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-700 | Cosets are Equivalent | All left cosets of a group $G$ with respect to a subgroup $H$ are equivalent.
That is, any two left cosets are in one-to-one correspondence.
The same applies to right cosets.
As a special case of this:
:$\forall x \in G: \order {x H} = \order H = \order {H x}$
where $H$ is a subgroup of $G$. | Let us set up mappings $\theta: H \to H x$ and $\phi: H x \to H$ as follows:
:$\forall u \in H: \map \theta u = u x$
:$\forall v \in H x: \map \phi v = v x^{-1}$
From Element in Right Coset iff Product with Inverse in Subgroup:
:$v \in H x \implies v x^{-1} \in H$
Now:
:$\forall v \in H x: \theta \circ \map \phi v = v ... | All [[Definition:Left Coset|left cosets]] of a [[Definition:Group|group]] $G$ with respect to a [[Definition:Subgroup|subgroup]] $H$ are [[Definition:Set Equivalence|equivalent]].
That is, any two [[Definition:Left Coset|left cosets]] are [[Definition:Bijection|in one-to-one correspondence]].
The same applies to [[De... | Let us set up [[Definition:Mapping|mappings]] $\theta: H \to H x$ and $\phi: H x \to H$ as follows:
:$\forall u \in H: \map \theta u = u x$
:$\forall v \in H x: \map \phi v = v x^{-1}$
From [[Element in Right Coset iff Product with Inverse in Subgroup]]:
:$v \in H x \implies v x^{-1} \in H$
Now:
:$\forall v \in H ... | Cosets are Equivalent/Proof 1 | https://proofwiki.org/wiki/Cosets_are_Equivalent | https://proofwiki.org/wiki/Cosets_are_Equivalent/Proof_1 | [
"Cosets",
"Cosets are Equivalent"
] | [
"Definition:Coset/Left Coset",
"Definition:Group",
"Definition:Subgroup",
"Definition:Set Equivalence",
"Definition:Coset/Left Coset",
"Definition:Bijection",
"Definition:Coset/Right Coset",
"Definition:Subgroup"
] | [
"Definition:Mapping",
"Element in Right Coset iff Product with Inverse in Subgroup",
"Definition:Identity Mapping",
"Definition:Bijection",
"Definition:Inverse Mapping",
"Definition:Set Equivalence",
"Set Equivalence behaves like Equivalence Relation",
"Definition:Mapping"
] |
proofwiki-701 | Cosets are Equivalent | All left cosets of a group $G$ with respect to a subgroup $H$ are equivalent.
That is, any two left cosets are in one-to-one correspondence.
The same applies to right cosets.
As a special case of this:
:$\forall x \in G: \order {x H} = \order H = \order {H x}$
where $H$ is a subgroup of $G$. | Follows directly from Set Equivalence of Regular Representations.
{{qed}} | All [[Definition:Left Coset|left cosets]] of a [[Definition:Group|group]] $G$ with respect to a [[Definition:Subgroup|subgroup]] $H$ are [[Definition:Set Equivalence|equivalent]].
That is, any two [[Definition:Left Coset|left cosets]] are [[Definition:Bijection|in one-to-one correspondence]].
The same applies to [[De... | Follows directly from [[Set Equivalence of Regular Representations]].
{{qed}} | Cosets are Equivalent/Proof 2 | https://proofwiki.org/wiki/Cosets_are_Equivalent | https://proofwiki.org/wiki/Cosets_are_Equivalent/Proof_2 | [
"Cosets",
"Cosets are Equivalent"
] | [
"Definition:Coset/Left Coset",
"Definition:Group",
"Definition:Subgroup",
"Definition:Set Equivalence",
"Definition:Coset/Left Coset",
"Definition:Bijection",
"Definition:Coset/Right Coset",
"Definition:Subgroup"
] | [
"Set Equivalence of Regular Representations"
] |
proofwiki-702 | Left and Right Coset Spaces are Equivalent | Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Let:
:$x H$ denote the left coset of $H$ by $x$
:$H y$ denote the right coset of $H$ by $y$.
Then:
:$\order {\set {x H: x \in G} } = \order {\set {H y: y \in G} }$ | Let there be exactly $r$ different left cosets of $H$ in $G$.
Let a complete repetition-free list of these left cosets be:
:$a_1 H, a_2 H, a_3 H \ldots, a_r H: a_1, a_2, \ldots, a_r \in G$
From Left Coset Space forms Partition, every element of $G$ is contained in exactly one of the left cosets.
Let $x \in G$.
Then, fo... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Let:
:$x H$ denote the [[Definition:Left Coset|left coset]] of $H$ by $x$
:$H y$ denote the [[Definition:Right Coset|right coset]] of $H$ by $y$.
Then:
:$\order {\set {x H: x \in G} } = \order {\set {H y... | Let there be exactly $r$ different [[Definition:Left Coset|left cosets]] of $H$ in $G$.
Let a complete repetition-free list of these [[Definition:Left Coset|left cosets]] be:
:$a_1 H, a_2 H, a_3 H \ldots, a_r H: a_1, a_2, \ldots, a_r \in G$
From [[Left Coset Space forms Partition]], every element of $G$ is contained... | Left and Right Coset Spaces are Equivalent/Proof 1 | https://proofwiki.org/wiki/Left_and_Right_Coset_Spaces_are_Equivalent | https://proofwiki.org/wiki/Left_and_Right_Coset_Spaces_are_Equivalent/Proof_1 | [
"Cosets",
"Left and Right Coset Spaces are Equivalent"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Coset/Left Coset",
"Definition:Coset/Right Coset"
] | [
"Definition:Coset/Left Coset",
"Definition:Coset/Left Coset",
"Left Coset Space forms Partition",
"Definition:Coset/Left Coset",
"Element in Right Coset iff Product with Inverse in Subgroup",
"Inverse of Group Inverse",
"Inverse of Group Inverse",
"Element in Left Coset iff Product with Inverse in Sub... |
proofwiki-703 | Left and Right Coset Spaces are Equivalent | Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Let:
:$x H$ denote the left coset of $H$ by $x$
:$H y$ denote the right coset of $H$ by $y$.
Then:
:$\order {\set {x H: x \in G} } = \order {\set {H y: y \in G} }$ | Let $G$ be a group and let $H \le G$.
Consider the mapping $\phi$ from the left coset space to the right coset space defined as:
:$\forall g \in G: \map \phi {g H} = H g^{-1}$
We need to show that $\phi$ is a bijection.
First we need to show that $\phi$ is well-defined.
That is, that $a H = b H \implies \map \phi {a H}... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Let:
:$x H$ denote the [[Definition:Left Coset|left coset]] of $H$ by $x$
:$H y$ denote the [[Definition:Right Coset|right coset]] of $H$ by $y$.
Then:
:$\order {\set {x H: x \in G} } = \order {\set {H y... | Let $G$ be a [[Definition:Group|group]] and let $H \le G$.
Consider the [[Definition:Mapping|mapping]] $\phi$ from the [[Definition:Left Coset Space|left coset space]] to the [[Definition:Right Coset Space|right coset space]] defined as:
:$\forall g \in G: \map \phi {g H} = H g^{-1}$
We need to show that $\phi$ is ... | Left and Right Coset Spaces are Equivalent/Proof 2 | https://proofwiki.org/wiki/Left_and_Right_Coset_Spaces_are_Equivalent | https://proofwiki.org/wiki/Left_and_Right_Coset_Spaces_are_Equivalent/Proof_2 | [
"Cosets",
"Left and Right Coset Spaces are Equivalent"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Coset/Left Coset",
"Definition:Coset/Right Coset"
] | [
"Definition:Group",
"Definition:Mapping",
"Definition:Coset Space/Left Coset Space",
"Definition:Coset Space/Right Coset Space",
"Definition:Bijection",
"Definition:Well-Defined/Mapping",
"Left Cosets are Equal iff Product with Inverse in Subgroup",
"Right Cosets are Equal iff Product with Inverse in ... |
proofwiki-704 | Cosets in Abelian Group | Let $G$ be an abelian group.
Then every right coset modulo $H$ is a left coset modulo $H$.
That is:
:$\forall x \in G: x H = H x$
In an abelian group, therefore, we can talk about '''congruence modulo $H$''' and not worry about whether it is left or right. | {{begin-eqn}}
{{eqn | o =
| r = \forall x, y \in G: x^{-1} y = y x^{-1}
| c =
}}
{{eqn | o = \leadsto
| r = \paren {x \equiv^l y \pmod H \iff y \equiv^r x \pmod H}
| c = {{Defof|Congruence Modulo Subgroup}}
}}
{{eqn | o = \leadsto
| r = \paren {x \equiv^l y \pmod H \iff x \equiv^r y \pmo... | Let $G$ be an [[Definition:Abelian Group|abelian group]].
Then every [[Definition:Right Coset|right coset]] modulo $H$ is a [[Definition:Left Coset|left coset]] modulo $H$.
That is:
:$\forall x \in G: x H = H x$
In an [[Definition:Abelian Group|abelian group]], therefore, we can talk about '''[[Definition:Congruenc... | {{begin-eqn}}
{{eqn | o =
| r = \forall x, y \in G: x^{-1} y = y x^{-1}
| c =
}}
{{eqn | o = \leadsto
| r = \paren {x \equiv^l y \pmod H \iff y \equiv^r x \pmod H}
| c = {{Defof|Congruence Modulo Subgroup}}
}}
{{eqn | o = \leadsto
| r = \paren {x \equiv^l y \pmod H \iff x \equiv^r y \pmo... | Cosets in Abelian Group | https://proofwiki.org/wiki/Cosets_in_Abelian_Group | https://proofwiki.org/wiki/Cosets_in_Abelian_Group | [
"Cosets",
"Abelian Groups"
] | [
"Definition:Abelian Group",
"Definition:Coset/Right Coset",
"Definition:Coset/Left Coset",
"Definition:Abelian Group",
"Definition:Congruence Modulo Subgroup",
"Definition:Congruence Modulo Subgroup/Left Congruence",
"Definition:Congruence Modulo Subgroup/Right Congruence"
] | [
"Congruence Modulo Subgroup is Equivalence Relation",
"Definition:Symmetric Relation"
] |
proofwiki-705 | Conjugate of Set by Identity | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $S \subseteq G$.
Then the conjugate of $S$ by $e$ is $S$:
:$S^e = S$ | {{begin-eqn}}
{{eqn | l = S^e
| r = \set {y \in G: \exists x \in S: y = e \circ x \circ e^{-1} }
| c = {{Defof|Conjugate of Group Subset}}
}}
{{eqn | r = \set {y \in G: \exists x \in S: y = x}
| c = {{Defof|Identity Element}}
}}
{{eqn | r = S
| c=
}}
{{end-eqn}}
{{Qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $S \subseteq G$.
Then the [[Definition:Conjugate of Group Subset|conjugate of $S$ by $e$]] is $S$:
:$S^e = S$ | {{begin-eqn}}
{{eqn | l = S^e
| r = \set {y \in G: \exists x \in S: y = e \circ x \circ e^{-1} }
| c = {{Defof|Conjugate of Group Subset}}
}}
{{eqn | r = \set {y \in G: \exists x \in S: y = x}
| c = {{Defof|Identity Element}}
}}
{{eqn | r = S
| c=
}}
{{end-eqn}}
{{Qed}} | Conjugate of Set by Identity | https://proofwiki.org/wiki/Conjugate_of_Set_by_Identity | https://proofwiki.org/wiki/Conjugate_of_Set_by_Identity | [
"Conjugacy"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Conjugate (Group Theory)/Subset"
] | [] |
proofwiki-706 | Conjugate of Set by Group Product | Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$.
Let $S^a$ denote the $G$-conjugate of $S$ by $a$ as:
:$S^a := \set {y \in G: \exists x \in S: y = a \circ x \circ a^{-1} } = a \circ S \circ a^{-1}$
Then:
:$\paren {S^a}^b = S^{b \circ a}$ | $S^a$ is defined as $a \circ S \circ a^{-1}$ from the definition of the conjugate of a set.
From the definition of subset product with a singleton, this can be seen to be the same thing as:
:$S^a = \set a \circ S \circ \set {a^{-1} }$.
Thus we can express $\paren {S^a}^b$ as $b \circ \paren {a \circ S \circ a^{-1} } \c... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $S \subseteq G$.
Let $S^a$ denote the [[Definition:Conjugate of Group Subset|$G$-conjugate of $S$ by $a$]] as:
:$S^a := \set {y \in G: \exists x \in S: y = a \circ x \circ a^{-1} } = a \circ S \circ a^{-1}$
Then:
:$\paren {S^a}^b = S^{b \circ a}$ | $S^a$ is defined as $a \circ S \circ a^{-1}$ from the definition of the [[Definition:Conjugate of Group Subset|conjugate of a set]].
From the definition of [[Definition:Subset Product with Singleton|subset product with a singleton]], this can be seen to be the same thing as:
:$S^a = \set a \circ S \circ \set {a^{-1} ... | Conjugate of Set by Group Product | https://proofwiki.org/wiki/Conjugate_of_Set_by_Group_Product | https://proofwiki.org/wiki/Conjugate_of_Set_by_Group_Product | [
"Conjugacy"
] | [
"Definition:Group",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Conjugate (Group Theory)/Subset"
] | [
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Subset Product/Singleton",
"Definition:Subset Product",
"Subset Product within Semigroup is Associative",
"Definition:Associative Operation",
"Subset Product within Semigroup is Associative",
"Inverse of Group Product",
"Subset Product within S... |
proofwiki-707 | Conjugate of Subgroup is Subgroup | Let $G$ be a group.
Let $H \le G$ be a subgroup of $G$.
Then the conjugate of $H$ by $a$ is a subgroup of $G$:
:$\forall H \le G, a \in G: H^a \le G$ | Let $H \le G$.
First, we show that $x, y \in H^a \implies x \circ y \in H^a$:
{{begin-eqn}}
{{eqn | l = x, y
| o = \in
| r = H^a
| c =
}}
{{eqn | ll= \leadsto
| l = a x a^{-1}, a y a^{-1}
| o = \in
| r = H
| c = {{Defof|Conjugate of Group Subset}}
}}
{{eqn | ll= \leadsto
... | Let $G$ be a [[Definition:Group|group]].
Let $H \le G$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then the [[Definition:Conjugate of Group Subset|conjugate]] of $H$ by $a$ is a [[Definition:Subgroup|subgroup]] of $G$:
:$\forall H \le G, a \in G: H^a \le G$ | Let $H \le G$.
First, we show that $x, y \in H^a \implies x \circ y \in H^a$:
{{begin-eqn}}
{{eqn | l = x, y
| o = \in
| r = H^a
| c =
}}
{{eqn | ll= \leadsto
| l = a x a^{-1}, a y a^{-1}
| o = \in
| r = H
| c = {{Defof|Conjugate of Group Subset}}
}}
{{eqn | ll= \leadsto
... | Conjugate of Subgroup is Subgroup/Proof 1 | https://proofwiki.org/wiki/Conjugate_of_Subgroup_is_Subgroup | https://proofwiki.org/wiki/Conjugate_of_Subgroup_is_Subgroup/Proof_1 | [
"Conjugacy",
"Subgroups",
"Conjugate of Subgroup is Subgroup"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Subgroup"
] | [
"Power of Conjugate equals Conjugate of Power",
"Two-Step Subgroup Test"
] |
proofwiki-708 | Conjugate of Subgroup is Subgroup | Let $G$ be a group.
Let $H \le G$ be a subgroup of $G$.
Then the conjugate of $H$ by $a$ is a subgroup of $G$:
:$\forall H \le G, a \in G: H^a \le G$ | Let $*: G \times G / H \to G / H$ be the group action on the (left) coset space:
:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
It is established in Action of Group on Coset Space is Group Action that $*$ is a group action.
Then from Stabilizer of Coset under Group Action on Coset Spac... | Let $G$ be a [[Definition:Group|group]].
Let $H \le G$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then the [[Definition:Conjugate of Group Subset|conjugate]] of $H$ by $a$ is a [[Definition:Subgroup|subgroup]] of $G$:
:$\forall H \le G, a \in G: H^a \le G$ | Let $*: G \times G / H \to G / H$ be the [[Definition:Group Action on Coset Space|group action on the (left) coset space]]:
:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
It is established in [[Action of Group on Coset Space is Group Action]] that $*$ is a [[Definition:Group Action|gr... | Conjugate of Subgroup is Subgroup/Proof 2 | https://proofwiki.org/wiki/Conjugate_of_Subgroup_is_Subgroup | https://proofwiki.org/wiki/Conjugate_of_Subgroup_is_Subgroup/Proof_2 | [
"Conjugacy",
"Subgroups",
"Conjugate of Subgroup is Subgroup"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Subgroup"
] | [
"Definition:Group Action on Coset Space",
"Action of Group on Coset Space is Group Action",
"Definition:Group Action",
"Stabilizer of Coset under Group Action on Coset Space",
"Definition:Stabilizer",
"Stabilizer is Subgroup"
] |
proofwiki-709 | Inner Automorphisms form Normal Subgroup of Automorphism Group | Let $G$ be a group.
Then the set $\Inn G$ of all inner automorphisms of $G$ is a normal subgroup of the automorphism group $\Aut G$ of $G$:
:$\Inn G \lhd \Aut G$ | Let $G$ be a group whose identity is $e$.
From Inner Automorphisms form Subgroup of Automorphism Group, $\Inn G$ forms a subgroup of $\Aut G$.
It remains to be shown that $\Inn G$ is normal in $\Aut G$.
Let $\kappa_x: G \to G$ be the inner automorphism defined as:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
Let ... | Let $G$ be a [[Definition:Group|group]].
Then the [[Definition:Set|set]] $\Inn G$ of all [[Definition:Inner Automorphism|inner automorphisms]] of $G$ is a [[Definition:Normal Subgroup|normal subgroup]] of the [[Definition:Automorphism Group of Group|automorphism group]] $\Aut G$ of $G$:
:$\Inn G \lhd \Aut G$ | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
From [[Inner Automorphisms form Subgroup of Automorphism Group]], $\Inn G$ forms a [[Definition:Subgroup|subgroup]] of $\Aut G$.
It remains to be shown that $\Inn G$ is [[Definition:Normal Subgroup|normal]] in $\Aut G$.
L... | Inner Automorphisms form Normal Subgroup of Automorphism Group | https://proofwiki.org/wiki/Inner_Automorphisms_form_Normal_Subgroup_of_Automorphism_Group | https://proofwiki.org/wiki/Inner_Automorphisms_form_Normal_Subgroup_of_Automorphism_Group | [
"Inner Automorphisms",
"Automorphism Groups",
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Set",
"Definition:Inner Automorphism",
"Definition:Normal Subgroup",
"Definition:Automorphism Group/Group"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Inner Automorphisms form Subgroup of Automorphism Group",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Inner Automorphism",
"Normal Subgroup Test",
"Definition:Group Homomorphism"
] |
proofwiki-710 | Subgroup of Abelian Group is Normal | Every subgroup of an abelian group is normal. | Let $G$ be an abelian group.
Let $H \le G$ be a subgroup of $G$.
Then for all $a \in G$:
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = H^a
| c = where $H^a$ is the conjugate of $H$ by $a$
}}
{{eqn | ll= \leadstoandfrom
| l = a y a^{-1}
| o = \in
| r = H
| c = {{Defof|Conjugate of... | Every [[Definition:Subgroup|subgroup]] of an [[Definition:Abelian Group|abelian group]] is [[Definition:Normal Subgroup|normal]]. | Let $G$ be an [[Definition:Abelian Group|abelian group]].
