
Let $t = \lcm \set {k... | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $\pi$ be a product of [[Definition:Disjoint Permutations|disjoint permutations]] of [[Definition:Order of Group Element|orders]] $k_1, k_2, \ldots, k_r$.
Then:
:$\order \pi = \lcm \set {k_1, k_2, \ldots, k_r}$
where:
... | Suppose $\pi$ is a [[Definition:Cyclic Permutation|cycle]].
Then from [[Order of Cycle is Length of Cycle]], $\order \pi$ is its [[Definition:Length of Cyclic Permutation|length]].
As the [[Definition:Lowest Common Multiple of Integers|LCM]] of $n \in \Z$ is $n$ itself, the result follows.
Let $\pi = \rho_1 \rho_2 ... | Order of Product of Disjoint Permutations | https://proofwiki.org/wiki/Order_of_Product_of_Disjoint_Permutations | https://proofwiki.org/wiki/Order_of_Product_of_Disjoint_Permutations | [
"Order of Product of Disjoint Permutations",
"Disjoint Permutations",
"Cycle Decompositions"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Disjoint Permutations",
"Definition:Order of Group Element",
"Definition:Order of Group Element",
"Definition:Lowest Common Multiple/Integers"
] | [
"Definition:Cyclic Permutation",
"Order of Cycle is Length of Cycle",
"Definition:Cyclic Permutation",
"Definition:Lowest Common Multiple/Integers",
"Definition:Order of Group Element",
"Definition:Disjoint Permutations",
"Disjoint Permutations Commute",
"Definition:Disjoint Permutations",
"Definiti... |
proofwiki-1191 | Group Action defines Permutation Representation | Let $\map \Gamma X$ be the set of permutations on a set $X$.
Let $G$ be a group.
Let $\phi: G \times X \to X$ be a group action.
For $g \in G$, let $\phi_g: X \to X$ be the mapping defined as:
:$\map {\phi_g} x = \map \phi {g, x}$
Let $\tilde \phi: G \to \map \Gamma X$ be the permutation representation associated to $\... | From Group Action determines Bijection:
:$\phi_g \in \map \Gamma X$
for $g \in G$.
Let $g, h \in G$.
{{Recall|Group Action}}
:$\forall \tuple {g, x} \in G \times X: \map \phi {g, x} \in X = g \wedge x \in X$
First we show that for all $x \in X$:
:$\map {\phi_g \circ \phi_h} x = \map {\phi_{g h} } x$
Thus:
{{begin-eqn}}... | Let $\map \Gamma X$ be the [[Definition:Set|set]] of [[Definition:Permutation|permutations]] on a [[Definition:Set|set]] $X$.
Let $G$ be a [[Definition:Group|group]].
Let $\phi: G \times X \to X$ be a [[Definition:Group Action|group action]].
For $g \in G$, let $\phi_g: X \to X$ be the [[Definition:Mapping|mapping]]... | From [[Group Action determines Bijection]]:
:$\phi_g \in \map \Gamma X$
for $g \in G$.
Let $g, h \in G$.
{{Recall|Group Action}}
:$\forall \tuple {g, x} \in G \times X: \map \phi {g, x} \in X = g \wedge x \in X$
First we show that for all $x \in X$:
:$\map {\phi_g \circ \phi_h} x = \map {\phi_{g h} } x$
Thus:
{{... | Group Action defines Permutation Representation | https://proofwiki.org/wiki/Group_Action_defines_Permutation_Representation | https://proofwiki.org/wiki/Group_Action_defines_Permutation_Representation | [
"Group Actions",
"Permutation Representations",
"Group Homomorphisms"
] | [
"Definition:Set",
"Definition:Permutation",
"Definition:Set",
"Definition:Group",
"Definition:Group Action",
"Definition:Mapping",
"Definition:Permutation Representation/Group Action",
"Definition:Group Homomorphism"
] | [
"Group Action determines Bijection",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Homomorphism"
] |
proofwiki-1192 | Group Action determines Bijection | Let $*$ be a group action of $G$ on $X$.
Then each $g \in G$ determines a bijection $\phi_g: X \to X$ given by:
:$\map {\phi_g} x = g * x$
Its inverse is:
:$\phi_{g^{-1} }: X \to X$.
These bijection are sometimes called '''transformations''' of $X$. | === Proof of Injectivity ===
Let $x, y \in X$
Then:
{{begin-eqn}}
{{eqn | l = \map {\phi_g} x
| r = \map {\phi_g} y
| c =
}}
{{eqn | ll= \leadsto
| l = g * x
| r = g * y
| c =
}}
{{eqn | ll= \leadsto
| l = g^{-1} * \paren {g * x}
| r = g^{-1} * \paren {g * y}
| c =
}}
... | Let $*$ be a [[Definition:Group Action|group action]] of $G$ on $X$.