Let $H \le G$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then for all $a \in G$:
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = H^a
| c = where $H^a$ is the [[Definition:Conjugate of Group Subset|conjugate of $H$ by $a$]]
}}
{{eqn | ll= \le... | Subgroup of Abelian Group is Normal | https://proofwiki.org/wiki/Subgroup_of_Abelian_Group_is_Normal | https://proofwiki.org/wiki/Subgroup_of_Abelian_Group_is_Normal | [
"Normal Subgroups",
"Abelian Groups"
] | [
"Definition:Subgroup",
"Definition:Abelian Group",
"Definition:Normal Subgroup"
] | [
"Definition:Abelian Group",
"Definition:Subgroup",
"Definition:Conjugate (Group Theory)/Subset",
"Conjugate of Commuting Elements",
"Definition:Abelian Group"
] |
proofwiki-711 | Subgroup is Normal iff it contains Product of Inverses | A subgroup $H$ of a group $G$ is normal {{iff}}:
:$\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$ | === Necessary Condition ===
Suppose $H$ is normal.
Let $a b \in H$.
{{begin-eqn}}
{{eqn | l = a b
| o = \in
| r = H
| c =
}}
{{eqn | ll= \leadsto
| l = a^{-1} \left({a b}\right) a
| o = \in
| r = H
| c = Subgroup is Normal iff Contains Conjugate Elements
}}
{{eqn | ll= \leadst... | A [[Definition:Subgroup|subgroup]] $H$ of a [[Definition:Group|group]] $G$ is [[Definition:Normal Subgroup|normal]] {{iff}}:
:$\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$ | === Necessary Condition ===
Suppose $H$ is [[Definition:Normal Subgroup|normal]].
Let $a b \in H$.
{{begin-eqn}}
{{eqn | l = a b
| o = \in
| r = H
| c =
}}
{{eqn | ll= \leadsto
| l = a^{-1} \left({a b}\right) a
| o = \in
| r = H
| c = [[Subgroup is Normal iff Contains Conju... | Subgroup is Normal iff it contains Product of Inverses | https://proofwiki.org/wiki/Subgroup_is_Normal_iff_it_contains_Product_of_Inverses | https://proofwiki.org/wiki/Subgroup_is_Normal_iff_it_contains_Product_of_Inverses | [
"Normal Subgroups"
] | [
"Definition:Subgroup",
"Definition:Group",
"Definition:Normal Subgroup"
] | [
"Definition:Normal Subgroup",
"Subgroup is Normal iff Contains Conjugate Elements",
"Definition:Group",
"Definition:Group"
] |
proofwiki-712 | Intersection with Normal Subgroup is Normal | Let $G$ be a group.
Let $H$ be a subgroup of $G$, and let $N$ be a normal subgroup of $G$.
Then $H \cap N$ is a normal subgroup of $H$. | By Intersection of Subgroups is Subgroup, $H \cap N$ is a subgroup of $N$.
It remains to be shown that $H \cap N$ is normal in $H$.
Let $g \in H$ and $x \in H \cap N$.
However, $H \le G$ and $ g \in H$ imply:
:$g \in G$
Because $N \lhd G$:
:$\forall n \in N: \forall g \in G: g n g^{-1} \in N$
Because $H \le G$ and the... | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$, and let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Then $H \cap N$ is a [[Definition:Normal Subgroup|normal subgroup]] of $H$. | By [[Intersection of Subgroups is Subgroup]], $H \cap N$ is a [[Definition:Subgroup|subgroup]] of $N$.
It remains to be shown that $H \cap N$ is [[Definition:Normal Subgroup|normal]] in $H$.
Let $g \in H$ and $x \in H \cap N$.
However, $H \le G$ and $ g \in H$ imply:
:$g \in G$
Because $N \lhd G$:
:$\forall n \in ... | Intersection with Normal Subgroup is Normal | https://proofwiki.org/wiki/Intersection_with_Normal_Subgroup_is_Normal | https://proofwiki.org/wiki/Intersection_with_Normal_Subgroup_is_Normal | [
"Normal Subgroups",
"Intersection with Normal Subgroup is Normal"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Normal Subgroup"
] | [
"Intersection of Subgroups is Subgroup",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Normal Subgroup",
"Definition:Subgroup",
"Definition:Normal Subgroup"
] |
proofwiki-713 | Kernel is Normal Subgroup of Domain | Let $\phi$ be a group homomorphism.
Then the kernel of $\phi$ is a normal subgroup of the domain of $\phi$:
:$\map \ker \phi \lhd \Dom \phi$ | Let $\phi: G_1 \to G_2$ be a group homomorphism, where the identities of $G_1$ and $G_2$ are $e_{G_1}$ and $e_{G_2}$ respectively.
By Kernel of Group Homomorphism is Subgroup:
:$\map \ker \phi \le \Dom \phi$
Let $k \in \map \ker \phi, x \in G_1$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {x k x^{-1} }
| r = \map ... | Let $\phi$ be a [[Definition:Group Homomorphism|group homomorphism]].
Then the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$ is a [[Definition:Normal Subgroup|normal subgroup]] of the [[Definition:Domain of Mapping|domain]] of $\phi$:
:$\map \ker \phi \lhd \Dom \phi$ | Let $\phi: G_1 \to G_2$ be a [[Definition:Group Homomorphism|group homomorphism]], where the [[Definition:Identity Element|identities]] of $G_1$ and $G_2$ are $e_{G_1}$ and $e_{G_2}$ respectively.
By [[Kernel of Group Homomorphism is Subgroup]]:
:$\map \ker \phi \le \Dom \phi$
Let $k \in \map \ker \phi, x \in G_1$.... | Kernel is Normal Subgroup of Domain | https://proofwiki.org/wiki/Kernel_is_Normal_Subgroup_of_Domain | https://proofwiki.org/wiki/Kernel_is_Normal_Subgroup_of_Domain | [
"Kernels of Group Homomorphisms",
"Normal Subgroups"
] | [
"Definition:Group Homomorphism",
"Definition:Kernel of Group Homomorphism",
"Definition:Normal Subgroup",
"Definition:Domain (Set Theory)/Mapping"
] | [
"Definition:Group Homomorphism",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Kernel of Group Homomorphism is Subgroup",
"Homomorphism to Group Preserves Inverses",
"Subgroup is Normal iff Contains Conjugate Elements",
"Definition:Normal Subgroup"
] |
proofwiki-714 | Coset Product is Well-Defined | Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is well-defined. | Let $N \lhd G$ where $G$ is a group.
Let $a, a', b, b' \in G$ such that:
:$a \circ N = a' \circ N$
and:
:$b \circ N = b' \circ N$
To show that the coset product is well-defined, we need to demonstrate that:
:$\paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$
So:
{{begin-eqn}}
{{eqn | l = a \circ N
| r = ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let $a, b \in G$.
Then the [[Definition:Coset Product|coset product]]:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is [[Definition:Well-Defined Operation|wel... | Let $N \lhd G$ where $G$ is a [[Definition:Group|group]].
Let $a, a', b, b' \in G$ such that:
:$a \circ N = a' \circ N$
and:
:$b \circ N = b' \circ N$
To show that the coset product is [[Definition:Well-Defined Operation|well-defined]], we need to demonstrate that:
:$\paren {a \circ b} \circ N = \paren {a' \circ b'} ... | Coset Product is Well-Defined/Proof 1 | https://proofwiki.org/wiki/Coset_Product_is_Well-Defined | https://proofwiki.org/wiki/Coset_Product_is_Well-Defined/Proof_1 | [
"Coset Product",
"Coset Product is Well-Defined"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Coset Product",
"Definition:Well-Defined/Operation"
] | [
"Definition:Group",
"Definition:Well-Defined/Operation",
"Cosets are Equal iff Product with Inverse in Subgroup",
"Definition:Normal Subgroup",
"Inverse of Group Product",
"Cosets are Equal iff Product with Inverse in Subgroup"
] |
proofwiki-715 | Coset Product is Well-Defined | Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is well-defined. | Let $N \lhd G$ where $G$ is a group.
Consider $\paren {a \circ N} \circ \paren {b \circ N}$ as a subset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \set {a \circ n_1 \circ b \circ n_2: n_1, n_2 \in N}$
This is justified by Coset Product of Normal Subgroup is Consistent with Subset Product Definition.
Since... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let $a, b \in G$.
Then the [[Definition:Coset Product|coset product]]:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is [[Definition:Well-Defined Operation|wel... | Let $N \lhd G$ where $G$ is a [[Definition:Group|group]].
Consider $\paren {a \circ N} \circ \paren {b \circ N}$ as a [[Definition:Subset Product|subset product]]:
:$\paren {a \circ N} \circ \paren {b \circ N} = \set {a \circ n_1 \circ b \circ n_2: n_1, n_2 \in N}$
This is justified by [[Coset Product of Normal Subg... | Coset Product is Well-Defined/Proof 2 | https://proofwiki.org/wiki/Coset_Product_is_Well-Defined | https://proofwiki.org/wiki/Coset_Product_is_Well-Defined/Proof_2 | [
"Coset Product",
"Coset Product is Well-Defined"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Coset Product",
"Definition:Well-Defined/Operation"
] | [
"Definition:Group",
"Definition:Subset Product",
"Coset Product of Normal Subgroup is Consistent with Subset Product Definition",
"Definition:Normal Subgroup",
"Definition:Conjugate (Group Theory)",
"Definition:Set Equality/Definition 2"
] |
proofwiki-716 | Coset Product is Well-Defined | Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is well-defined. | Let $N \lhd G$ where $G$ is a group.
Let $a, a', b, b' \in G$ such that:
:$N \circ a = N \circ a'$
and:
:$N \circ b = N \circ b'$
We need to show that:
:$N \circ \paren {a \circ b} = N \circ \paren {a' \circ b'}$
So:
{{begin-eqn}}
{{eqn | l = N \circ \paren {a \circ b}
| r = \paren {N \circ a} \circ b
| c =... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let $a, b \in G$.
Then the [[Definition:Coset Product|coset product]]:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is [[Definition:Well-Defined Operation|wel... | Let $N \lhd G$ where $G$ is a [[Definition:Group|group]].
Let $a, a', b, b' \in G$ such that:
:$N \circ a = N \circ a'$
and:
:$N \circ b = N \circ b'$
We need to show that:
:$N \circ \paren {a \circ b} = N \circ \paren {a' \circ b'}$
So:
{{begin-eqn}}
{{eqn | l = N \circ \paren {a \circ b}
| r = \paren {N \c... | Coset Product is Well-Defined/Proof 3 | https://proofwiki.org/wiki/Coset_Product_is_Well-Defined | https://proofwiki.org/wiki/Coset_Product_is_Well-Defined/Proof_3 | [
"Coset Product",
"Coset Product is Well-Defined"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Coset Product",
"Definition:Well-Defined/Operation"
] | [
"Definition:Group"
] |
proofwiki-717 | Coset Product is Well-Defined | Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is well-defined. | Let $N \lhd G$ where $G$ is a group.
By Left Congruence Class Modulo Subgroup is Left Coset, it can be shown that the left congruence modulo $N$ is an equivalence relation.
Let $\RR^l_N$ denote the equivalence relation on $G$
:$\RR^l_N := \set {\tuple {x, y} \in G \times G: x^{-1} y \in N}$
Let $\struct {G / N, \circ_N... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let $a, b \in G$.
Then the [[Definition:Coset Product|coset product]]:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is [[Definition:Well-Defined Operation|wel... | Let $N \lhd G$ where $G$ is a [[Definition:Group|group]].
By [[Left Congruence Class Modulo Subgroup is Left Coset]], it can be shown that the [[Definition:Left Congruence Modulo Subgroup|left congruence modulo $N$]] is an [[Definition:Equivalence Relation|equivalence relation]].
Let $\RR^l_N$ denote the [[Definition... | Coset Product is Well-Defined/Proof 4 | https://proofwiki.org/wiki/Coset_Product_is_Well-Defined | https://proofwiki.org/wiki/Coset_Product_is_Well-Defined/Proof_4 | [
"Coset Product",
"Coset Product is Well-Defined"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Coset Product",
"Definition:Well-Defined/Operation"
] | [
"Definition:Group",
"Left Congruence Class Modulo Subgroup is Left Coset",
"Definition:Congruence Modulo Subgroup/Left Congruence",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Quotient Group",
"Definition:Operation Induced on Quotient Set",
"Left Congruence Class ... |
proofwiki-718 | Coset Product is Well-Defined | Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is well-defined. | Let $N \lhd G$ where $G$ is a group.
Let $a, a', b, b' \in G$ such that:
:$a \circ N = a' \circ N$
and:
:$b \circ N = b' \circ N$
To show that the coset product is well-defined, we need to demonstrate that:
:$\paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$
So:
{{begin-eqn}}
{{eqn | l = a \circ N
| r = ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let $a, b \in G$.
Then the [[Definition:Coset Product|coset product]]:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is [[Definition:Well-Defined Operation|wel... | Let $N \lhd G$ where $G$ is a [[Definition:Group|group]].
Let $a, a', b, b' \in G$ such that:
:$a \circ N = a' \circ N$
and:
:$b \circ N = b' \circ N$
To show that the coset product is [[Definition:Well-Defined Operation|well-defined]], we need to demonstrate that:
:$\paren {a \circ b} \circ N = \paren {a' \circ b'} ... | Coset Product is Well-Defined/Proof 5 | https://proofwiki.org/wiki/Coset_Product_is_Well-Defined | https://proofwiki.org/wiki/Coset_Product_is_Well-Defined/Proof_5 | [
"Coset Product",
"Coset Product is Well-Defined"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Coset Product",
"Definition:Well-Defined/Operation"
] | [
"Definition:Group",
"Definition:Well-Defined/Operation",
"Definition:Normal Subgroup"
] |
proofwiki-719 | Quotient Group of Abelian Group is Abelian | Let $G$ be an abelian group.
Let $N \le G$.
Then the quotient group $G / N$ is abelian. | First we note that because $G$ is abelian, from Subgroup of Abelian Group is Normal we have $N \lhd G$.
Thus $G / N$ exists for ''all'' subgroups of $G$.
Let $X = x N, Y = y N$ where $x, y \in G$.
From the definition of coset product:
{{begin-eqn}}
{{eqn | l = X Y
| r = \paren {x N} \paren {y N}
| c =
}}
{... | Let $G$ be an [[Definition:Abelian Group|abelian group]].
Let $N \le G$.
Then the [[Definition:Quotient Group|quotient group]] $G / N$ is [[Definition:Abelian Group|abelian]]. | First we note that because $G$ is [[Definition:Abelian Group|abelian]], from [[Subgroup of Abelian Group is Normal]] we have $N \lhd G$.
Thus $G / N$ exists for ''all'' [[Definition:Subgroup|subgroups]] of $G$.
Let $X = x N, Y = y N$ where $x, y \in G$.
From the definition of [[Definition:Coset Product|coset produc... | Quotient Group of Abelian Group is Abelian | https://proofwiki.org/wiki/Quotient_Group_of_Abelian_Group_is_Abelian | https://proofwiki.org/wiki/Quotient_Group_of_Abelian_Group_is_Abelian | [
"Quotient Groups"
] | [
"Definition:Abelian Group",
"Definition:Quotient Group",
"Definition:Abelian Group"
] | [
"Definition:Abelian Group",
"Subgroup of Abelian Group is Normal",
"Definition:Subgroup",
"Definition:Coset Product",
"Definition:Abelian Group",
"Category:Quotient Groups"
] |
proofwiki-720 | Quotient Theorem for Group Epimorphisms | Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be groups.
Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a group epimorphism.
Let $e_G$ and $e_H$ be the identities of $G$ and $H$ respectively.
Let $K = \map \ker \phi$ be the kernel of $\phi$.
There exists one and only one group isomorphism $\psi: G / K \... | Let $\RR_\phi$ be the equivalence on $G$ defined by $\phi$.
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = e_G
| o = \RR_\phi
| r = x
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \map \phi x
| r = \map \phi {e_G}
| c = Definition of $\RR_\phi$
}}
{{eqn | ll= \leadstoandfrom
... | Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be [[Definition:Group|groups]].
Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a [[Definition:Group Epimorphism|group epimorphism]].
Let $e_G$ and $e_H$ be the [[Definition:Identity Element|identities]] of $G$ and $H$ respectively.
Let $K = \map \ker \ph... | Let $\RR_\phi$ be the [[Definition:Equivalence Relation Induced by Mapping|equivalence on $G$ defined by $\phi$]].
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = e_G
| o = \RR_\phi
| r = x
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \map \phi x
| r = \map \phi {e_G}
| c = Def... | Quotient Theorem for Group Epimorphisms | https://proofwiki.org/wiki/Quotient_Theorem_for_Group_Epimorphisms | https://proofwiki.org/wiki/Quotient_Theorem_for_Group_Epimorphisms | [
"Quotient Groups",
"Group Epimorphisms",
"Quotient Epimorphisms",
"Group Isomorphisms",
"Quotient Theorems",
"Named Theorems"
] | [
"Definition:Group",
"Definition:Group Epimorphism",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Kernel of Group Homomorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Quotient Epimorphism"
] | [
"Definition:Equivalence Relation Induced by Mapping",
"Homomorphism to Group Preserves Identity",
"Quotient Theorem for Epimorphisms",
"Kernel is Normal Subgroup of Domain",
"Congruence Relation induces Normal Subgroup",
"Definition:Equivalence Relation Induced by Mapping",
"Quotient Theorem for Epimorp... |
proofwiki-721 | Congruence Relation induces Normal Subgroup | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\RR$ be a congruence relation for $\circ$.
Let $H = \eqclass e \RR$, where $\eqclass e \RR$ is the equivalence class of $e$ under $\RR$.
Then:
:$(1): \quad \struct {H, \circ \restriction_H}$ is a normal subgroup of $G$
:$(2): \quad \RR$ is the equivalence ... | === Proof of Normal Subgroup ===
{{:Congruence Relation on Group induces Normal Subgroup}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\RR$ be a [[Definition:Congruence Relation|congruence relation]] for $\circ$.
Let $H = \eqclass e \RR$, where $\eqclass e \RR$ is the [[Definition:Equivalence Class|equivalence class]] of $e$ under $\R... | === [[Congruence Relation on Group induces Normal Subgroup|Proof of Normal Subgroup]] ===
{{:Congruence Relation on Group induces Normal Subgroup}} | Congruence Relation induces Normal Subgroup | https://proofwiki.org/wiki/Congruence_Relation_induces_Normal_Subgroup | https://proofwiki.org/wiki/Congruence_Relation_induces_Normal_Subgroup | [
"Normal Subgroups",
"Congruence Relations"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Congruence Relation",
"Definition:Equivalence Class",
"Definition:Normal Subgroup",
"Congruence Modulo Subgroup is Equivalence Relation",
"Definition:Subgroup",
"Definition:Semigroup"
] | [
"Congruence Relation on Group induces Normal Subgroup"
] |
proofwiki-722 | Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal | Let $G$ be a group.
Let $H \lhd G$ where $\lhd$ denotes that $H$ is a normal subgroup of $G$.
Let $K \lhd G / H$.
Let $L = q_H^{-1} \sqbrk K$, where:
:$q_H: G \to G / H$ is the quotient epimorphism from $G$ to the quotient group $G / H$
:$q_H^{-1} \sqbrk K$ is the preimage of $K$ under $q_H$.
Then:
:$L \lhd G$
and ther... | By Quotient Mapping on Structure is Epimorphism, both $q_K$ and $q_H$ are epimorphisms.
From Composite of Group Epimorphisms is Epimorphism we have that $q_K \circ q_H: G \to \paren {G / H} / K$ is also an epimorphism.
Now:
:$\forall x \in G: x \in \map \ker {q_K \circ q_H} \iff \map {q_K} {\map {q_H} x} = K = e_{G / H... | Let $G$ be a [[Definition:Group|group]].
Let $H \lhd G$ where $\lhd$ denotes that $H$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let $K \lhd G / H$.
Let $L = q_H^{-1} \sqbrk K$, where:
:$q_H: G \to G / H$ is the [[Definition:Quotient Group Epimorphism|quotient epimorphism]] from $G$ to the [[Definit... | By [[Quotient Mapping on Structure is Epimorphism]], both $q_K$ and $q_H$ are [[Definition:Group Epimorphism|epimorphisms]].
From [[Composite of Group Epimorphisms is Epimorphism]] we have that $q_K \circ q_H: G \to \paren {G / H} / K$ is also an [[Definition:Group Epimorphism|epimorphism]].
Now:
:$\forall x \in G: ... | Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal/Proof 1 | https://proofwiki.org/wiki/Preimage_of_Normal_Subgroup_of_Quotient_Group_under_Quotient_Epimorphism_is_Normal | https://proofwiki.org/wiki/Preimage_of_Normal_Subgroup_of_Quotient_Group_under_Quotient_Epimorphism_is_Normal/Proof_1 | [
"Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal",
"Quotient Groups",
"Group Isomorphisms",
"Group Epimorphisms"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Quotient Epimorphism/Group",
"Definition:Quotient Group",
"Definition:Preimage/Mapping/Subset",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Quotient Mapping on Structure is Epimorphism",
"Definition:Group Epimorphism",
"Composite of Group Epimorphisms is Epimorphism",
"Definition:Group Epimorphism",
"Kernel is Normal Subgroup of Domain",
"Quotient Mapping on Structure is Epimorphism",
"Definition:Group Epimorphism",
"Composite of Group E... |
proofwiki-723 | Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal | Let $G$ be a group.
Let $H \lhd G$ where $\lhd$ denotes that $H$ is a normal subgroup of $G$.
Let $K \lhd G / H$.
Let $L = q_H^{-1} \sqbrk K$, where:
:$q_H: G \to G / H$ is the quotient epimorphism from $G$ to the quotient group $G / H$
:$q_H^{-1} \sqbrk K$ is the preimage of $K$ under $q_H$.
Then:
:$L \lhd G$
and ther... | Let $e$ be the identity element of $G$.
Let $\RR$ be the congruence relation defined by $H$ in $G$.
Let $\SS$ be the congruence relation defined by $K$ in $G / H$.
Let $\TT$ be the relation on $G$ defined as:
:$\forall x, y \in G: x \mathrel \TT y \iff x H \mathrel \SS y H$
From Equivalence Relation induced by Congruen... | Let $G$ be a [[Definition:Group|group]].
Let $H \lhd G$ where $\lhd$ denotes that $H$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let $K \lhd G / H$.
Let $L = q_H^{-1} \sqbrk K$, where:
:$q_H: G \to G / H$ is the [[Definition:Quotient Group Epimorphism|quotient epimorphism]] from $G$ to the [[Definit... | Let $e$ be the [[Definition:Identity Element|identity element]] of $G$.
Let $\RR$ be the [[Definition:Congruence Modulo Subgroup|congruence relation defined by $H$]] in $G$.
Let $\SS$ be the [[Definition:Congruence Modulo Subgroup|congruence relation defined by $K$]] in $G / H$.
Let $\TT$ be the [[Definition:Relati... | Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal/Proof 2 | https://proofwiki.org/wiki/Preimage_of_Normal_Subgroup_of_Quotient_Group_under_Quotient_Epimorphism_is_Normal | https://proofwiki.org/wiki/Preimage_of_Normal_Subgroup_of_Quotient_Group_under_Quotient_Epimorphism_is_Normal/Proof_2 | [
"Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal",
"Quotient Groups",
"Group Isomorphisms",
"Group Epimorphisms"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Quotient Epimorphism/Group",
"Definition:Quotient Group",
"Definition:Preimage/Mapping/Subset",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Congruence Modulo Subgroup",
"Definition:Congruence Modulo Subgroup",
"Definition:Relation",
"Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence",
"Definition:Congruence Relation",
"Congruence R... |
proofwiki-724 | Trivial Quotient Group is Quotient Group | Let $G$ be a group.
Then the trivial quotient group:
:$G / \set {e_G} \cong G$
where:
:$\cong$ denotes group isomorphism
:$e_G$ denotes the identity element of $G$
is a quotient group. | From Trivial Subgroup is Normal:
:$\set {e_G} \lhd G$
Let $x \in G$.
Then:
:$x \set {e_G} = \set {x e_G} = \set x$
So each (left) coset of $G$ modulo $\set {e_G}$ has one element.
Now we set up the quotient epimorphism $\psi: G \to G / \set {e_G}$:
:$\forall x \in G: \map \phi x = x \set {e_G}$
which is of course a sur... | Let $G$ be a [[Definition:Group|group]].
Then the [[Definition:Trivial Quotient Group|trivial quotient group]]:
:$G / \set {e_G} \cong G$
where:
:$\cong$ denotes [[Definition:Group Isomorphism|group isomorphism]]
:$e_G$ denotes the [[Definition:Identity Element|identity element]] of $G$
is a [[Definition:Quotient G... | From [[Trivial Subgroup is Normal]]:
:$\set {e_G} \lhd G$
Let $x \in G$.
Then:
:$x \set {e_G} = \set {x e_G} = \set x$
So each [[Definition:Left Coset|(left) coset]] of $G$ modulo $\set {e_G}$ has one [[Definition:Element|element]].
Now we set up the [[Definition:Quotient Group Epimorphism|quotient epimorphism]] $... | Trivial Quotient Group is Quotient Group | https://proofwiki.org/wiki/Trivial_Quotient_Group_is_Quotient_Group | https://proofwiki.org/wiki/Trivial_Quotient_Group_is_Quotient_Group | [
"Examples of Quotient Groups",
"Group Isomorphisms"
] | [
"Definition:Group",
"Definition:Trivial Quotient Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Quotient Group"
] | [
"Trivial Subgroup is Normal",
"Definition:Coset/Left Coset",
"Definition:Element",
"Definition:Quotient Epimorphism/Group",
"Definition:Surjection",
"Definition:Injection",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] |
proofwiki-725 | Correspondence Theorem (Group Theory) | Let $G$ be a group.
Let $N \lhd G$ be a normal subgroup of $G$.
Then every subgroup of the quotient group $G / N$ is of the form $H / N = \set {h N: h \in H}$, where $N \le H \le G$.
Conversely, if $N \le H \le G$ then $H / N \le G / N$.
The correspondence between subgroups of $G / N$ and subgroups of $G$ containing $N... | Let $H'$ be a subgroup of $G / N$, so that it consists of a certain set $\set {h N}$ of left cosets of $N$ in $G$.
Let us define the subset $\map \beta {H'} \subseteq G$:
:$\map \beta {H'} = \set {g \in G: g N \in H'}$
Then clearly:
:$N \subseteq \map \beta {H'}$
Also:
:$e_G \in N$
so:
:$e_G \in \map \beta {H'}$
Let $x... | Let $G$ be a [[Definition:Group|group]].
Let $N \lhd G$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Then every [[Definition:Subgroup|subgroup]] of the [[Definition:Quotient Group|quotient group]] $G / N$ is of the form $H / N = \set {h N: h \in H}$, where $N \le H \le G$.
Conversely, if $N \le H \le... | Let $H'$ be a [[Definition:Subgroup|subgroup]] of $G / N$, so that it consists of a certain [[Definition:Set|set]] $\set {h N}$ of [[Definition:Left Coset|left cosets]] of $N$ in $G$.
Let us define the [[Definition:Subset|subset]] $\map \beta {H'} \subseteq G$:
:$\map \beta {H'} = \set {g \in G: g N \in H'}$
Then cl... | Correspondence Theorem (Group Theory) | https://proofwiki.org/wiki/Correspondence_Theorem_(Group_Theory) | https://proofwiki.org/wiki/Correspondence_Theorem_(Group_Theory) | [
"Normal Subgroups",
"Quotient Groups",
"Named Theorems"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Subgroup",
"Definition:Quotient Group",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Bijection",
"Definition:Bijection",
"Definition:Normal Subgroup",
"Definition:Normal Subgroup"
] | [
"Definition:Subgroup",
"Definition:Set",
"Definition:Coset/Left Coset",
"Definition:Subset",
"Quotient Group is Group",
"Two-Step Subgroup Test",
"Definition:Set",
"Definition:Subgroup",
"Definition:Set",
"Definition:Subgroup",
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"De... |
proofwiki-726 | Centralizer of Group Element is Subgroup | Let $\struct {G, \circ}$ be a group and let $a \in G$.
Then $\map {C_G} a$, the centralizer of $a$ in $G$, is a subgroup of $G$. | Let $\struct {G, \circ}$ be a group.
We have that:
:$\forall a \in G: e \circ a = a \circ e \implies e \in C_G \paren a$
Thus $C_G \paren a \ne \O$.
Let $x, y \in C_G \paren a$.
Then:
{{begin-eqn}}
{{eqn | l = x \circ a
| r = a \circ x
| c =
}}
{{eqn | l = y \circ a
| r = a \circ y
| c =
}}
{{... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] and let $a \in G$.
Then $\map {C_G} a$, the [[Definition:Centralizer of Group Element|centralizer]] of $a$ in $G$, is a [[Definition:Subgroup|subgroup]] of $G$. | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
We have that:
:$\forall a \in G: e \circ a = a \circ e \implies e \in C_G \paren a$
Thus $C_G \paren a \ne \O$.
Let $x, y \in C_G \paren a$.
Then:
{{begin-eqn}}
{{eqn | l = x \circ a
| r = a \circ x
| c =
}}
{{eqn | l = y \circ a
| r = a... | Centralizer of Group Element is Subgroup/Proof 1 | https://proofwiki.org/wiki/Centralizer_of_Group_Element_is_Subgroup | https://proofwiki.org/wiki/Centralizer_of_Group_Element_is_Subgroup/Proof_1 | [
"Centralizer of Group Element is Subgroup",
"Centralizers",
"Subgroups"
] | [
"Definition:Group",
"Definition:Centralizer/Group Element",
"Definition:Subgroup"
] | [
"Definition:Group",
"Two-Step Subgroup Test"
] |
proofwiki-727 | Centralizer of Group Element is Subgroup | Let $\struct {G, \circ}$ be a group and let $a \in G$.
Then $\map {C_G} a$, the centralizer of $a$ in $G$, is a subgroup of $G$. | Let $\struct {G, \circ}$ be a group.
We have that:
:$\forall a \in G: e \circ a = a \circ e \implies e \in \map {C_G} a$
Thus $\map {C_G} a \ne \O$.
Let $x, y \in \map {C_G} a$.
Then from Commutation with Group Elements implies Commuation with Product with Inverse:
:$a \circ x \circ y^{-1} = x \circ y^{-1} \circ a$
so:... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] and let $a \in G$.
Then $\map {C_G} a$, the [[Definition:Centralizer of Group Element|centralizer]] of $a$ in $G$, is a [[Definition:Subgroup|subgroup]] of $G$. | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
We have that:
:$\forall a \in G: e \circ a = a \circ e \implies e \in \map {C_G} a$
Thus $\map {C_G} a \ne \O$.
Let $x, y \in \map {C_G} a$.
Then from [[Commutation with Group Elements implies Commuation with Product with Inverse]]:
:$a \circ x \circ y^{-1}... | Centralizer of Group Element is Subgroup/Proof 2 | https://proofwiki.org/wiki/Centralizer_of_Group_Element_is_Subgroup | https://proofwiki.org/wiki/Centralizer_of_Group_Element_is_Subgroup/Proof_2 | [
"Centralizer of Group Element is Subgroup",
"Centralizers",
"Subgroups"
] | [
"Definition:Group",
"Definition:Centralizer/Group Element",
"Definition:Subgroup"
] | [
"Definition:Group",
"Commutation with Group Elements implies Commuation with Product with Inverse",
"One-Step Subgroup Test"
] |
proofwiki-728 | Centralizer in Subgroup is Intersection | Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then:
:$\forall x \in G: \map {C_H} x = \map {C_G} x \cap H$
That is, the centralizer of an element in a subgroup is the intersection of that subgroup with the centralizer of the element in the group. | It is clear that:
:$g \in \map {C_H} x \iff g \in \map {C_G} x \land g \in H$
The result follows by definition of set intersection.
{{qed}} | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then:
:$\forall x \in G: \map {C_H} x = \map {C_G} x \cap H$
That is, the [[Definition:Centralizer of Group Element|centralizer]] of an [[Definition:Element|element]] in a [[Definition:Subgroup|subgroup]] is the [[Defini... | It is clear that:
:$g \in \map {C_H} x \iff g \in \map {C_G} x \land g \in H$
The result follows by definition of [[Definition:Set Intersection|set intersection]].
{{qed}} | Centralizer in Subgroup is Intersection | https://proofwiki.org/wiki/Centralizer_in_Subgroup_is_Intersection | https://proofwiki.org/wiki/Centralizer_in_Subgroup_is_Intersection | [
"Subgroups",
"Centralizers",
"Set Intersection"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Centralizer/Group Element",
"Definition:Element",
"Definition:Subgroup",
"Definition:Set Intersection",
"Definition:Subgroup",
"Definition:Centralizer/Group Element",
"Definition:Element",
"Definition:Group"
] | [
"Definition:Set Intersection"
] |
proofwiki-729 | Kernel of Inner Automorphism Group is Center | Let the mapping $\kappa: G \to \Inn G$ from a group $G$ to its inner automorphism group $\Inn G$ be defined as:
:$\map \kappa a = \kappa_a$
where $\kappa_a$ is the inner automorphism of $G$ given by $a$.
Then $\kappa$ is a group epimorphism, and its kernel is the center of $G$:
:$\map \ker \kappa = \map Z G$ | Let $\kappa: G \to \Aut G$ be a mapping defined by $\map \kappa x = \kappa_x$.
It is clear that $\Img \kappa = \Inn G$.
It is also clear that $\kappa$ is a homomorphism:
{{begin-eqn}}
{{eqn | l = \map \kappa x \map \kappa y
| r = \kappa_x \circ \kappa_y
| c =
}}
{{eqn | r = \kappa_{x y}
| c = Inner A... | Let the [[Definition:Mapping|mapping]] $\kappa: G \to \Inn G$ from a [[Definition:Group|group]] $G$ to its [[Definition:Inner Automorphism Group|inner automorphism group]] $\Inn G$ be defined as:
:$\map \kappa a = \kappa_a$
where $\kappa_a$ is the [[Definition:Inner Automorphism|inner automorphism]] of $G$ given by $a$... | Let $\kappa: G \to \Aut G$ be a [[Definition:Mapping|mapping]] defined by $\map \kappa x = \kappa_x$.
It is clear that $\Img \kappa = \Inn G$.
It is also clear that $\kappa$ is a [[Definition:Group Homomorphism|homomorphism]]:
{{begin-eqn}}
{{eqn | l = \map \kappa x \map \kappa y
| r = \kappa_x \circ \kappa_y
... | Kernel of Inner Automorphism Group is Center | https://proofwiki.org/wiki/Kernel_of_Inner_Automorphism_Group_is_Center | https://proofwiki.org/wiki/Kernel_of_Inner_Automorphism_Group_is_Center | [
"Inner Automorphisms",
"Centers of Groups"
] | [
"Definition:Mapping",
"Definition:Group",
"Definition:Inner Automorphism Group",
"Definition:Inner Automorphism",
"Definition:Group Epimorphism",
"Definition:Kernel of Group Homomorphism",
"Definition:Center (Abstract Algebra)/Group"
] | [
"Definition:Mapping",
"Definition:Group Homomorphism",
"Inner Automorphism is Automorphism",
"Definition:Surjection",
"Definition:Group Epimorphism",
"Definition:Kernel of Group Homomorphism",
"Equality of Mappings",
"Definition:Kernel of Group Homomorphism",
"Definition:Center (Abstract Algebra)/Gr... |
proofwiki-730 | Normalizer is Subgroup | Let $G$ be a group.
The normalizer of a subset $S \subseteq G$ is a subgroup of $G$.
:$S \subseteq G \implies \map {N_G} S \le G$ | Let $a, b \in \map {N_G} S$.
Then:
{{begin-eqn}}
{{eqn | l = S^{a b}
| r = \paren {S^b}^a
| c = Conjugate of Set by Group Product
}}
{{eqn | r = S^a
| c = {{Defof|Normalizer}}
}}
{{eqn | r = S
| c = {{Defof|Normalizer}}
}}
{{end-eqn}}
Therefore $a b \in \map {N_G} S$.
Now let $a \in \map {N_G} S... | Let $G$ be a [[Definition:Group|group]].
The [[Definition:Normalizer|normalizer]] of a [[Definition:Subset|subset]] $S \subseteq G$ is a [[Definition:Subgroup|subgroup]] of $G$.
:$S \subseteq G \implies \map {N_G} S \le G$ | Let $a, b \in \map {N_G} S$.
Then:
{{begin-eqn}}
{{eqn | l = S^{a b}
| r = \paren {S^b}^a
| c = [[Conjugate of Set by Group Product]]
}}
{{eqn | r = S^a
| c = {{Defof|Normalizer}}
}}
{{eqn | r = S
| c = {{Defof|Normalizer}}
}}
{{end-eqn}}
Therefore $a b \in \map {N_G} S$.
Now let $a \in \ma... | Normalizer is Subgroup | https://proofwiki.org/wiki/Normalizer_is_Subgroup | https://proofwiki.org/wiki/Normalizer_is_Subgroup | [
"Normalizers"
] | [
"Definition:Group",
"Definition:Normalizer",
"Definition:Subset",
"Definition:Subgroup"
] | [
"Conjugate of Set by Group Product",
"Two-Step Subgroup Test"
] |
proofwiki-731 | Subgroup is Normal Subgroup of Normalizer | Let $G$ be a group.
A subgroup $H \le G$ is a normal subgroup of its normalizer:
:$H \le G \implies H \lhd \map {N_G} H$ | From Subgroup is Subgroup of Normalizer we have that $H \le \map {N_G} H$.
It remains to show that $H$ is normal in $\map {N_G} H$.
Let $a \in H$ and $b \in \map {N_G} H$.
By the definition of normalizer:
:$b a b^{-1} \in H$
Thus $H$ is normal in $\map {N_G} H$.
{{qed}} | Let $G$ be a [[Definition:Group|group]].
A [[Definition:Subgroup|subgroup]] $H \le G$ is a [[Definition:Normal Subgroup|normal subgroup]] of its [[Definition:Normalizer|normalizer]]:
:$H \le G \implies H \lhd \map {N_G} H$ | From [[Subgroup is Subgroup of Normalizer]] we have that $H \le \map {N_G} H$.
It remains to show that $H$ is [[Definition:Normal Subgroup|normal]] in $\map {N_G} H$.
Let $a \in H$ and $b \in \map {N_G} H$.
By the definition of [[Definition:Normalizer|normalizer]]:
:$b a b^{-1} \in H$
Thus $H$ is [[Definition:Norma... | Subgroup is Normal Subgroup of Normalizer | https://proofwiki.org/wiki/Subgroup_is_Normal_Subgroup_of_Normalizer | https://proofwiki.org/wiki/Subgroup_is_Normal_Subgroup_of_Normalizer | [
"Normal Subgroups",
"Normalizers",
"Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Normalizer"
] | [
"Subgroup is Subgroup of Normalizer",
"Definition:Normal Subgroup",
"Definition:Normalizer",
"Definition:Normal Subgroup"
] |
proofwiki-732 | Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup | Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then $\map {N_G} H$, the normalizer of $H$ in $G$, is the largest subgroup of $G$ containing $H$ as a normal subgroup. | From Subgroup is Subgroup of Normalizer, we have that $H \le \map {N_G} H$.
Now we need to show that $H \lhd \map {N_G} H$.