Then each $g \in G$ determines a [[Definition:Bijection|bijection]] $\phi_g: X \to X$ given by:
:$\map {\phi_g} x = g * x$
Its [[Definition:Inverse Mapping|inverse]] is:
:$\phi_{g^{-1} }: X \to X$.
These [[Definition:Bijection|bijection]] are some... | === Proof of Injectivity ===
Let $x, y \in X$
Then:
{{begin-eqn}}
{{eqn | l = \map {\phi_g} x
| r = \map {\phi_g} y
| c =
}}
{{eqn | ll= \leadsto
| l = g * x
| r = g * y
| c =
}}
{{eqn | ll= \leadsto
| l = g^{-1} * \paren {g * x}
| r = g^{-1} * \paren {g * y}
| c =
... | Group Action determines Bijection | https://proofwiki.org/wiki/Group_Action_determines_Bijection | https://proofwiki.org/wiki/Group_Action_determines_Bijection | [
"Group Actions"
] | [
"Definition:Group Action",
"Definition:Bijection",
"Definition:Inverse Mapping",
"Definition:Bijection"
] | [
"Definition:Injection",
"Definition:Injection"
] |
proofwiki-1193 | Group Action Induces Equivalence Relation | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $X$ be a set.
Let $*: G \times S \to S$ be a group action.
Let $\RR_G$ be the relation induced by $G$, that is:
:$\forall x, y \in X: x \mathrel {\RR_G} y \iff y \in \Orb x$
where:
:$\Orb x$ denotes the orbit of $x \in X$.
Then:
:$\RR_G$ is an equivalence r... | Let $x \mathrel {\RR_G} y \iff y \in \Orb x$.
Checking in turn each of the criteria for equivalence: | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $X$ be a [[Definition:Set|set]].
Let $*: G \times S \to S$ be a [[Definition:Group Action|group action]].
Let $\RR_G$ be the [[Definition:Equivalence Relation Induced by Group Action|relation induced]]... | Let $x \mathrel {\RR_G} y \iff y \in \Orb x$.
Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Group Action Induces Equivalence Relation | https://proofwiki.org/wiki/Group_Action_Induces_Equivalence_Relation | https://proofwiki.org/wiki/Group_Action_Induces_Equivalence_Relation | [
"Group Actions",
"Equivalence Relations"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Set",
"Definition:Group Action",
"Definition:Equivalence Relation Induced by Group Action",
"Definition:Orbit (Group Theory)",
"Definition:Equivalence Relation",
"Definition:Equivalence Class",
"Definition:E... | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-1194 | Partition Equation | Let group $G$ act on a finite set $X$.
Let the distinct orbits of $X$ under the action of $G$ be:
:$\Orb {x_1}, \Orb {x_2}, \ldots, \Orb {x_s}$
Then:
:$\card X = \card {\Orb {x_1} } + \card {\Orb {x_2} } + \cdots + \card {\Orb {x_s} }$ | Follows trivially from the fact that the Group Action Induces Equivalence Relation.
{{explain|cite a result about equiv rels and partitions, and one about cardinality and partitions}}
{{qed}}
Category:Group Actions
Category:Named Theorems
pwvgvk2c6p89r84hq9vudcd3u47firc | Let [[Definition:Group|group]] $G$ [[Definition:Group Action|act on]] a [[Definition:Finite Set|finite set]] $X$.
Let the distinct [[Definition:Orbit (Group Theory)|orbits]] of $X$ under the [[Definition:Group Action|action]] of $G$ be:
:$\Orb {x_1}, \Orb {x_2}, \ldots, \Orb {x_s}$
Then:
:$\card X = \card {\Orb {x_1... | Follows trivially from the fact that the [[Group Action Induces Equivalence Relation]].
{{explain|cite a result about equiv rels and partitions, and one about cardinality and partitions}}
{{qed}}
[[Category:Group Actions]]
[[Category:Named Theorems]]
pwvgvk2c6p89r84hq9vudcd3u47firc | Partition Equation | https://proofwiki.org/wiki/Partition_Equation | https://proofwiki.org/wiki/Partition_Equation | [
"Group Actions",
"Named Theorems"
] | [
"Definition:Group",
"Definition:Group Action",
"Definition:Finite Set",
"Definition:Orbit (Group Theory)",
"Definition:Group Action"
] | [
"Group Action Induces Equivalence Relation",
"Category:Group Actions",
"Category:Named Theorems"
] |
proofwiki-1195 | Stabilizer is Subgroup | Let $\struct {G, \circ}$ be a group which acts on a set $X$.