For $a \in \map {N_G} H$, the conjugate of $H$ by $a$ in $\map {N_G} H$ is:
{{begin-eqn}}
{{eqn | l = H^a
| r = \set {x \in \map {N_G} H: a x a^{-1} \in H}
| c = {{Defof|Conjugate of G... | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $\map {N_G} H$, the [[Definition:Normalizer|normalizer]] of $H$ in $G$, is the largest [[Definition:Subgroup|subgroup]] of $G$ containing $H$ as a [[Definition:Normal Subgroup|normal subgroup]]. | From [[Subgroup is Subgroup of Normalizer]], we have that $H \le \map {N_G} H$.
Now we need to show that $H \lhd \map {N_G} H$.
For $a \in \map {N_G} H$, the [[Definition:Conjugate of Group Subset|conjugate]] of $H$ by $a$ in $\map {N_G} H$ is:
{{begin-eqn}}
{{eqn | l = H^a
| r = \set {x \in \map {N_G} H: a x... | Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup | https://proofwiki.org/wiki/Normalizer_of_Subgroup_is_Largest_Subgroup_containing_that_Subgroup_as_Normal_Subgroup | https://proofwiki.org/wiki/Normalizer_of_Subgroup_is_Largest_Subgroup_containing_that_Subgroup_as_Normal_Subgroup | [
"Normalizers"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Normalizer",
"Definition:Subgroup",
"Definition:Normal Subgroup"
] | [
"Subgroup is Subgroup of Normalizer",
"Definition:Conjugate (Group Theory)/Subset",
"Intersection with Subset is Subset",
"Definition:Normal Subgroup",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Subset",
"Definition:Conjugate (Group Theory)/Subset",
"De... |
proofwiki-733 | Normal Subgroup iff Normalizer is Group | Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then $H$ is normal in $G$ {{iff}} the normalizer of $H$ is equal to $G$:
:$H \lhd G \iff \map {N_G} H = G$ | === Sufficient Condition ===
Let $H$ be normal in $G$.
Then $G$ is trivially the largest subgroup of $G$ in which $H$ is normal.
Thus from Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:
:$\map {N_G} H = G$
{{qed|lemma}} | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $H$ is [[Definition:Normal Subgroup|normal]] in $G$ {{iff}} the [[Definition:Normalizer|normalizer]] of $H$ is equal to $G$:
:$H \lhd G \iff \map {N_G} H = G$ | === Sufficient Condition ===
Let $H$ be [[Definition:Normal Subgroup|normal]] in $G$.
Then $G$ is trivially the largest [[Definition:Subgroup|subgroup]] of $G$ in which $H$ is [[Definition:Normal Subgroup|normal]].
Thus from [[Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup]]:
... | Normal Subgroup iff Normalizer is Group | https://proofwiki.org/wiki/Normal_Subgroup_iff_Normalizer_is_Group | https://proofwiki.org/wiki/Normal_Subgroup_iff_Normalizer_is_Group | [
"Normal Subgroups",
"Normalizers"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Normalizer"
] | [
"Definition:Normal Subgroup",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup",
"Definition:Normal Subgroup",
"Definition:Normal Subgroup"
] |
proofwiki-734 | Normalizer of Conjugate is Conjugate of Normalizer | Let $G$ be a group.
Let $a \mathop \in G$.
Let $S$ be a subset of $G$.
Let $S^a$ denote the conjugate of $S$ by $a$.
Let $\map {N_G} S$ denote the normalizer of $S$ in $G$.
Then:
:$\map {N_G} {S^a} = \paren {\map {N_G} S}^a$
That is, the normalizer of a conjugate is the conjugate of the normalizer: | From the definition of conjugate:
:$S^a = \set {y \in G: \exists x \in S: y = a x a^{-1} } = a S a^{-1}$
From the definition of normalizer:
:$\map {N_G} S = \set {x \in G: S^x = S}$
Thus:
{{begin-eqn}}
{{eqn | l = \map {N_G} {S^a}
| r = \set {x \in G: \paren {S^a}^x = S^a}
| c =
}}
{{eqn | r = \set {x \in ... | Let $G$ be a [[Definition:Group|group]].
Let $a \mathop \in G$.
Let $S$ be a [[Definition:Subset|subset]] of $G$.
Let $S^a$ denote the [[Definition:Conjugate of Group Subset|conjugate]] of $S$ by $a$.
Let $\map {N_G} S$ denote the [[Definition:Normalizer|normalizer]] of $S$ in $G$.
Then:
:$\map {N_G} {S^a} = \par... | From the definition of [[Definition:Conjugate of Group Subset|conjugate]]:
:$S^a = \set {y \in G: \exists x \in S: y = a x a^{-1} } = a S a^{-1}$
From the definition of [[Definition:Normalizer|normalizer]]:
:$\map {N_G} S = \set {x \in G: S^x = S}$
Thus:
{{begin-eqn}}
{{eqn | l = \map {N_G} {S^a}
| r = \set {x... | Normalizer of Conjugate is Conjugate of Normalizer | https://proofwiki.org/wiki/Normalizer_of_Conjugate_is_Conjugate_of_Normalizer | https://proofwiki.org/wiki/Normalizer_of_Conjugate_is_Conjugate_of_Normalizer | [
"Normalizers",
"Conjugacy"
] | [
"Definition:Group",
"Definition:Subset",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Normalizer",
"Definition:Normalizer",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Normalizer"
] | [
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Normalizer",
"Power of Conjugate equals Conjugate of Power",
"Power of Conjugate equals Conjugate of Power",
"Definition:Set Equality/Definition 2"
] |
proofwiki-735 | Normalizer of Center is Group | Let $G$ be a group.
Let $\map Z G$ be the center of $G$.
Let $x \in G$.
Let $\map {N_G} x$ be the normalizer of $x$ in $G$.
Then:
:$\map Z G = \set {x \in G: \map {N_G} x = G}$
That is, the center of a group $G$ is the set of elements $x$ of $G$ such that the normalizer of $x$ is the whole of $G$.
Thus:
:$x \in \map Z ... | $\map {N_G} x$ is the normalizer of the set $\set x$.
Thus:
{{begin-eqn}}
{{eqn | l = \map {N_G} x
| r = G
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall a \in G
| l = \set x^a
| r = \set x
| c = {{Defof|Normalizer}}
}}
{{eqn | ll= \leadstoandfrom
| q = \forall a \in G
... | Let $G$ be a [[Definition:Group|group]].
Let $\map Z G$ be the [[Definition:Center of Group|center]] of $G$.
Let $x \in G$.
Let $\map {N_G} x$ be the [[Definition:Normalizer|normalizer of $x$ in $G$]].
Then:
:$\map Z G = \set {x \in G: \map {N_G} x = G}$
That is, the [[Definition:Center of Group|center]] of a [[... | $\map {N_G} x$ is the [[Definition:Normalizer|normalizer]] of the [[Definition:Set|set]] $\set x$.
Thus:
{{begin-eqn}}
{{eqn | l = \map {N_G} x
| r = G
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall a \in G
| l = \set x^a
| r = \set x
| c = {{Defof|Normalizer}}
}}
{{eqn | ll=... | Normalizer of Center is Group | https://proofwiki.org/wiki/Normalizer_of_Center_is_Group | https://proofwiki.org/wiki/Normalizer_of_Center_is_Group | [
"Normalizers",
"Centers of Groups"
] | [
"Definition:Group",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Normalizer",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Group",
"Definition:Set",
"Definition:Element",
"Definition:Normalizer",
"Definition:Index of Subgroup"
] | [
"Definition:Normalizer",
"Definition:Set"
] |
proofwiki-736 | Quotient Group of Ideal is Coset Space | Let $\struct {R, +, \circ}$ be a ring.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +}$ be the quotient group of $\struct {R, +}$ by $\struct {J, +}$.
Then each element of $\struct {R / J, +}$ is a coset of $J$ in $R$, that is, is of the form $x + J = \set {x + j: j \in J}$ for some $x \in R$.
The rule of addition ... | From Ideal is Additive Normal Subgroup that $J$ is a normal subgroup of $R$ under $+$.
From Quotient Ring Addition is Well-Defined, $+$ is a well-defined operation.
The rest follows directly from the definition of quotient group.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
Let $\struct {R / J, +}$ be the [[Definition:Quotient Group|quotient group]] of $\struct {R, +}$ by $\struct {J, +}$.
Then each element of $\struct {R / J, +}$ is a [[Definition:Cos... | From [[Ideal is Additive Normal Subgroup]] that $J$ is a [[Definition:Normal Subgroup|normal subgroup]] of $R$ under $+$.
From [[Quotient Ring Addition is Well-Defined]], $+$ is a [[Definition:Well-Defined Operation|well-defined operation]].
The rest follows directly from the definition of [[Definition:Quotient Group... | Quotient Group of Ideal is Coset Space | https://proofwiki.org/wiki/Quotient_Group_of_Ideal_is_Coset_Space | https://proofwiki.org/wiki/Quotient_Group_of_Ideal_is_Coset_Space | [
"Ideal Theory",
"Quotient Groups",
"Cosets"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ideal of Ring",
"Definition:Quotient Group",
"Definition:Coset",
"Definition:Coset",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Ideal is Additive Normal Subgroup",
"Definition:Normal Subgroup",
"Quotient Ring is Ring/Quotient Ring Addition is Well-Defined",
"Definition:Well-Defined/Operation",
"Definition:Quotient Group"
] |
proofwiki-737 | Quotient Ring is Ring/Quotient Ring Product is Well-Defined | Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring of $R$ by $J$.
Then $\circ$ is well-defined on $R / J$, that is:
:$x_1 + J = x_2 + J, y_1 + J = y_2 + J \implies x_1 \circ y_1 + J = x_2 \circ y_2 + J$ | From Left Cosets are Equal iff Product with Inverse in Subgroup, we have:
{{begin-eqn}}
{{eqn | l = x_1 + J
| r = x_2 + J
| c =
}}
{{eqn | ll= \leadsto
| l = x_1 + \paren {-x_2}
| o = \in
| r = J
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = y_1 + J
| r = y_2 + J
... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
Let $\struct {R / J, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]] of... | From [[Left Cosets are Equal iff Product with Inverse in Subgroup]], we have:
{{begin-eqn}}
{{eqn | l = x_1 + J
| r = x_2 + J
| c =
}}
{{eqn | ll= \leadsto
| l = x_1 + \paren {-x_2}
| o = \in
| r = J
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = y_1 + J
| r = y_2 + J... | Quotient Ring is Ring/Quotient Ring Product is Well-Defined | https://proofwiki.org/wiki/Quotient_Ring_is_Ring/Quotient_Ring_Product_is_Well-Defined | https://proofwiki.org/wiki/Quotient_Ring_is_Ring/Quotient_Ring_Product_is_Well-Defined | [
"Quotient Rings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Well-Defined/Operation"
] | [
"Left Cosets are Equal iff Product with Inverse in Subgroup",
"Definition:Ideal of Ring",
"Definition:Group",
"Definition:Ring (Abstract Algebra)",
"Left Cosets are Equal iff Product with Inverse in Subgroup"
] |
proofwiki-738 | Quotient Ring of Commutative Ring is Commutative | Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$.
If $\struct {R, +, \circ}$ is a commutative ring, then so is $\struct {R / J, +, \circ}$. | Let $\struct {R, +, \circ}$ be a commutative ring
That means $\circ$ is commutative on $R$.
Thus:
{{begin-eqn}}
{{eqn | q = \forall x, y \in R
| l = \paren {x + J} \circ \paren {y + J}
| r = x \circ y + J
| c = Definition of $\circ$ in $R / J$
}}
{{eqn | r = y \circ x + J
| c = Commutativity of ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
Let $\struct {R / J, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]] de... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]]
That means $\circ$ is [[Definition:Commutative Operation|commutative]] on $R$.
Thus:
{{begin-eqn}}
{{eqn | q = \forall x, y \in R
| l = \paren {x + J} \circ \paren {y + J}
| r = x \circ y + J
| c = [[Definition:Quotie... | Quotient Ring of Commutative Ring is Commutative | https://proofwiki.org/wiki/Quotient_Ring_of_Commutative_Ring_is_Commutative | https://proofwiki.org/wiki/Quotient_Ring_of_Commutative_Ring_is_Commutative | [
"Quotient Rings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Commutative Ring"
] | [
"Definition:Commutative Ring",
"Definition:Commutative/Operation",
"Definition:Quotient Ring",
"Definition:Commutative/Operation",
"Definition:Quotient Ring"
] |
proofwiki-739 | Quotient Ring of Ring with Unity is Ring with Unity | Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$.
Then $\struct {R / J, +, \circ}$ is a ring with unity, and its unity is $1_R + J$. | Let $\struct {R, +, \circ}$ be a ring with unity.
First, let $J \subsetneq R$.
By {{Corollary|Ideal of Unit is Whole Ring}}:
:$1_R \in J \implies J = R$
So $1_R \notin J$.
Thus $1_R + J \ne J$, so $1_R + J \ne 0_{R/J}$.
Now let $x \in R$.
{{begin-eqn}}
{{eqn | l = \paren {1_R + J} \circ \paren {x + J}
| r = 1_R \... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
Let $\struct {R / J, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]]... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
First, let $J \subsetneq R$.
By {{Corollary|Ideal of Unit is Whole Ring}}:
:$1_R \in J \implies J = R$
So $1_R \notin J$.
Thus $1_R + J \ne J$, so $1_R + J \ne 0_{R/J}$.
Now let $x \in R$.
{{begin-eqn}}
{{eqn | l = \paren {1_R + J}... | Quotient Ring of Ring with Unity is Ring with Unity | https://proofwiki.org/wiki/Quotient_Ring_of_Ring_with_Unity_is_Ring_with_Unity | https://proofwiki.org/wiki/Quotient_Ring_of_Ring_with_Unity_is_Ring_with_Unity | [
"Quotient Rings",
"Rings with Unity"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Ring with Unity",
"Definition:Unity (Abstract Algebra)/Ring"
] | [
"Definition:Ring with Unity",
"Definition:Quotient Ring",
"Definition:Quotient Ring",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Null Ring",
"Definition:Null Ring"
] |
proofwiki-740 | Ring Epimorphism with Trivial Kernel is Isomorphism | $\phi$ is an isomorphism {{iff}} $K = \set {0_{R_1} }$. | From Kernel is Trivial iff Monomorphism, $\phi$ is a ring monomorphism {{iff}} $K = \set {0_{R_1} }$.
As $\phi$ is also an epimorphism, the result follows.
{{qed}} | $\phi$ is an [[Definition:Ring Isomorphism|isomorphism]] {{iff}} $K = \set {0_{R_1} }$. | From [[Kernel is Trivial iff Monomorphism]], $\phi$ is a [[Definition:Ring Monomorphism|ring monomorphism]] {{iff}} $K = \set {0_{R_1} }$.
As $\phi$ is also an [[Definition:Ring Epimorphism|epimorphism]], the result follows.
{{qed}} | Ring Epimorphism with Trivial Kernel is Isomorphism | https://proofwiki.org/wiki/Ring_Epimorphism_with_Trivial_Kernel_is_Isomorphism | https://proofwiki.org/wiki/Ring_Epimorphism_with_Trivial_Kernel_is_Isomorphism | [
"Ring Epimorphisms",
"Ring Isomorphisms"
] | [
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism"
] | [
"Kernel is Trivial iff Monomorphism",
"Definition:Ring Monomorphism",
"Definition:Ring Epimorphism"
] |
proofwiki-741 | Quotient Epimorphism is Epimorphism/Ring | Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$.
Let $\phi: R \to R / J$ be the quotient (ring) epimorphism from $R$ to $R / J$:
:$x \in R: \map \phi x = x + J$
Then $\phi$ is a ring epimo... | Let $x, y \in R$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {x + y}
| r = \paren {x + y} + J
| c = Definition of $\phi$
}}
{{eqn | r = \paren {x + J} + \paren {y + J}
| c = Quotient Ring Addition is Well-Defined
}}
{{eqn | r = \map \phi x + \map \phi y
| c = Definition of $\phi$
}}
{{end-eqn}}
a... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
Let $\struct {R / J, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]] de... | Let $x, y \in R$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {x + y}
| r = \paren {x + y} + J
| c = Definition of $\phi$
}}
{{eqn | r = \paren {x + J} + \paren {y + J}
| c = [[Quotient Ring Addition is Well-Defined]]
}}
{{eqn | r = \map \phi x + \map \phi y
| c = Definition of $\phi$
}}
{{end-e... | Quotient Epimorphism is Epimorphism/Ring | https://proofwiki.org/wiki/Quotient_Epimorphism_is_Epimorphism/Ring | https://proofwiki.org/wiki/Quotient_Epimorphism_is_Epimorphism/Ring | [
"Quotient Epimorphism is Epimorphism",
"Quotient Rings",
"Quotient Epimorphisms",
"Kernels of Ring Homomorphisms"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Quotient Epimorphism/Ring",
"Definition:Ring Epimorphism",
"Definition:Kernel of Ring Homomorphism"
] | [
"Quotient Ring is Ring/Quotient Ring Addition is Well-Defined",
"Quotient Ring is Ring/Quotient Ring Product is Well-Defined",
"Definition:Ring Homomorphism",
"Definition:Surjection",
"Definition:Ring Epimorphism",
"Definition:Quotient Ring",
"Left Coset Equals Subgroup iff Element in Subgroup"
] |
proofwiki-742 | Ring Homomorphism Preserves Subrings | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S$ be a subring of $R_1$.
Then $\phi \sqbrk S$ is a subring of $R_2$. | Since $S \ne \O$, $\phi \sqbrk S \ne \O$.
From Group Homomorphism Preserves Subgroups, $\struct {\phi \sqbrk S, +_2}$ is a subgroup of $\struct {R_2, +_2}$.
From Homomorphism Preserves Subsemigroups, $\struct {\phi \sqbrk S, \circ_2}$ is a subsemigroup of $\struct {R_2, \circ_2}$.
Thus, as $\struct {R_2, +_2}$ is a gro... | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Let $S$ be a [[Definition:Subring|subring]] of $R_1$.
Then $\phi \sqbrk S$ is a [[Definition:Subring|subring]] of $R_2$. | Since $S \ne \O$, $\phi \sqbrk S \ne \O$.
From [[Group Homomorphism Preserves Subgroups]], $\struct {\phi \sqbrk S, +_2}$ is a [[Definition:Subgroup|subgroup]] of $\struct {R_2, +_2}$.
From [[Homomorphism Preserves Subsemigroups]], $\struct {\phi \sqbrk S, \circ_2}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\... | Ring Homomorphism Preserves Subrings/Proof 1 | https://proofwiki.org/wiki/Ring_Homomorphism_Preserves_Subrings | https://proofwiki.org/wiki/Ring_Homomorphism_Preserves_Subrings/Proof_1 | [
"Ring Homomorphism Preserves Subrings",
"Ring Homomorphisms",
"Subrings"
] | [
"Definition:Ring Homomorphism",
"Definition:Subring",
"Definition:Subring"
] | [
"Group Homomorphism Preserves Subgroups",
"Definition:Subgroup",
"Homomorphism Preserves Subsemigroups",
"Definition:Subsemigroup",
"Definition:Group",
"Definition:Semigroup"
] |
proofwiki-743 | Ring Homomorphism Preserves Subrings | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S$ be a subring of $R_1$.
Then $\phi \sqbrk S$ is a subring of $R_2$. | From Morphism Property Preserves Closure, $\phi \sqbrk {R_1}$ is a closed algebraic structure.
From Epimorphism Preserves Rings, $\phi \sqbrk S$ is a ring.
Hence the result, from the definition of subring.
{{qed}} | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Let $S$ be a [[Definition:Subring|subring]] of $R_1$.
Then $\phi \sqbrk S$ is a [[Definition:Subring|subring]] of $R_2$. | From [[Morphism Property Preserves Closure]], $\phi \sqbrk {R_1}$ is a [[Definition:Closed Algebraic Structure|closed algebraic structure]].
From [[Epimorphism Preserves Rings]], $\phi \sqbrk S$ is a [[Definition:Ring (Abstract Algebra)|ring]].
Hence the result, from the definition of [[Definition:Subring|subring]].
... | Ring Homomorphism Preserves Subrings/Proof 2 | https://proofwiki.org/wiki/Ring_Homomorphism_Preserves_Subrings | https://proofwiki.org/wiki/Ring_Homomorphism_Preserves_Subrings/Proof_2 | [
"Ring Homomorphism Preserves Subrings",
"Ring Homomorphisms",
"Subrings"
] | [
"Definition:Ring Homomorphism",
"Definition:Subring",
"Definition:Subring"
] | [
"Morphism Property Preserves Closure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Epimorphism Preserves Rings",
"Definition:Ring (Abstract Algebra)",
"Definition:Subring"
] |
proofwiki-744 | Ring Homomorphism Preserves Subrings | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S$ be a subring of $R_1$.
Then $\phi \sqbrk S$ is a subring of $R_2$. | Let $S$ be a subring of $R_1$.
Since $S \ne \O$ it follows that $\phi \sqbrk S \ne \O$.
Let $x, y \in \phi \sqbrk S$.
Then:
:$\exists s, t \in S: x = \map \phi s, y = \map \phi t$
So:
{{begin-eqn}}
{{eqn | l = x +_2 \paren {-y}
| r = \map \phi s +_2 \paren {-\map \phi t}
| c =
}}
{{eqn | r = \map \phi {s +... | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Let $S$ be a [[Definition:Subring|subring]] of $R_1$.
Then $\phi \sqbrk S$ is a [[Definition:Subring|subring]] of $R_2$. | Let $S$ be a [[Definition:Subring|subring]] of $R_1$.
Since $S \ne \O$ it follows that $\phi \sqbrk S \ne \O$.
Let $x, y \in \phi \sqbrk S$.
Then:
:$\exists s, t \in S: x = \map \phi s, y = \map \phi t$
So:
{{begin-eqn}}
{{eqn | l = x +_2 \paren {-y}
| r = \map \phi s +_2 \paren {-\map \phi t}
| c =
}... | Ring Homomorphism Preserves Subrings/Proof 3 | https://proofwiki.org/wiki/Ring_Homomorphism_Preserves_Subrings | https://proofwiki.org/wiki/Ring_Homomorphism_Preserves_Subrings/Proof_3 | [
"Ring Homomorphism Preserves Subrings",
"Ring Homomorphisms",
"Subrings"
] | [
"Definition:Ring Homomorphism",
"Definition:Subring",
"Definition:Subring"
] | [
"Definition:Subring",
"Definition:Ring Homomorphism",
"Definition:Subring",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Ring Negative",
"Definition:Ring Homomorphism",
"Definition:Subring",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Subring Test"
] |
proofwiki-745 | Kernel of Ring Homomorphism is Subring | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then the kernel of $\phi$ is a subring of $R_1$. | From Ring Homomorphism of Addition is Group Homomorphism and Kernel of Group Homomorphism is Subgroup:
:$\struct {\map \ker \phi, +_1} \le \struct {R_1, +_1}$
where $\le$ denotes subgroup.
Let $x, y \in \map \ker \phi$.
{{begin-eqn}}
{{eqn | l = \map \phi {x \circ_1 y}
| r = \map \phi x \circ_2 \map \phi y
... | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Then the [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$ is a [[Definition:Subring|subring]] of $R_1$. | From [[Ring Homomorphism of Addition is Group Homomorphism]] and [[Kernel of Group Homomorphism is Subgroup]]:
:$\struct {\map \ker \phi, +_1} \le \struct {R_1, +_1}$
where $\le$ denotes [[Definition:Subgroup|subgroup]].
Let $x, y \in \map \ker \phi$.
{{begin-eqn}}
{{eqn | l = \map \phi {x \circ_1 y}
| r = \m... | Kernel of Ring Homomorphism is Subring | https://proofwiki.org/wiki/Kernel_of_Ring_Homomorphism_is_Subring | https://proofwiki.org/wiki/Kernel_of_Ring_Homomorphism_is_Subring | [
"Ring Homomorphisms",
"Subrings"
] | [
"Definition:Ring Homomorphism",
"Definition:Kernel of Ring Homomorphism",
"Definition:Subring"
] | [
"Ring Homomorphism of Addition is Group Homomorphism",
"Kernel of Group Homomorphism is Subgroup",
"Definition:Subgroup",
"Definition:Morphism Property",
"Definition:Kernel of Ring Homomorphism",
"Subring Test",
"Definition:Subring"
] |
proofwiki-746 | Kernel of Ring Homomorphism is Ideal | The kernel of $\phi$ is an ideal of $R_1$. | By Kernel of Ring Homomorphism is Subring, $\map \ker \phi$ is a subring of $R_1$.
Let $s \in \map \ker \phi$, so $\map \phi s = 0_{R_2}$.
Suppose $x \in R_1$. Then:
{{begin-eqn}}
{{eqn | l = \map \phi {x \circ_1 s}
| r = \map \phi x \circ_2 \map \phi s
| c = {{Defof|Morphism Property}}
}}
{{eqn | r = \map ... | The [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$ is an [[Definition:Ideal of Ring|ideal]] of $R_1$. | By [[Kernel of Ring Homomorphism is Subring]], $\map \ker \phi$ is a [[Definition:Subring|subring]] of $R_1$.
Let $s \in \map \ker \phi$, so $\map \phi s = 0_{R_2}$.
Suppose $x \in R_1$. Then:
{{begin-eqn}}
{{eqn | l = \map \phi {x \circ_1 s}
| r = \map \phi x \circ_2 \map \phi s
| c = {{Defof|Morphism P... | Kernel of Ring Homomorphism is Ideal | https://proofwiki.org/wiki/Kernel_of_Ring_Homomorphism_is_Ideal | https://proofwiki.org/wiki/Kernel_of_Ring_Homomorphism_is_Ideal | [
"Ring Homomorphisms",
"Ideal Theory"
] | [
"Definition:Kernel of Ring Homomorphism",
"Definition:Ideal of Ring"
] | [
"Kernel of Ring Homomorphism is Subring",
"Definition:Subring"
] |
proofwiki-747 | Ideals of Division Ring | Let $\struct {R, +, \circ}$ be a division ring whose zero is $0_R$.
The only ideals of $\struct {R, +, \circ}$ are $\set {0_R}$ and $R$ itself.
That is, $\struct {R, +, \circ}$ has no non-null proper ideals. | From Null Ring is Ideal, $\set {0_R}$ is an ideal of $\struct {R, +, \circ}$, as $\struct {R, +, \circ}$, being a division ring, is also a ring.
By definition, every non-zero element of a division ring is a unit.
So $S \ne \set {0_R} \implies \exists x \in S: x \ne 0_R$ such that $r$ is a unit of $R$.
The result follow... | Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
The only [[Definition:Ideal of Ring|ideals]] of $\struct {R, +, \circ}$ are $\set {0_R}$ and $R$ itself.
That is, $\struct {R, +, \circ}$ has no non-[[Definition:Null Ideal|null]] [[Definition:Pr... | From [[Null Ring is Ideal]], $\set {0_R}$ is an [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$, as $\struct {R, +, \circ}$, being a [[Definition:Division Ring|division ring]], is also a [[Definition:Ring (Abstract Algebra)|ring]].
By definition, every non-[[Definition:Ring Zero|zero]] [[Definition:Eleme... | Ideals of Division Ring | https://proofwiki.org/wiki/Ideals_of_Division_Ring | https://proofwiki.org/wiki/Ideals_of_Division_Ring | [
"Ideal Theory"
] | [
"Definition:Division Ring",
"Definition:Ring Zero",
"Definition:Ideal of Ring",
"Definition:Null Ideal",
"Definition:Ideal of Ring/Proper Ideal"
] | [
"Null Ring is Ideal",
"Definition:Ideal of Ring",
"Definition:Division Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Element",
"Definition:Division Ring",
"Definition:Unit of Ring",
"Definition:Unit of Ring",
"Ideal of Unit is Whole Ring"
] |
proofwiki-748 | Quotient Ring of Kernel of Ring Epimorphism | There exists a unique ring isomorphism $g: R_1 / K \to R_2$ such that:
:$g \circ q_K = \phi$ | From the Quotient Theorem for Epimorphisms, there is one and only one isomorphism that satisfies the conditions for each of the operations on $R_1$.
Hence the result.
{{qed}} | There exists a [[Definition:Unique|unique]] [[Definition:Ring Isomorphism|ring isomorphism]] $g: R_1 / K \to R_2$ such that:
:$g \circ q_K = \phi$ | From the [[Quotient Theorem for Epimorphisms]], there is [[Definition:Unique|one and only one]] [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] that satisfies the conditions for each of the operations on $R_1$.
Hence the result.
{{qed}} | Quotient Ring of Kernel of Ring Epimorphism | https://proofwiki.org/wiki/Quotient_Ring_of_Kernel_of_Ring_Epimorphism | https://proofwiki.org/wiki/Quotient_Ring_of_Kernel_of_Ring_Epimorphism | [
"Ring Epimorphisms",
"Ring Isomorphisms",
"Quotient Rings"
] | [
"Definition:Unique",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism"
] | [
"Quotient Theorem for Epimorphisms",
"Definition:Unique",
"Definition:Isomorphism (Abstract Algebra)"
] |
proofwiki-749 | Ring Epimorphism Preserves Ideals | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $J$ be an ideal of $R_1$.
Then $\phi \sqbrk J$ is an ideal of $R_2$. | $J$ is an ideal of $R_1$, so it is also a subring of $R_1$.
From Ring Homomorphism Preserves Subrings, it follows that $\phi \sqbrk J$ is a subring of $R_2$.
Now suppose $u \in \phi \sqbrk J$.
Let $v \in R_2$.
Then $\exists x \in J, y \in R_1$ such that $\map \phi x = u, \map \phi y = v$.
Thus, by the morphism property... | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Epimorphism|ring epimorphism]].
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R_1$.
Then $\phi \sqbrk J$ is an [[Definition:Ideal of Ring|ideal]] of $R_2$. | $J$ is an [[Definition:Ideal of Ring|ideal]] of $R_1$, so it is also a [[Definition:Subring|subring]] of $R_1$.
From [[Ring Homomorphism Preserves Subrings]], it follows that $\phi \sqbrk J$ is a [[Definition:Subring|subring]] of $R_2$.
Now suppose $u \in \phi \sqbrk J$.
Let $v \in R_2$.
Then $\exists x \in J, y \... | Ring Epimorphism Preserves Ideals | https://proofwiki.org/wiki/Ring_Epimorphism_Preserves_Ideals | https://proofwiki.org/wiki/Ring_Epimorphism_Preserves_Ideals | [
"Ideal Theory",
"Ring Epimorphisms"
] | [
"Definition:Ring Epimorphism",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring"
] | [
"Definition:Ideal of Ring",
"Definition:Subring",
"Ring Homomorphism Preserves Subrings",
"Definition:Subring",
"Definition:Morphism Property"
] |
proofwiki-750 | Preimage of Image of Subring under Ring Homomorphism | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $K = \map \ker \phi$ be the kernel of $\phi$.
Let $J$ be a subring of $R_1$.
Then:
:$\phi^{-1} \sqbrk {\phi \sqbrk J} = J + K$ | Let $x \in \phi^{-1} \sqbrk {\phi \sqbrk J}$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \phi^{-1} \sqbrk {\phi \sqbrk J}
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi x
| o = \in
| r = \phi \sqbrk J
| c = {{Defof|Preimage of Element under Mapping}}
}}
{{eqn | ll= \lead... | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Let $K = \map \ker \phi$ be the [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$.
Let $J$ be a [[Definition:Subring|subring]] of $R_1$.
Then:
:$\phi^{-1} \sqbrk {\phi \sqbrk ... | Let $x \in \phi^{-1} \sqbrk {\phi \sqbrk J}$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \phi^{-1} \sqbrk {\phi \sqbrk J}
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi x
| o = \in
| r = \phi \sqbrk J
| c = {{Defof|Preimage of Element under Mapping}}
}}
{{eqn | ll= \lea... | Preimage of Image of Subring under Ring Homomorphism | https://proofwiki.org/wiki/Preimage_of_Image_of_Subring_under_Ring_Homomorphism | https://proofwiki.org/wiki/Preimage_of_Image_of_Subring_under_Ring_Homomorphism | [
"Ring Homomorphisms",
"Subrings"
] | [
"Definition:Ring Homomorphism",
"Definition:Kernel of Ring Homomorphism",
"Definition:Subring"
] | [
"Definition:Ring Homomorphism",
"Definition:Set Equality/Definition 2"
] |
proofwiki-751 | Preimage of Subring under Ring Homomorphism is Subring | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S_2$ be a subring of $R_2$.
Then $S_1 = \phi^{-1} \sqbrk {S_2}$ is a subring of $R_1$ such that $\map \ker \phi \subseteq S_1$. | Let $K = \map \ker \phi$ be the kernel of $R_1$.
We have that $0_{R_2} \in S_2$ and so $\set {0_{R_2} } \subseteq S_2$.
From Subset Maps to Subset:
:$\phi^{-1} \sqbrk {\set {0_{R_2} } } \subseteq \phi^{-1} \sqbrk {S_2} = S_1$
But by definition, $K = \phi^{-1} \sqbrk {\set {0_{R_2} } }$
and so $S_1$ is a subset of $R_1$... | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Let $S_2$ be a [[Definition:Subring|subring]] of $R_2$.
Then $S_1 = \phi^{-1} \sqbrk {S_2}$ is a [[Definition:Subring|subring]] of $R_1$ such that $\map \ker \phi \subseteq S_1$. | Let $K = \map \ker \phi$ be the [[Definition:Kernel of Ring Homomorphism|kernel]] of $R_1$.
We have that $0_{R_2} \in S_2$ and so $\set {0_{R_2} } \subseteq S_2$.
From [[Subset Maps to Subset]]:
:$\phi^{-1} \sqbrk {\set {0_{R_2} } } \subseteq \phi^{-1} \sqbrk {S_2} = S_1$
But by definition, $K = \phi^{-1} \sqbrk {\... | Preimage of Subring under Ring Homomorphism is Subring | https://proofwiki.org/wiki/Preimage_of_Subring_under_Ring_Homomorphism_is_Subring | https://proofwiki.org/wiki/Preimage_of_Subring_under_Ring_Homomorphism_is_Subring | [
"Ring Homomorphisms",
"Subrings"
] | [
"Definition:Ring Homomorphism",
"Definition:Subring",
"Definition:Subring"
] | [
"Definition:Kernel of Ring Homomorphism",
"Image of Subset under Mapping is Subset of Image",
"Definition:Subset",
"Definition:Subring",
"Definition:Ring Homomorphism",
"Definition:Subring",
"Group Homomorphism Preserves Inverses",
"Definition:Subring",
"Definition:Ring Homomorphism",
"Definition:... |
proofwiki-752 | Preimage of Ideal under Ring Homomorphism is Ideal | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S_2$ be an ideal of $R_2$.
Then $S_1 = \phi^{-1} \sqbrk {S_2}$ is an ideal of $R_1$ such that $\map \ker \phi \subseteq S_1$. | From Preimage of Subring under Ring Homomorphism is Subring we have that $S_1 = \phi^{-1} \sqbrk {S_2}$ is a subring of $R_1$ such that $\map \ker \phi \subseteq S_1$.
We now need to show that $S_1$ is an ideal of $R_1$.
Let $s_1 \in S_1, r_1 \in R_1$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {r_1 \circ_1 s_1}
|... | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Let $S_2$ be an [[Definition:Ideal of Ring|ideal]] of $R_2$.
Then $S_1 = \phi^{-1} \sqbrk {S_2}$ is an [[Definition:Ideal of Ring|ideal]] of $R_1$ such that $\map \ker \phi \subseteq S_1$. | From [[Preimage of Subring under Ring Homomorphism is Subring]] we have that $S_1 = \phi^{-1} \sqbrk {S_2}$ is a [[Definition:subring|subring]] of $R_1$ such that $\map \ker \phi \subseteq S_1$.
We now need to show that $S_1$ is an [[Definition:Ideal of Ring|ideal]] of $R_1$.
Let $s_1 \in S_1, r_1 \in R_1$.
Then:
... | Preimage of Ideal under Ring Homomorphism is Ideal | https://proofwiki.org/wiki/Preimage_of_Ideal_under_Ring_Homomorphism_is_Ideal | https://proofwiki.org/wiki/Preimage_of_Ideal_under_Ring_Homomorphism_is_Ideal | [
"Ring Homomorphisms",
"Ideal Theory"
] | [
"Definition:Ring Homomorphism",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring"
] | [
"Preimage of Subring under Ring Homomorphism is Subring",
"Definition:subring",
"Definition:Ideal of Ring",
"Definition:Ring Homomorphism",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring"
] |
proofwiki-753 | Image of Preimage of Subring under Ring Epimorphism | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $S_2$ be a subring of $R_2$.
Then:
:$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$ | As $\phi$ is an epimorphism, it is a surjection, and so:
:$\Img \phi = R_2$
So:
:$S_2 \subseteq \Img {R_1}$
The result then follows from Image of Preimage under Mapping.
{{qed}} | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Epimorphism|ring epimorphism]].
Let $S_2$ be a [[Definition:Subring|subring]] of $R_2$.
Then:
:$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$ | As $\phi$ is an [[Definition:Ring Epimorphism|epimorphism]], it is a [[Definition:Surjection|surjection]], and so:
:$\Img \phi = R_2$
So:
:$S_2 \subseteq \Img {R_1}$
The result then follows from [[Image of Preimage under Mapping]].
{{qed}} | Image of Preimage of Subring under Ring Epimorphism | https://proofwiki.org/wiki/Image_of_Preimage_of_Subring_under_Ring_Epimorphism | https://proofwiki.org/wiki/Image_of_Preimage_of_Subring_under_Ring_Epimorphism | [
"Ring Epimorphisms",
"Subrings"
] | [
"Definition:Ring Epimorphism",
"Definition:Subring"
] | [
"Definition:Ring Epimorphism",
"Definition:Surjection",
"Image of Preimage under Mapping"
] |
proofwiki-754 | Diagonal Relation is Universally Congruent | The diagonal relation $\Delta_S$ on a set $S$ is universally congruent on $S$. | We have that the diagonal relation is an equivalence relation.
Let $\struct {S, \circ}$ be any algebraic structure.
{{begin-eqn}}
{{eqn | o =
| r = x_1 \mathrel {\Delta_S} x_2 \land y_1 \mathrel {\Delta_S} y_2
| c =
}}
{{eqn | o = \leadsto
| r = x_1 = x_2 \land y_1 = y_2
| c = {{Defof|Diagonal... | The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ on a [[Definition:Set|set]] $S$ is [[Definition:Universally Congruent|universally congruent]] on $S$. | We have that the [[Diagonal Relation is Equivalence|diagonal relation is an equivalence relation]].
Let $\struct {S, \circ}$ be any [[Definition:Algebraic Structure with One Operation|algebraic structure]].
{{begin-eqn}}
{{eqn | o =
| r = x_1 \mathrel {\Delta_S} x_2 \land y_1 \mathrel {\Delta_S} y_2
| c... | Diagonal Relation is Universally Congruent | https://proofwiki.org/wiki/Diagonal_Relation_is_Universally_Congruent | https://proofwiki.org/wiki/Diagonal_Relation_is_Universally_Congruent | [
"Examples of Congruence Relations"
] | [
"Definition:Diagonal Relation",
"Definition:Set",
"Definition:Universally Congruent"
] | [
"Diagonal Relation is Equivalence",
"Definition:Algebraic Structure/One Operation",
"Definition:Universally Congruent"
] |
proofwiki-755 | Positive Elements of Ordered Ring | Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $P$ be the set of positive elements of $R$ , that is, $P = R_{\ge 0}$.