Let $\Stab x$ be the stabilizer of $x$ by $G$.
Then for each $x \in X$, $\Stab x$ is a subgroup of $G$. | From the {{GroupActionAxiom|2}}:
:$e * x = x \implies e \in \Stab x$
and so $\Stab x$ cannot be empty.
Let $g, h \in \Stab x$.
{{begin-eqn}}
{{eqn | l = g, h
| o = \in
| r = \Stab x
| c =
}}
{{eqn | ll= \leadsto
| l = g * x
| r = x
| c = {{Defof|Stabilizer}} of $x$ by $G$
}}
{{eqn |... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] which [[Definition:Group Action|acts on]] a [[Definition:Set|set]] $X$.
Let $\Stab x$ be the [[Definition:Stabilizer|stabilizer of $x$ by $G$]].
Then for each $x \in X$, $\Stab x$ is a [[Definition:Subgroup|subgroup]] of $G$. | From the {{GroupActionAxiom|2}}:
:$e * x = x \implies e \in \Stab x$
and so $\Stab x$ cannot be [[Definition:Empty Set|empty]].
Let $g, h \in \Stab x$.
{{begin-eqn}}
{{eqn | l = g, h
| o = \in
| r = \Stab x
| c =
}}
{{eqn | ll= \leadsto
| l = g * x
| r = x
| c = {{Defof|Stabilize... | Stabilizer is Subgroup | https://proofwiki.org/wiki/Stabilizer_is_Subgroup | https://proofwiki.org/wiki/Stabilizer_is_Subgroup | [
"Group Actions",
"Stabilizers",
"Stabilizer is Subgroup"
] | [
"Definition:Group",
"Definition:Group Action",
"Definition:Set",
"Definition:Stabilizer",
"Definition:Subgroup"
] | [
"Definition:Empty Set",
"Two-Step Subgroup Test"
] |
proofwiki-1196 | Orbit-Stabilizer Theorem | Let $G$ be a group which acts on a finite set $X$.
Let $x \in X$.
Let $\Orb x$ denote the orbit of $x$.
Let $\Stab x$ denote the stabilizer of $x$ by $G$.
Let $\index G {\Stab x}$ denote the index of $\Stab x$ in $G$.
Then:
:$\order {\Orb x} = \index G {\Stab x} = \dfrac {\order G} {\order {\Stab x} }$ | Let us define the mapping:
:$\phi: G \to \Orb x$
such that:
:$\map \phi g = g * x$
where $*$ denotes the group action.
It is clear that $\phi$ is surjective, because from the definition $x$ was acted on by all the elements of $G$.
Next, from Stabilizer is Subgroup: Corollary:
:$\map \phi g = \map \phi h \iff g^{-1} h \... | Let $G$ be a [[Definition:Group|group]] which [[Definition:Group Action|acts on]] a [[Definition:Finite Set|finite set]] $X$.
Let $x \in X$.
Let $\Orb x$ denote the [[Definition:Orbit (Group Theory)|orbit]] of $x$.
Let $\Stab x$ denote the [[Definition:Stabilizer|stabilizer of $x$ by $G$]].
Let $\index G {\Stab x}... | Let us define the [[Definition:Mapping|mapping]]:
:$\phi: G \to \Orb x$
such that:
:$\map \phi g = g * x$
where $*$ denotes the [[Definition:Group Action|group action]].
It is clear that $\phi$ is [[Definition:Surjection|surjective]], because from the definition $x$ was acted on by all the [[Definition:Element|elemen... | Orbit-Stabilizer Theorem/Proof 1 | https://proofwiki.org/wiki/Orbit-Stabilizer_Theorem | https://proofwiki.org/wiki/Orbit-Stabilizer_Theorem/Proof_1 | [
"Group Actions",
"Stabilizers",
"Named Theorems",
"Orbit-Stabilizer Theorem"
] | [
"Definition:Group",
"Definition:Group Action",
"Definition:Finite Set",
"Definition:Orbit (Group Theory)",
"Definition:Stabilizer",
"Definition:Index of Subgroup"
] | [
"Definition:Mapping",
"Definition:Group Action",
"Definition:Surjection",
"Definition:Element",
"Stabilizer is Subgroup/Corollary",
"Definition:Bijection"
] |
proofwiki-1197 | Orbit-Stabilizer Theorem | Let $G$ be a group which acts on a finite set $X$.
Let $x \in X$.
Let $\Orb x$ denote the orbit of $x$.
Let $\Stab x$ denote the stabilizer of $x$ by $G$.
Let $\index G {\Stab x}$ denote the index of $\Stab x$ in $G$.
Then:
:$\order {\Orb x} = \index G {\Stab x} = \dfrac {\order G} {\order {\Stab x} }$ | Let $x \in X$.