Then:
: $(1): \quad P + P \subseteq P$
: $(2): \quad P \cap \paren {-P} = \set {0_R}$
: $(3): \quad P \circ P \subseteq P$
If $\le$ is a... | === Necessary Condition ===
First, suppose that $\le$ is compatible with the ring structure of $R$.
Hence:
:$(OR1): \quad \le$ is compatible with $+$
:$(OR2): \quad \forall x, y \in R: 0_R \le x, 0_R \le y \implies 0_R \le x \circ y$.
$(1)$: Let $x, y \in R: 0_R \le x, 0_R \le y$.
Then $0_R + 0_R \le x + y$ by the fact... | Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]] [[Definition:Ring with Unity|with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $P$ be the set of [[Definition:Positive|positive]] elements of $R$ , that is, $P = R_{\ge 0}$... | === Necessary Condition ===
First, suppose that $\le$ is [[Definition:Ordering Compatible with Ring Structure|compatible with the ring structure]] of $R$.
Hence:
:$(OR1): \quad \le$ is [[Definition:Relation Compatible with Operation|compatible]] with $+$
:$(OR2): \quad \forall x, y \in R: 0_R \le x, 0_R \le y \implie... | Positive Elements of Ordered Ring | https://proofwiki.org/wiki/Positive_Elements_of_Ordered_Ring | https://proofwiki.org/wiki/Positive_Elements_of_Ordered_Ring | [
"Ordered Rings"
] | [
"Definition:Ordered Ring",
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Positive",
"Definition:Total Ordering",
"Definition:Totally Ordered Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Ordering",
"Definition:Ordering Comp... | [
"Definition:Ordering Compatible with Ring Structure",
"Definition:Relation Compatible with Operation",
"Definition:Relation Compatible with Operation",
"Properties of Ordered Ring",
"Definition:Ordering",
"Definition:Ordering Compatible with Ring Structure",
"Definition:Total Ordering",
"Properties of... |
proofwiki-756 | Symmetric Difference with Intersection forms Ring | Let $S$ be a set.
Let:
:$\symdif$ denote the symmetric difference operation
:$\cap$ denote the set intersection operation
:$\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cap}$ is a commutative ring with unity, in which the unity is $S$.
This ring is not an integral domain. | From Symmetric Difference on Power Set forms Abelian Group, $\struct {\powerset S, \symdif}$ is an abelian group, where $\O$ is the identity and each element is self-inverse.
From Power Set with Intersection is Monoid, $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$.
Also Intersection Distri... | Let $S$ be a [[Definition:Set|set]].
Let:
:$\symdif$ denote the [[Definition:Symmetric Difference|symmetric difference operation]]
:$\cap$ denote the [[Definition:Set Intersection|set intersection operation]]
:$\powerset S$ denote the [[Definition:Power Set|power set]] of $S$.
Then $\struct {\powerset S, \symdif, \c... | From [[Symmetric Difference on Power Set forms Abelian Group]], $\struct {\powerset S, \symdif}$ is an [[Definition:Abelian Group|abelian group]], where $\O$ is the [[Definition:Identity Element|identity]] and each [[Definition:Element|element]] is [[Definition:Self-Inverse Element|self-inverse]].
From [[Power Set wit... | Symmetric Difference with Intersection forms Ring/Proof 1 | https://proofwiki.org/wiki/Symmetric_Difference_with_Intersection_forms_Ring | https://proofwiki.org/wiki/Symmetric_Difference_with_Intersection_forms_Ring/Proof_1 | [
"Commutative Algebra",
"Set Intersection",
"Symmetric Difference",
"Power Set",
"Symmetric Difference with Intersection forms Ring",
"Examples of Commutative and Unitary Rings"
] | [
"Definition:Set",
"Definition:Symmetric Difference",
"Definition:Set Intersection",
"Definition:Power Set",
"Definition:Commutative and Unitary Ring",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Integral Domain"
] | [
"Symmetric Difference on Power Set forms Abelian Group",
"Definition:Abelian Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Element",
"Definition:Self-Inverse Element",
"Power Set with Intersection is Commutative Monoid",
"Definition:Commutative Monoid",
"Definition:I... |
proofwiki-757 | Symmetric Difference with Intersection forms Ring | Let $S$ be a set.
Let:
:$\symdif$ denote the symmetric difference operation
:$\cap$ denote the set intersection operation
:$\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cap}$ is a commutative ring with unity, in which the unity is $S$.
This ring is not an integral domain. | From Power Set is Closed under Symmetric Difference and Power Set is Closed under Intersection, we have that both $\struct {\powerset S, \symdif}$ and $\struct {\powerset S, \cap}$ are closed.
Hence $\powerset S$ is a ring of sets, and hence a commutative ring.
From Intersection with Subset is Subset, we have $A \subse... | Let $S$ be a [[Definition:Set|set]].
Let:
:$\symdif$ denote the [[Definition:Symmetric Difference|symmetric difference operation]]
:$\cap$ denote the [[Definition:Set Intersection|set intersection operation]]
:$\powerset S$ denote the [[Definition:Power Set|power set]] of $S$.
Then $\struct {\powerset S, \symdif, \c... | From [[Power Set is Closed under Symmetric Difference]] and [[Power Set is Closed under Intersection]], we have that both $\struct {\powerset S, \symdif}$ and $\struct {\powerset S, \cap}$ are [[Definition:Closed Algebraic Structure|closed]].
Hence $\powerset S$ is a [[Definition:Ring of Sets|ring of sets]], and hence... | Symmetric Difference with Intersection forms Ring/Proof 2 | https://proofwiki.org/wiki/Symmetric_Difference_with_Intersection_forms_Ring | https://proofwiki.org/wiki/Symmetric_Difference_with_Intersection_forms_Ring/Proof_2 | [
"Commutative Algebra",
"Set Intersection",
"Symmetric Difference",
"Power Set",
"Symmetric Difference with Intersection forms Ring",
"Examples of Commutative and Unitary Rings"
] | [
"Definition:Set",
"Definition:Symmetric Difference",
"Definition:Set Intersection",
"Definition:Power Set",
"Definition:Commutative and Unitary Ring",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Integral Domain"
] | [
"Power Set is Closed under Symmetric Difference",
"Power Set is Closed under Intersection",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Ring of Sets",
"Ring of Sets is Commutative Ring",
"Intersection with Subset is Subset",
"Definition:Unity (Abstract Algebra)/Ring",
"Pow... |
proofwiki-758 | Field of Quotients of Subdomain | Let $\struct {F, +, \circ}$ be a field whose unity is $1_F$.
Let $\struct {D, +, \circ}$ be a subdomain of $\struct {F, +, \circ}$ whose unity is $1_D$.
Let:
:$K = \set {\dfrac x y: x \in D, y \in D^*}$
where $\dfrac x y$ is the division product of $x$ by $y$.
Then $\struct {K, +, \circ}$ is a field of quotients of $\s... | $1_D = 1_F$ by Subdomain Test.
The sum and product of two elements of $K$ are also in $K$ by Addition of Division Products and Product of Division Products.
The additive and product inverses of $K$ are also in $K$ by Negative of Division Product and Inverse of Division Product.
Thus by Subfield Test, $\struct {K, +, \c... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Unity of Ring|unity]] is $1_F$.
Let $\struct {D, +, \circ}$ be a [[Definition:Subdomain|subdomain]] of $\struct {F, +, \circ}$ whose [[Definition:Unity of Ring|unity]] is $1_D$.
Let:
:$K = \set {\dfrac x y: x \in D, y \... | $1_D = 1_F$ by [[Subdomain Test]].
The sum and product of two elements of $K$ are also in $K$ by [[Addition of Division Products]] and [[Product of Division Products]].
The additive and product inverses of $K$ are also in $K$ by [[Negative of Division Product]] and [[Inverse of Division Product]].
Thus by [[Subfield... | Field of Quotients of Subdomain | https://proofwiki.org/wiki/Field_of_Quotients_of_Subdomain | https://proofwiki.org/wiki/Field_of_Quotients_of_Subdomain | [
"Fields of Quotients"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Subdomain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Division Product",
"Definition:Field of Quotients"
] | [
"Subdomain Test",
"Addition of Division Products",
"Product of Division Products",
"Negative of Division Product",
"Inverse of Division Product",
"Subfield Test",
"Definition:Subfield"
] |
proofwiki-759 | Existence of Field of Quotients | Let $\struct {D, +, \circ}$ be an integral domain.
Then there exists a field of quotients of $\struct {D, +, \circ}$. | Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$. | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
Then there exists a [[Definition:Field of Quotients|field of quotients]] of $\struct {D, +, \circ}$. | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$. | Existence of Field of Quotients | https://proofwiki.org/wiki/Existence_of_Field_of_Quotients | https://proofwiki.org/wiki/Existence_of_Field_of_Quotients | [
"Fields of Quotients",
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Field of Quotients"
] | [
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Integral Domain"
] |
proofwiki-760 | Quotient Theorem for Monomorphisms | Let $K, L$ be fields of quotients of integral domains $\struct {R, +_R, \circ_R}, \struct {S, +_S, \circ_S}$ respectively.
Let $\phi: R \to S$ be a monomorphism.
Then there is one and only one monomorphism $\psi: K \to L$ extending $\phi$, and:
:$\forall x \in R, y \in R^*: \map \psi {\dfrac x y} = \dfrac {\map \phi x}... | By definition, $\struct {K, \circ_R}$ and $\struct {L, \circ_S}$ are inverse completions of $\struct {R, \circ_R}$ and $\struct {S, \circ_S}$ respectively.
{{Questionable|this is not the definition}}
So by the Extension Theorem for Homomorphisms, there is one and only one monomorphism $\psi: \struct {K, \circ_R} \to \s... | Let $K, L$ be [[Definition:Field of Quotients|fields of quotients]] of [[Definition:Integral Domain|integral domains]] $\struct {R, +_R, \circ_R}, \struct {S, +_S, \circ_S}$ respectively.
Let $\phi: R \to S$ be a [[Definition:Ring Monomorphism|monomorphism]].
Then there is one and only one [[Definition:Ring Monomorp... | By definition, $\struct {K, \circ_R}$ and $\struct {L, \circ_S}$ are [[Definition:Inverse Completion|inverse completions]] of $\struct {R, \circ_R}$ and $\struct {S, \circ_S}$ respectively.
{{Questionable|this is not the definition}}
So by the [[Extension Theorem for Homomorphisms]], there is one and only one [[Defin... | Quotient Theorem for Monomorphisms | https://proofwiki.org/wiki/Quotient_Theorem_for_Monomorphisms | https://proofwiki.org/wiki/Quotient_Theorem_for_Monomorphisms | [
"Fields of Quotients",
"Integral Domains",
"Ring Monomorphisms",
"Named Theorems"
] | [
"Definition:Field of Quotients",
"Definition:Integral Domain",
"Definition:Ring Monomorphism",
"Definition:Ring Monomorphism",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism"
] | [
"Definition:Inverse Completion",
"Extension Theorem for Homomorphisms",
"Definition:Ring Monomorphism",
"Extension Theorem for Isomorphisms",
"Definition:Isomorphism (Abstract Algebra)",
"Addition of Division Products",
"Definition:Morphism Property",
"Addition of Division Products",
"Definition:Rin... |
proofwiki-761 | Field of Quotients is Unique | Let $\struct {D, +, \circ}$ be an integral domain.
Let $K, L$ be field of quotients of $\struct {D, +, \circ}$.
Then there is one and only one (field) isomorphism $\phi: K \to L$ satisfying:
:$\forall x \in D: \map \phi x = x$ | Follows directly from the Quotient Theorem for Monomorphisms.
{{qed}} | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
Let $K, L$ be [[Definition:Field of Quotients|field of quotients]] of $\struct {D, +, \circ}$.
Then there is [[Definition:Exactly One|one and only one]] [[Definition:Field Isomorphism|(field) isomorphism]] $\phi: K \to L$ satisfying:
... | Follows directly from the [[Quotient Theorem for Monomorphisms]].
{{qed}} | Field of Quotients is Unique | https://proofwiki.org/wiki/Field_of_Quotients_is_Unique | https://proofwiki.org/wiki/Field_of_Quotients_is_Unique | [
"Fields of Quotients",
"Field Isomorphisms",
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Field of Quotients",
"Definition:Unique",
"Definition:Isomorphism (Abstract Algebra)/Field Isomorphism"
] | [
"Quotient Theorem for Monomorphisms"
] |
proofwiki-762 | Divided by Positive Element of Field of Quotients | Let $\struct {K, +, \circ}$ be the field of quotients of a totally ordered integral domain $\struct {D, +, \circ, \le}$.
Then:
:$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{>0}$ | By definition of field of quotients:
:$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{\ne 0}$
Suppose $z = x' / y'$ such that $y' \notin D_{>0}$.
Then $y' < 0$ as $D$ is totally ordered.
Then:
{{begin-eqn}}
{{eqn | l = x' / y'
| r = x' \circ \paren {y'}^{-1}
| c = {{Defof|Division over Field}... | Let $\struct {K, +, \circ}$ be the [[Definition:Field of Quotients|field of quotients]] of a [[Definition:Totally Ordered Ring|totally ordered]] [[Definition:Integral Domain|integral domain]] $\struct {D, +, \circ, \le}$.
Then:
:$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{>0}$ | By definition of [[Definition:Field of Quotients|field of quotients]]:
:$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{\ne 0}$
Suppose $z = x' / y'$ such that $y' \notin D_{>0}$.
Then $y' < 0$ as $D$ is [[Definition:Totally Ordered Ring|totally ordered]].
Then:
{{begin-eqn}}
{{eqn | l = x' / y'
... | Divided by Positive Element of Field of Quotients | https://proofwiki.org/wiki/Divided_by_Positive_Element_of_Field_of_Quotients | https://proofwiki.org/wiki/Divided_by_Positive_Element_of_Field_of_Quotients | [
"Fields of Quotients",
"Integral Domains"
] | [
"Definition:Field of Quotients",
"Definition:Totally Ordered Ring",
"Definition:Integral Domain"
] | [
"Definition:Field of Quotients",
"Definition:Totally Ordered Ring",
"Product of Ring Negatives",
"Negative of Product Inverse",
"Properties of Ordered Ring",
"Category:Fields of Quotients",
"Category:Integral Domains"
] |
proofwiki-763 | Total Ordering on Field of Quotients is Unique | Let $\struct {K, +, \circ}$ be a field of quotients of an ordered integral domain $\struct {D, +, \circ, \le}$.
Then there is one and only one total ordering $\le'$ on $K$ which is compatible with its ring structure and induces on $D$ its given total ordering $\le$.
That ordering is the one defined by:
:$P = \set {\dfr... | First, note that from Divided by Positive Element of Field of Quotients:
:$\forall z \in K: \exists x, y \in R: z = \dfrac x y, y \in R_+^*$
Now we show that $P$ satistfies conditions $(1)$ to $(4)$ of Positive Elements of Ordered Ring.
From Addition of Division Products and Product of Division Products, it is clear th... | Let $\struct {K, +, \circ}$ be a [[Definition:Field of Quotients|field of quotients]] of an [[Definition:Ordered Integral Domain|ordered integral domain]] $\struct {D, +, \circ, \le}$.
Then there is one and only one [[Definition:Total Ordering|total ordering]] $\le'$ on $K$ which is [[Definition:Ordering Compatible w... | First, note that from [[Divided by Positive Element of Field of Quotients]]:
:$\forall z \in K: \exists x, y \in R: z = \dfrac x y, y \in R_+^*$
Now we show that $P$ satistfies conditions $(1)$ to $(4)$ of [[Positive Elements of Ordered Ring]].
From [[Addition of Division Products]] and [[Product of Division Produ... | Total Ordering on Field of Quotients is Unique | https://proofwiki.org/wiki/Total_Ordering_on_Field_of_Quotients_is_Unique | https://proofwiki.org/wiki/Total_Ordering_on_Field_of_Quotients_is_Unique | [
"Fields of Quotients",
"Integral Domains"
] | [
"Definition:Field of Quotients",
"Definition:Ordered Integral Domain",
"Definition:Total Ordering",
"Definition:Ordering Compatible with Ring Structure",
"Definition:Total Ordering"
] | [
"Divided by Positive Element of Field of Quotients",
"Positive Elements of Ordered Ring",
"Addition of Division Products",
"Product of Division Products",
"Positive Elements of Ordered Ring",
"Definition:Total Ordering",
"Definition:Ordering Compatible with Ring Structure",
"Intersection is Largest Su... |
proofwiki-764 | Order Embedding between Quotient Fields is Unique | Let $\struct {R_1, +_1, \circ_1, \le_1}$ and $\struct {S, +_2, \circ_2, \le_2}$ be totally ordered integral domains.
Let $K, L$ be totally ordered fields of quotients of $\struct {R_1, +_1, \circ_1, \le_1}$ and $\struct {S, +_2, \circ_2, \le_2}$ respectively.
Let $\phi: R \to S$ be a order embedding.
Then there exists ... | By Field of Quotients is Unique, all we need to show is:
:$\forall x_1, x_2 \in R, y_1, y_2 \in R_{> 0}: \dfrac {x_1} {y_1} \le \dfrac {x_2} {y_2} \iff \dfrac {\map \phi {x_1}} {\map \phi {y_1} } \le \dfrac {\map \phi {x_2} } {\map \phi {y_2} }$
Let $x_1 / y_1 \le x_2 / y_2$, where $y_1, y_2 \in R_{> 0}$.
As $y_1, y_2 ... | Let $\struct {R_1, +_1, \circ_1, \le_1}$ and $\struct {S, +_2, \circ_2, \le_2}$ be [[Definition:Totally Ordered Ring|totally ordered]] [[Definition:Integral Domain|integral domains]].
Let $K, L$ be [[Definition:Totally Ordered Ring|totally ordered]] [[Definition:Field of Quotients|fields of quotients]] of $\struct {R_... | By [[Field of Quotients is Unique]], all we need to show is:
:$\forall x_1, x_2 \in R, y_1, y_2 \in R_{> 0}: \dfrac {x_1} {y_1} \le \dfrac {x_2} {y_2} \iff \dfrac {\map \phi {x_1}} {\map \phi {y_1} } \le \dfrac {\map \phi {x_2} } {\map \phi {y_2} }$
Let $x_1 / y_1 \le x_2 / y_2$, where $y_1, y_2 \in R_{> 0}$.
As $y... | Order Embedding between Quotient Fields is Unique | https://proofwiki.org/wiki/Order_Embedding_between_Quotient_Fields_is_Unique | https://proofwiki.org/wiki/Order_Embedding_between_Quotient_Fields_is_Unique | [
"Ordered Rings",
"Fields of Quotients",
"Total Orderings",
"Order Embeddings"
] | [
"Definition:Totally Ordered Ring",
"Definition:Integral Domain",
"Definition:Totally Ordered Ring",
"Definition:Field of Quotients",
"Definition:Order Embedding",
"Definition:Unique",
"Definition:Order Embedding",
"Definition:Order Isomorphism"
] | [
"Field of Quotients is Unique",
"Definition:Order Embedding"
] |
proofwiki-765 | Strict Ordering Preserved under Product with Cancellable Element | Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $x, y, z \in S$ be such that:
:$(1): \quad z$ is cancellable for $\circ$
:$(2): \quad x \prec y$
Then:
:$x \circ z \prec y \circ z$
:$z \circ x \prec z \circ y$ | Let $z$ be cancellable and $x \prec y$.
Then by the definition of ordered semigroup:
:$x \circ z \preceq y \circ z$
From the fact that $z$ is cancellable:
:$x \circ z = y \circ z \iff x = y$
Thus as $x \circ z \ne y \circ z$ it follows from Strictly Precedes is Strict Ordering that:
:$x \circ z \prec y \circ z$
Similar... | Let $\struct {S, \circ, \preceq}$ be an [[Definition:Ordered Semigroup|ordered semigroup]].
Let $x, y, z \in S$ be such that:
:$(1): \quad z$ is [[Definition:Cancellable Element|cancellable]] for $\circ$
:$(2): \quad x \prec y$
Then:
:$x \circ z \prec y \circ z$
:$z \circ x \prec z \circ y$ | Let $z$ be [[Definition:Cancellable Element|cancellable]] and $x \prec y$.
Then by the definition of [[Definition:Ordered Semigroup|ordered semigroup]]:
:$x \circ z \preceq y \circ z$
From the fact that $z$ is [[Definition:Cancellable Element|cancellable]]:
:$x \circ z = y \circ z \iff x = y$
Thus as $x \circ z \ne ... | Strict Ordering Preserved under Product with Cancellable Element | https://proofwiki.org/wiki/Strict_Ordering_Preserved_under_Product_with_Cancellable_Element | https://proofwiki.org/wiki/Strict_Ordering_Preserved_under_Product_with_Cancellable_Element | [
"Ordered Semigroups"
] | [
"Definition:Ordered Semigroup",
"Definition:Cancellable Element"
] | [
"Definition:Cancellable Element",
"Definition:Ordered Semigroup",
"Definition:Cancellable Element",
"Strictly Precedes is Strict Ordering"
] |
proofwiki-766 | Ordering of Inverses in Ordered Monoid | Let $\struct {S, \circ, \preceq}$ be an ordered monoid whose identity is $e$.
Let $x, y \in S$ be invertible.
Then:
:$x \prec y \iff y^{-1} \prec x^{-1}$ | === Necessary Condition ===
{{begin-eqn}}
{{eqn | l = x
| o = \prec
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = e
| r = x^{-1} \circ x \prec x^{-1} \circ y
| c = Strict Ordering Preserved under Product with Cancellable Element
}}
{{eqn | ll= \leadsto
| l = y^{-1}
| r = e... | Let $\struct {S, \circ, \preceq}$ be an [[Definition:Ordered Monoid|ordered monoid]] whose [[Definition:Identity Element|identity]] is $e$.
Let $x, y \in S$ be [[Definition:Invertible Element|invertible]].
Then:
:$x \prec y \iff y^{-1} \prec x^{-1}$ | === Necessary Condition ===
{{begin-eqn}}
{{eqn | l = x
| o = \prec
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = e
| r = x^{-1} \circ x \prec x^{-1} \circ y
| c = [[Strict Ordering Preserved under Product with Cancellable Element]]
}}
{{eqn | ll= \leadsto
| l = y^{-1}
| ... | Ordering of Inverses in Ordered Monoid | https://proofwiki.org/wiki/Ordering_of_Inverses_in_Ordered_Monoid | https://proofwiki.org/wiki/Ordering_of_Inverses_in_Ordered_Monoid | [
"Order Theory",
"Monoids"
] | [
"Definition:Ordered Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Invertible Element"
] | [
"Strict Ordering Preserved under Product with Cancellable Element"
] |
proofwiki-767 | Monomorphism from Total Ordering | Let the following conditions hold:
:$(1): \quad$ Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.
:$(2): \quad$ Let $\phi: S \to T$ be a mapping.
:$(3): \quad$ Let $\preceq$ be a total ordering on $S$.
Then $\phi \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ is... | This follows:
:$(1): \quad$ As a direct consequence of Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing
:$(2): \quad$ From the definition of monomorphism as a homomorphism which is an injection.
{{Qed}} | Let the following conditions hold:
:$(1): \quad$ Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be [[Definition:Ordered Semigroup|ordered semigroups]].
:$(2): \quad$ Let $\phi: S \to T$ be a [[Definition:Mapping|mapping]].
:$(3): \quad$ Let $\preceq$ be a [[Definition:Total Ordering|total orderin... | This follows:
:$(1): \quad$ As a direct consequence of [[Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing]]
:$(2): \quad$ From the definition of [[Definition:Monomorphism (Abstract Algebra)|monomorphism]] as a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is an [[Definitio... | Monomorphism from Total Ordering | https://proofwiki.org/wiki/Monomorphism_from_Total_Ordering | https://proofwiki.org/wiki/Monomorphism_from_Total_Ordering | [
"Total Orderings",
"Monomorphisms (Abstract Algebra)"
] | [
"Definition:Ordered Semigroup",
"Definition:Mapping",
"Definition:Total Ordering",
"Definition:Ordered Structure Monomorphism",
"Definition:Strictly Increasing",
"Definition:Homomorphism (Abstract Algebra)"
] | [
"Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing",
"Definition:Monomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Injection"
] |
proofwiki-768 | Extension Theorem for Total Orderings | Let the following conditions be fulfilled:
:$(1):\quad$ Let $\struct {S, \circ, \preceq}$ be a totally ordered commutative semigroup
:$(2):\quad$ Let all the elements of $\struct {S, \circ, \preceq}$ be cancellable
:$(3):\quad$ Let $\struct {T, \circ}$ be an inverse completion of $\struct {S, \circ}$.
Then:
:$(1):\quad... | By Inverse Completion is Commutative Semigroup:
:every element of $T$ is of the form $x \circ y^{-1}$ where $x, y \in S$. | Let the following conditions be fulfilled:
:$(1):\quad$ Let $\struct {S, \circ, \preceq}$ be a [[Definition:Totally Ordered Commutative Semigroup|totally ordered commutative semigroup]]
:$(2):\quad$ Let all the [[Definition:Element|elements]] of $\struct {S, \circ, \preceq}$ be [[Definition:Cancellable Element|cancel... | By [[Inverse Completion is Commutative Semigroup]]:
:every [[Definition:Element|element]] of $T$ is of the form $x \circ y^{-1}$ where $x, y \in S$. | Extension Theorem for Total Orderings | https://proofwiki.org/wiki/Extension_Theorem_for_Total_Orderings | https://proofwiki.org/wiki/Extension_Theorem_for_Total_Orderings | [
"Total Orderings",
"Semigroups",
"Named Theorems"
] | [
"Definition:Totally Ordered Commutative Semigroup",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:Inverse Completion",
"Definition:Relation",
"Definition:Well-Defined/Relation",
"Definition:Total Ordering",
"Definition:Relation Compatible with Operation",
"Definition:Total Orde... | [
"Inverse Completion is Commutative Semigroup",
"Definition:Element",
"Definition:Element"
] |
proofwiki-769 | Strict Lower Closure of Sum with One | Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Then:
:$\forall n \in \struct {S, \circ, \preceq}: \paren {n \circ 1}^\prec = n^\prec \cup \set n$
where $n^\prec$ is defined as the strict lower closure of $n$, that is, the set of elements strictly preceding $n$. | First note that as $\struct {S, \circ, \preceq}$ is well-ordered and hence totally ordered, the Trichotomy Law applies.
Thus:
{{begin-eqn}}
{{eqn | q = \forall m \in S
| o =
| r = m \notin n^\prec
| c =
}}
{{eqn | o = \leadstoandfrom
| r = \neg \ m \prec n
| c = {{Defof|Strict Lower Clos... | Let $\struct {S, \circ, \preceq}$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
Then:
:$\forall n \in \struct {S, \circ, \preceq}: \paren {n \circ 1}^\prec = n^\prec \cup \set n$
where $n^\prec$ is defined as the [[Definition:Strict Lower Closure of Element|strict lower closure]] of $n... | First note that as $\struct {S, \circ, \preceq}$ is [[Definition:Naturally Ordered Semigroup|well-ordered]] and hence [[Definition:Totally Ordered Set|totally ordered]], the [[Trichotomy Law (Ordering)|Trichotomy Law]] applies.
Thus:
{{begin-eqn}}
{{eqn | q = \forall m \in S
| o =
| r = m \notin n^\prec... | Strict Lower Closure of Sum with One | https://proofwiki.org/wiki/Strict_Lower_Closure_of_Sum_with_One | https://proofwiki.org/wiki/Strict_Lower_Closure_of_Sum_with_One | [
"Naturally Ordered Semigroup"
] | [
"Definition:Naturally Ordered Semigroup",
"Definition:Strict Lower Closure/Element",
"Definition:Strictly Precede"
] | [
"Definition:Naturally Ordered Semigroup",
"Definition:Totally Ordered Set",
"Trichotomy Law (Ordering)",
"Trichotomy Law (Ordering)",
"Sum with One is Immediate Successor in Naturally Ordered Semigroup",
"Definition:Relative Complement",
"Relative Complement of Relative Complement",
"Relative Compleme... |
proofwiki-770 | Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor | Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Then:
:$\forall m, n \in \struct {S, \circ, \preceq}: m \preceq n \implies \closedint m {n \circ 1} = \closedint m n \cup \set {n \circ 1}$
where $\closedint m n$ is the closed interval between $m$ and $n$. | Let $m \preceq n$. Then:
{{begin-eqn}}
{{eqn | o =
| r = x \in \closedint m {n \circ 1}
| c =
}}
{{eqn | o = \leadstoandfrom
| r = m \preceq x \land x \preceq \paren {n \circ 1}
| c = {{Defof|Closed Interval}}
}}
{{eqn | o = \leadstoandfrom
| r = m \preceq x \land \paren {x \prec n \circ... | Let $\struct {S, \circ, \preceq}$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
Then:
:$\forall m, n \in \struct {S, \circ, \preceq}: m \preceq n \implies \closedint m {n \circ 1} = \closedint m n \cup \set {n \circ 1}$
where $\closedint m n$ is the [[Definition:Closed Interval|closed i... | Let $m \preceq n$. Then:
{{begin-eqn}}
{{eqn | o =
| r = x \in \closedint m {n \circ 1}
| c =
}}
{{eqn | o = \leadstoandfrom
| r = m \preceq x \land x \preceq \paren {n \circ 1}
| c = {{Defof|Closed Interval}}
}}
{{eqn | o = \leadstoandfrom
| r = m \preceq x \land \paren {x \prec n \cir... | Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor | https://proofwiki.org/wiki/Closed_Interval_of_Naturally_Ordered_Semigroup_with_Successor_equals_Union_with_Successor | https://proofwiki.org/wiki/Closed_Interval_of_Naturally_Ordered_Semigroup_with_Successor_equals_Union_with_Successor | [
"Naturally Ordered Semigroup"
] | [
"Definition:Naturally Ordered Semigroup",
"Definition:Interval/Ordered Set/Closed"
] | [] |
proofwiki-771 | Principle of Induction applied to Interval of Naturally Ordered Semigroup | Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Let $\closedint p q$ be a closed interval of $\struct {S, \circ, \preceq}$.
Let $T \subseteq \closedint p q$ such that the minimal element of $\closedint p q$ is in $T$.
Let:
:$x \in T: x \prec q \implies x \circ 1 \in T$
Then:
:$T = \closedint p q$ | Let $T' = T \cup \set {x \in S: q \prec x}$.
Then $T'$ satisfies the conditions of the Principle of Mathematical Induction.
Therefore:
:$T' = \set {x \in S: p \preceq x}$
Therefore:
:$T = \closedint p q$
{{Qed}} | Let $\struct {S, \circ, \preceq}$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
Let $\closedint p q$ be a [[Definition:Closed Interval|closed interval]] of $\struct {S, \circ, \preceq}$.
Let $T \subseteq \closedint p q$ such that the [[Definition:Minimal Element|minimal element]] of $\c... | Let $T' = T \cup \set {x \in S: q \prec x}$.
Then $T'$ satisfies the conditions of the [[Principle of Mathematical Induction for Naturally Ordered Semigroup|Principle of Mathematical Induction]].
Therefore:
:$T' = \set {x \in S: p \preceq x}$
Therefore:
:$T = \closedint p q$
{{Qed}} | Principle of Induction applied to Interval of Naturally Ordered Semigroup | https://proofwiki.org/wiki/Principle_of_Induction_applied_to_Interval_of_Naturally_Ordered_Semigroup | https://proofwiki.org/wiki/Principle_of_Induction_applied_to_Interval_of_Naturally_Ordered_Semigroup | [
"Naturally Ordered Semigroup"
] | [
"Definition:Naturally Ordered Semigroup",
"Definition:Interval/Ordered Set/Closed",
"Definition:Minimal/Element"
] | [
"Principle of Mathematical Induction/Naturally Ordered Semigroup"
] |
proofwiki-772 | Naturally Ordered Semigroup is Unique | Let $\struct {S, \circ, \preceq}$ and $\struct {S', \circ', \preceq'}$ be naturally ordered semigroups.
Let:
:$0'$ be the smallest element of $S'$
:$1'$ be the smallest element of $S' \setminus \set {0'} = S'^*$.
Then the mapping $g: S \to S'$ defined as:
:$\forall a \in S: \map g a = \circ'^a 1'$
is an isomorphism fro... | === Proof that Mapping is Isomorphism ===
{{:Naturally Ordered Semigroup is Unique/Existence of Isomorphism}}{{Qed|lemma}} | Let $\struct {S, \circ, \preceq}$ and $\struct {S', \circ', \preceq'}$ be [[Definition:Naturally Ordered Semigroup|naturally ordered semigroups]].
Let:
:$0'$ be the [[Definition:Smallest Element|smallest element]] of $S'$
:$1'$ be the [[Definition:Smallest Element|smallest element]] of $S' \setminus \set {0'} = S'^*$... | === [[Naturally Ordered Semigroup is Unique/Existence of Isomorphism|Proof that Mapping is Isomorphism]] ===
{{:Naturally Ordered Semigroup is Unique/Existence of Isomorphism}}{{Qed|lemma}} | Naturally Ordered Semigroup is Unique | https://proofwiki.org/wiki/Naturally_Ordered_Semigroup_is_Unique | https://proofwiki.org/wiki/Naturally_Ordered_Semigroup_is_Unique | [
"Naturally Ordered Semigroup is Unique",
"Naturally Ordered Semigroup",
"Isomorphisms (Abstract Algebra)"
] | [
"Definition:Naturally Ordered Semigroup",
"Definition:Smallest Element",
"Definition:Smallest Element",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Naturally Ordered Semigroup"
... | [
"Naturally Ordered Semigroup is Unique/Existence of Isomorphism"
] |
proofwiki-773 | Consecutive Subsets of N | Let $\N_k$ denote the initial segment of the natural numbers determined by $k$:
:$\N_k = \left\{{0, 1, 2, 3, \ldots, k - 1}\right\}$
Then:
:$\N_k = \N_{k + 1} \setminus \left\{{k}\right\}$
In particular:
:$\N_{k - 1} = \N_k \setminus \left\{{k - 1}\right\}$ | The result follows as a direct application of Strict Lower Closure of Sum with One.
{{qed}}
Category:Natural Numbers
ebnlj9h3zjmkiqhic6nj6grj7lm6oxj | Let $\N_k$ denote the [[Definition:Initial Segment of Natural Numbers|initial segment of the natural numbers]] determined by $k$:
:$\N_k = \left\{{0, 1, 2, 3, \ldots, k - 1}\right\}$
Then:
:$\N_k = \N_{k + 1} \setminus \left\{{k}\right\}$
In particular:
:$\N_{k - 1} = \N_k \setminus \left\{{k - 1}\right\}$ | The result follows as a direct application of [[Strict Lower Closure of Sum with One]].
{{qed}}
[[Category:Natural Numbers]]
ebnlj9h3zjmkiqhic6nj6grj7lm6oxj | Consecutive Subsets of N | https://proofwiki.org/wiki/Consecutive_Subsets_of_N | https://proofwiki.org/wiki/Consecutive_Subsets_of_N | [
"Natural Numbers"
] | [
"Definition:Initial Segment of Natural Numbers"
] | [
"Strict Lower Closure of Sum with One",
"Category:Natural Numbers"
] |
proofwiki-774 | Well-Ordering Principle | Every non-empty subset of $\N$ has a smallest (or '''first''') element.
That is, the relational structure $\struct {\N, \le}$ on the set of natural numbers $\N$ under the usual ordering $\le$ forms a well-ordered set.
This is called the '''well-ordering principle'''. | Let $S$ be a non-empty subset of the set of natural numbers $\N$.
We take as axiomatic that $\N$ is itself a subset of the set of real numbers $\R$.
Thus $S \subseteq \R$.
By definition:
:$\forall n \in \N: n \ge 0$
and so:
:$\forall n \in S: n \ge 0$
Hence $0$ is a lower bound of $S$.
This establishes the fact that $S... | Every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\N$ has a [[Definition:Smallest Element|smallest (or '''first''') element]].
That is, the [[Definition:Relational Structure|relational structure]] $\struct {\N, \le}$ on the [[Definition:Natural Numbers|set of natural numbers]] $\N$ under th... | Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|set of natural numbers]] $\N$.
We take as [[Definition:Axiom|axiomatic]] that $\N$ is itself a [[Definition:Subset|subset]] of the [[Definition:Real Number|set of real numbers]] $\R$.
Thus $S \subseteq... | Well-Ordering Principle/Proof by Restriction of Real Numbers | https://proofwiki.org/wiki/Well-Ordering_Principle | https://proofwiki.org/wiki/Well-Ordering_Principle/Proof_by_Restriction_of_Real_Numbers | [
"Number Theory",
"Natural Numbers",
"Named Theorems",
"Well-Orderings",
"Well-Ordering Principle",
"Ordering on Natural Numbers"
] | [
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Smallest Element",
"Definition:Relational Structure",
"Definition:Natural Numbers",
"Definition:Usual Ordering",
"Definition:Well-Ordered Set"
] | [
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Natural Numbers",
"Definition:Axiom",
"Definition:Subset",
"Definition:Real Number",
"Definition:Lower Bound of Set/Real Numbers",
"Definition:Bounded Below Set/Real Numbers",
"Continuum Property",
"Definition:Infimum of Set/Real Numbers... |
proofwiki-775 | Well-Ordering Principle | Every non-empty subset of $\N$ has a smallest (or '''first''') element.
That is, the relational structure $\struct {\N, \le}$ on the set of natural numbers $\N$ under the usual ordering $\le$ forms a well-ordered set.
This is called the '''well-ordering principle'''. | Consider the natural numbers $\N$ defined as the naturally ordered semigroup $\struct {S, \circ, \preceq}$.
From its definition, $\struct {S, \circ, \preceq}$ is well-ordered by $\preceq$.
The result follows.
As $\N_{\ne 0} = \N \setminus \set 0$, by Set Difference is Subset $\N_{\ne 0} \subseteq \N$.
As $\N$ is well-o... | Every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\N$ has a [[Definition:Smallest Element|smallest (or '''first''') element]].
That is, the [[Definition:Relational Structure|relational structure]] $\struct {\N, \le}$ on the [[Definition:Natural Numbers|set of natural numbers]] $\N$ under th... | Consider the [[Definition:Natural Numbers|natural numbers]] $\N$ defined as the [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]] $\struct {S, \circ, \preceq}$.
From its definition, $\struct {S, \circ, \preceq}$ is [[Definition:Well-Ordered Set|well-ordered]] by $\preceq$.
The result follows.
A... | Well-Ordering Principle/Proof using Naturally Ordered Semigroup | https://proofwiki.org/wiki/Well-Ordering_Principle | https://proofwiki.org/wiki/Well-Ordering_Principle/Proof_using_Naturally_Ordered_Semigroup | [
"Number Theory",
"Natural Numbers",
"Named Theorems",
"Well-Orderings",
"Well-Ordering Principle",
"Ordering on Natural Numbers"
] | [
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Smallest Element",
"Definition:Relational Structure",
"Definition:Natural Numbers",
"Definition:Usual Ordering",
"Definition:Well-Ordered Set"
] | [
"Definition:Natural Numbers",
"Definition:Naturally Ordered Semigroup",
"Definition:Well-Ordered Set",
"Set Difference is Subset",
"Definition:Well-Ordered Set",
"Definition:Smallest Element"
] |
proofwiki-776 | Well-Ordering Principle | Every non-empty subset of $\N$ has a smallest (or '''first''') element.