Let $\phi: \Orb x \to G \mathbin / \Stab x$ be a mapping from the orbit of $x$ to the left coset space of $\Stab x$ defined as:
:$\forall g \in G: \map \phi {g * x} = g \, \Stab x$
where $*$ is the group action.
Note: this is ''not'' a homomorphism because $\Orb x$ is not a group.
Suppose $g * x = h * x$... | Let $G$ be a [[Definition:Group|group]] which [[Definition:Group Action|acts on]] a [[Definition:Finite Set|finite set]] $X$.
Let $x \in X$.
Let $\Orb x$ denote the [[Definition:Orbit (Group Theory)|orbit]] of $x$.
Let $\Stab x$ denote the [[Definition:Stabilizer|stabilizer of $x$ by $G$]].
Let $\index G {\Stab x}... | Let $x \in X$.
Let $\phi: \Orb x \to G \mathbin / \Stab x$ be a [[Definition:Mapping|mapping]] from the [[Definition:Orbit (Group Theory)|orbit]] of $x$ to the [[Definition:Left Coset Space|left coset space]] of $\Stab x$ defined as:
:$\forall g \in G: \map \phi {g * x} = g \, \Stab x$
where $*$ is the [[Definition:Gr... | Orbit-Stabilizer Theorem/Proof 2 | https://proofwiki.org/wiki/Orbit-Stabilizer_Theorem | https://proofwiki.org/wiki/Orbit-Stabilizer_Theorem/Proof_2 | [
"Group Actions",
"Stabilizers",
"Named Theorems",
"Orbit-Stabilizer Theorem"
] | [
"Definition:Group",
"Definition:Group Action",
"Definition:Finite Set",
"Definition:Orbit (Group Theory)",
"Definition:Stabilizer",
"Definition:Index of Subgroup"
] | [
"Definition:Mapping",
"Definition:Orbit (Group Theory)",
"Definition:Coset Space/Left Coset Space",
"Definition:Group Action",
"Definition:Group Homomorphism",
"Definition:Group",
"Left Coset Space forms Partition",
"Definition:Well-Defined/Mapping",
"Left Coset Space forms Partition",
"Definition... |
proofwiki-1198 | Group Acts on Itself | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then $\struct {G, \circ}$ acts on itself by the rule:
:$\forall g, h \in G: g * h = g \circ h$ | Follows directly from the group axioms and the definition of a group action.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Then $\struct {G, \circ}$ [[Definition:Group Action|acts on]] itself by the rule:
:$\forall g, h \in G: g * h = g \circ h$ | Follows directly from the [[Axiom:Group Axioms|group axioms]] and the definition of a [[Definition:Group Action|group action]].
{{qed}} | Group Acts on Itself | https://proofwiki.org/wiki/Group_Acts_on_Itself | https://proofwiki.org/wiki/Group_Acts_on_Itself | [
"Group Actions"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Action"
] | [
"Axiom:Group Axioms",
"Definition:Group Action"
] |
proofwiki-1199 | Action of Group on Coset Space is Group Action | Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Let $*: G \times G / H \to G / H$ be the action on the (left) coset space:
:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
Then $G$ is a group action. | {{begin-eqn}}
{{eqn | l = a * \paren {b * g' H}
| r = a * \paren {\paren {b g'} H}
| c = Definition of $*$
}}
{{eqn | r = \paren {a \paren {b g'} } H
| c = Definition of $*$
}}
{{eqn | r = \paren {\paren {a b } g' } H
| c = {{Group-axiom|1}}
}}
{{eqn | r = \paren {a b} \paren{g' H }
| c = ... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Let $*: G \times G / H \to G / H$ be the [[Definition:Group Action on Coset Space|action on the (left) coset space]]:
:$\forall g \in G, \forall g' H \in G / H: g * \pa... | {{begin-eqn}}
{{eqn | l = a * \paren {b * g' H}
| r = a * \paren {\paren {b g'} H}
| c = Definition of $*$
}}
{{eqn | r = \paren {a \paren {b g'} } H
| c = Definition of $*$
}}
{{eqn | r = \paren {\paren {a b } g' } H
| c = {{Group-axiom|1}}
}}
{{eqn | r = \paren {a b} \paren{g' H }
| c = ... | Action of Group on Coset Space is Group Action | https://proofwiki.org/wiki/Action_of_Group_on_Coset_Space_is_Group_Action | https://proofwiki.org/wiki/Action_of_Group_on_Coset_Space_is_Group_Action | [
"Cosets",
"Group Action on Coset Space"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Group Action on Coset Space",
"Definition:Group Action"
] | [
"Subset Product within Semigroup is Associative/Corollary"
] |
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