That is, the relational structure $\struct {\N, \le}$ on the set of natural numbers $\N$ under the usual ordering $\le$ forms a well-ordered set.
This is called the '''well-ordering principle'''. | From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is a minimally inductive class under the successor mapping.
From Successor Mapping on Natural Numbers is Progressing, this successor mapping is a progressing mapping.
The result is a direct application of Minimally Inductive Class under P... | Every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\N$ has a [[Definition:Smallest Element|smallest (or '''first''') element]].
That is, the [[Definition:Relational Structure|relational structure]] $\struct {\N, \le}$ on the [[Definition:Natural Numbers|set of natural numbers]] $\N$ under th... | From [[Von Neumann Construction of Natural Numbers is Minimally Inductive]], $\omega$ is a [[Definition:Minimally Inductive Class under General Mapping|minimally inductive class]] under the [[Definition:Successor Mapping on Von Neumann Construction|successor mapping]].
From [[Successor Mapping on Natural Numbers is Pr... | Well-Ordering Principle/Proof using Von Neumann Construction | https://proofwiki.org/wiki/Well-Ordering_Principle | https://proofwiki.org/wiki/Well-Ordering_Principle/Proof_using_Von_Neumann_Construction | [
"Number Theory",
"Natural Numbers",
"Named Theorems",
"Well-Orderings",
"Well-Ordering Principle",
"Ordering on Natural Numbers"
] | [
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Smallest Element",
"Definition:Relational Structure",
"Definition:Natural Numbers",
"Definition:Usual Ordering",
"Definition:Well-Ordered Set"
] | [
"Von Neumann Construction of Natural Numbers is Minimally Inductive",
"Definition:Minimally Inductive Class under General Mapping",
"Definition:Natural Numbers/Von Neumann Construction/Successor Mapping",
"Successor Mapping on Natural Numbers is Progressing",
"Definition:Natural Numbers/Von Neumann Construc... |
proofwiki-777 | Equality of Natural Numbers | Let $m, n \in \N$.
Then:
:$\N_m \sim \N_n \iff m = n$
where $\sim$ denotes set equivalence and $\N_n$ denotes the set of all natural numbers less than $n$. | By Set Equivalence behaves like Equivalence Relation, we have that:
:$m = n \implies \N_m \sim \N_n$
It remains to show that:
:$m \ne n \implies \N_m \nsim \N_n$.
Since the naturals are totally ordered, it will be sufficient to show that:
:$m \in \N_n \implies \N_m \nsim \N_n$
Let $S = \set {n \in \N: \forall m \in \N_... | Let $m, n \in \N$.
Then:
:$\N_m \sim \N_n \iff m = n$
where $\sim$ denotes [[Definition:Set Equivalence|set equivalence]] and $\N_n$ denotes the [[Definition:Initial Segment of Natural Numbers|set of all natural numbers less than $n$]]. | By [[Set Equivalence behaves like Equivalence Relation]], we have that:
:$m = n \implies \N_m \sim \N_n$
It remains to show that:
:$m \ne n \implies \N_m \nsim \N_n$.
Since the [[Definition:Natural Number|naturals]] are [[Definition:Totally Ordered Set|totally ordered]], it will be sufficient to show that:
:$m \in... | Equality of Natural Numbers | https://proofwiki.org/wiki/Equality_of_Natural_Numbers | https://proofwiki.org/wiki/Equality_of_Natural_Numbers | [
"Natural Numbers",
"Proofs by Induction"
] | [
"Definition:Set Equivalence",
"Definition:Initial Segment of Natural Numbers"
] | [
"Set Equivalence behaves like Equivalence Relation",
"Definition:Natural Numbers",
"Definition:Totally Ordered Set",
"Definition:Set",
"Definition:Natural Numbers",
"Principle of Mathematical Induction",
"Initial Segment of Natural Numbers determined by Zero is Empty",
"Definition:Basis for the Induct... |
proofwiki-778 | Principle of Counting | Let $T$ be a set such that $T \sim \N_n$.
Then:
:$\forall m \in \N: n \ne m \implies T \nsim \N_m$ | The result follows directly from Equality of Natural Numbers and the fact that set equivalence is transitive.
{{qed}} | Let $T$ be a [[Definition:Set|set]] such that $T \sim \N_n$.
Then:
:$\forall m \in \N: n \ne m \implies T \nsim \N_m$ | The result follows directly from [[Equality of Natural Numbers]] and the fact that [[Set Equivalence behaves like Equivalence Relation|set equivalence is transitive]].
{{qed}} | Principle of Counting | https://proofwiki.org/wiki/Principle_of_Counting | https://proofwiki.org/wiki/Principle_of_Counting | [
"Natural Numbers",
"Named Theorems"
] | [
"Definition:Set"
] | [
"Equality of Natural Numbers",
"Set Equivalence behaves like Equivalence Relation"
] |
proofwiki-779 | Cardinality Less One | Let $S$ be a finite set.
Let:
:$\card S = n + 1$
where $\card S$ is the cardinality of $S$.
Let $a \in S$.
Then:
:$\card {S \setminus \set a} = n$
where $\setminus$ denotes set difference. | This follows as an immediate consequence of Set Equivalence Less One Element.
{{qed}} | Let $S$ be a [[Definition:Finite Set|finite set]].
Let:
:$\card S = n + 1$
where $\card S$ is the [[Definition:Cardinality|cardinality]] of $S$.
Let $a \in S$.
Then:
:$\card {S \setminus \set a} = n$
where $\setminus$ denotes [[Definition:Set Difference|set difference]]. | This follows as an immediate consequence of [[Set Equivalence Less One Element]].
{{qed}} | Cardinality Less One | https://proofwiki.org/wiki/Cardinality_Less_One | https://proofwiki.org/wiki/Cardinality_Less_One | [
"Set Theory"
] | [
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Set Difference"
] | [
"Set Equivalence Less One Element"
] |
proofwiki-780 | Cardinality of Empty Set | :$\card S = 0 \iff S = \O$
That is, the empty set is finite, and has a cardinality of zero. | Zero is defined as the cardinal of the empty set.
The result follows from Finite Cardinals and Ordinals are Equivalent. | :$\card S = 0 \iff S = \O$
That is, the [[Definition:Empty Set|empty set]] is [[Definition:Finite Set|finite]], and has a [[Definition:Cardinality|cardinality]] of [[Definition:Zero (Number)|zero]]. | [[Definition:Zero (Number)|Zero]] is defined as the [[Definition:Cardinal|cardinal]] of the [[Definition:Empty Set|empty set]].
The result follows from [[Finite Cardinals and Ordinals are Equivalent]]. | Cardinality of Empty Set | https://proofwiki.org/wiki/Cardinality_of_Empty_Set | https://proofwiki.org/wiki/Cardinality_of_Empty_Set | [
"Empty Set",
"Cardinality",
"Finite Sets"
] | [
"Definition:Empty Set",
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Zero (Number)"
] | [
"Definition:Zero (Number)",
"Definition:Cardinal",
"Definition:Empty Set",
"Finite Cardinals and Ordinals are Equivalent"
] |
proofwiki-781 | Cardinality of Subset of Finite Set | Let $A$ and $B$ be finite sets such that $A \subseteq B$.
Let
:$\card B = n$
where $\card {\, \cdot \,}$ denotes cardinality.
Then $\card A \le n$. | Let $A \subseteq B$.
There are two cases:
$(1): \quad A \ne B$
In this case:
:$A \subsetneqq B$
and from Cardinality of Proper Subset of Finite Set:
:$\card A < n$
$(2): \quad A = B$
In this case:
:$\card A = \card B$
and so:
:$\card A = n$
In both cases:
:$\card A \le n$
Hence the result.
{{qed}} | Let $A$ and $B$ be [[Definition:Finite Set|finite sets]] such that $A \subseteq B$.
Let
:$\card B = n$
where $\card {\, \cdot \,}$ denotes [[Definition:Cardinality|cardinality]].
Then $\card A \le n$. | Let $A \subseteq B$.
There are two cases:
$(1): \quad A \ne B$
In this case:
:$A \subsetneqq B$
and from [[Cardinality of Proper Subset of Finite Set]]:
:$\card A < n$
$(2): \quad A = B$
In this case:
:$\card A = \card B$
and so:
:$\card A = n$
In both cases:
:$\card A \le n$
Hence the result.
{{qed}} | Cardinality of Subset of Finite Set | https://proofwiki.org/wiki/Cardinality_of_Subset_of_Finite_Set | https://proofwiki.org/wiki/Cardinality_of_Subset_of_Finite_Set | [
"Subsets",
"Finite Sets",
"Cardinality"
] | [
"Definition:Finite Set",
"Definition:Cardinality"
] | [
"Cardinality of Proper Subset of Finite Set"
] |
proofwiki-782 | Cardinality of Codomain of Surjection | Let $S$ be a set.
Let:
:$\card S = n$
where $\card S$ denotes the cardinality of $S$.
Let $f: S \to T$ be a surjection.
Then $\card T \le n$.
The equality:
:$\card T = n$
occurs {{iff}} $f$ is a bijection. | We have that $\card S = n$.
Then by definition of cardinality:
:there is a surjection from $S$ to $T$
{{iff}}:
:there is a surjection from $\N_n$ to $T$.
So we need consider the case only when $S = \N_n$.
By definition of surjection:
:$\forall x \in T: f^{-1} \sqbrk {\set x} \ne \O$
where $f^{-1} \sqbrk {\set x}$ denot... | Let $S$ be a [[Definition:Set|set]].
Let:
:$\card S = n$
where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$.
Let $f: S \to T$ be a [[Definition:Surjection|surjection]].
Then $\card T \le n$.
The equality:
:$\card T = n$
occurs {{iff}} $f$ is a [[Definition:Bijection|bijection]]. | We have that $\card S = n$.
Then by definition of [[Definition:Cardinality|cardinality]]:
:there is a [[Definition:Surjection|surjection]] from $S$ to $T$
{{iff}}:
:there is a [[Definition:Surjection|surjection]] from $\N_n$ to $T$.
So we need consider the case only when $S = \N_n$.
By definition of [[Definition:Su... | Cardinality of Codomain of Surjection | https://proofwiki.org/wiki/Cardinality_of_Codomain_of_Surjection | https://proofwiki.org/wiki/Cardinality_of_Codomain_of_Surjection | [
"Surjections",
"Bijections",
"Cardinality"
] | [
"Definition:Set",
"Definition:Cardinality",
"Definition:Surjection",
"Definition:Bijection"
] | [
"Definition:Cardinality",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Preimage/Mapping/Subset",
"Well-Ordering Principle",
"Definition:Well-Ordered Set",
"Subset of Well-Ordered Set is Well-Ordered",
"Definition:Well-Ordered Set",
"Definition:Well-Ordered... |
proofwiki-783 | Equivalence of Mappings between Finite Sets of Same Cardinality | Let $S$ and $T$ be finite sets such that $\card S = \card T$.
Let $f: S \to T$ be a mapping.
Then the following statements are equivalent:
:$(1): \quad f$ is bijective
:$(2): \quad f$ is injective
:$(3): \quad f$ is surjective. | $(2)$ implies $(3)$:
Let $f$ be an injection.
Then by Cardinality of Image of Injection:
:$\card S = \card {f \sqbrk S}$
where $f \sqbrk S$ denotes the image of $S$ under $f$.
Therefore the subset $f \sqbrk S$ of $T$ has the same number of elements as $T$.
Therefore:
:$f \sqbrk S = T$
and so $f$ is a surjection.
{{qed|... | Let $S$ and $T$ be [[Definition:Finite|finite]] sets such that $\card S = \card T$.
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then the following statements are equivalent:
:$(1): \quad f$ is [[Definition:Bijection|bijective]]
:$(2): \quad f$ is [[Definition:Injection|injective]]
:$(3): \quad f$ is [[Def... | $(2)$ implies $(3)$:
Let $f$ be an [[Definition:Injection|injection]].
Then by [[Cardinality of Image of Injection]]:
:$\card S = \card {f \sqbrk S}$
where $f \sqbrk S$ denotes the [[Definition:Image of Subset under Mapping|image]] of $S$ under $f$.
Therefore the [[Definition:Subset|subset]] $f \sqbrk S$ of $T$ has ... | Equivalence of Mappings between Finite Sets of Same Cardinality | https://proofwiki.org/wiki/Equivalence_of_Mappings_between_Finite_Sets_of_Same_Cardinality | https://proofwiki.org/wiki/Equivalence_of_Mappings_between_Finite_Sets_of_Same_Cardinality | [
"Injections",
"Surjections",
"Bijections"
] | [
"Definition:Finite",
"Definition:Mapping",
"Definition:Bijection",
"Definition:Injection",
"Definition:Surjection"
] | [
"Definition:Injection",
"Cardinality of Image of Injection",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Subset",
"Definition:Element",
"Definition:Surjection",
"Definition:Surjection",
"Cardinality of Codomain of Surjection",
"Definition:Bijection",
"Definition:Bijection",
"Defi... |
proofwiki-784 | Natural Numbers are Infinite | The set $\N$ of natural numbers is infinite. | Let the mapping $s: \N \to \N$ be defined as:
:$\forall n \in \N: \map s n = n + 1$
$s$ is clearly an injection.
{{AimForCont}} $\N$ were finite.
By Equivalence of Mappings between Finite Sets of Same Cardinality it follows that $s$ is a surjection.
But:
:$\forall n \in \N: \map s n \ge 0 + 1 > 0$
So:
:$0 \notin \Img s... | The [[Definition:Set|set]] $\N$ of [[Definition:Natural Numbers|natural numbers]] is [[Definition:Infinite Set|infinite]]. | Let the [[Definition:Mapping|mapping]] $s: \N \to \N$ be defined as:
:$\forall n \in \N: \map s n = n + 1$
$s$ is clearly an [[Definition:Injection|injection]].
{{AimForCont}} $\N$ were [[Definition:Finite Set|finite]].
By [[Equivalence of Mappings between Finite Sets of Same Cardinality]] it follows that $s$ is a [... | Natural Numbers are Infinite | https://proofwiki.org/wiki/Natural_Numbers_are_Infinite | https://proofwiki.org/wiki/Natural_Numbers_are_Infinite | [
"Natural Numbers",
"Infinite Sets"
] | [
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Infinite Set"
] | [
"Definition:Mapping",
"Definition:Injection",
"Definition:Finite Set",
"Equivalence of Mappings between Finite Sets of Same Cardinality",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Contradiction",
"Definition:Finite Set",
"Definition:Infinite Set"
] |
proofwiki-785 | Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements | Let $\struct {S, \preceq}$ be a totally ordered set.
Then every finite $T$ such that $\O \subset T \subseteq S$ has both a smallest and a greatest element. | Let $A \subseteq \N_{>0}$ such that every $B \subseteq S$ such that $\card B = n$ has a greatest and a smallest element.
Any $B \subseteq S$ such that $\card B = 1$ has $1$ element, $b \in B$ say.
Then $b$ is both the greatest and smallest element of $B$.
So $1 \in A$.
Let $n \in A$.
Let $B \subseteq S$ such that $\car... | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Then every [[Definition:Finite Set|finite]] $T$ such that $\O \subset T \subseteq S$ has both a [[Definition:Smallest Element|smallest]] and a [[Definition:Greatest Element|greatest]] element. | Let $A \subseteq \N_{>0}$ such that every $B \subseteq S$ such that $\card B = n$ has a [[Definition:Greatest Element|greatest]] and a [[Definition:Smallest Element|smallest]] element.
Any $B \subseteq S$ such that $\card B = 1$ has $1$ [[Definition:Element|element]], $b \in B$ say.
Then $b$ is both the [[Definition... | Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements/Proof 1 | https://proofwiki.org/wiki/Finite_Non-Empty_Subset_of_Totally_Ordered_Set_has_Smallest_and_Greatest_Elements | https://proofwiki.org/wiki/Finite_Non-Empty_Subset_of_Totally_Ordered_Set_has_Smallest_and_Greatest_Elements/Proof_1 | [
"Total Orderings",
"Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements"
] | [
"Definition:Totally Ordered Set",
"Definition:Finite Set",
"Definition:Smallest Element",
"Definition:Greatest Element"
] | [
"Definition:Greatest Element",
"Definition:Smallest Element",
"Definition:Element",
"Definition:Greatest Element",
"Definition:Smallest Element",
"Cardinality Less One",
"Definition:Induction Hypothesis",
"Definition:Greatest Element",
"Definition:Smallest Element",
"Definition:Totally Ordered Set... |
proofwiki-786 | Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements | Let $\struct {S, \preceq}$ be a totally ordered set.
Then every finite $T$ such that $\O \subset T \subseteq S$ has both a smallest and a greatest element. | The result follows from:
: Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements
: Minimal Element of Chain is Smallest Element
: Maximal Element of Chain is Greatest Element
{{qed}} | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Then every [[Definition:Finite Set|finite]] $T$ such that $\O \subset T \subseteq S$ has both a [[Definition:Smallest Element|smallest]] and a [[Definition:Greatest Element|greatest]] element. | The result follows from:
: [[Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements]]
: [[Minimal Element of Chain is Smallest Element]]
: [[Maximal Element of Chain is Greatest Element]]
{{qed}} | Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements/Proof 2 | https://proofwiki.org/wiki/Finite_Non-Empty_Subset_of_Totally_Ordered_Set_has_Smallest_and_Greatest_Elements | https://proofwiki.org/wiki/Finite_Non-Empty_Subset_of_Totally_Ordered_Set_has_Smallest_and_Greatest_Elements/Proof_2 | [
"Total Orderings",
"Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements"
] | [
"Definition:Totally Ordered Set",
"Definition:Finite Set",
"Definition:Smallest Element",
"Definition:Greatest Element"
] | [
"Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements",
"Minimal Element of Chain is Smallest Element",
"Maximal Element of Chain is Greatest Element"
] |
proofwiki-787 | Unique Isomorphism between Equivalent Finite Totally Ordered Sets | Let $S$ and $T$ be finite sets of the same cardinality.
That is:
:$\card S = \card T$
Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be totally ordered sets.
Then there is exactly one order isomorphism from $\struct {S, \preceq}$ to $\struct {T, \preccurlyeq}$. | It is sufficient to consider the case where $\struct {T, \preccurlyeq}$ is $\struct {\N_n, \le}$ for some $n \in \N$.
Let $A$ be the set of all $n \in \N$ such that if:
:$(1): \quad S$ is any set such that $\card S = n$, and
:$(2): \quad \preceq$ is any total ordering on $S$
then there is exactly one isomorphism from $... | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]] of the same [[Definition:Cardinality|cardinality]].
That is:
:$\card S = \card T$
Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be [[Definition:Totally Ordered Set|totally ordered sets]].
Then there is exactly one [[Definition:Order Isomorphism|o... | It is sufficient to consider the case where $\struct {T, \preccurlyeq}$ is $\struct {\N_n, \le}$ for some $n \in \N$.
Let $A$ be the set of all $n \in \N$ such that if:
:$(1): \quad S$ is any set such that $\card S = n$, and
:$(2): \quad \preceq$ is any [[Definition:Total Ordering|total ordering]] on $S$
then there ... | Unique Isomorphism between Equivalent Finite Totally Ordered Sets | https://proofwiki.org/wiki/Unique_Isomorphism_between_Equivalent_Finite_Totally_Ordered_Sets | https://proofwiki.org/wiki/Unique_Isomorphism_between_Equivalent_Finite_Totally_Ordered_Sets | [
"Total Orderings",
"Order Isomorphisms"
] | [
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Totally Ordered Set",
"Definition:Order Isomorphism"
] | [
"Definition:Total Ordering",
"Definition:Order Isomorphism",
"Empty Mapping is Mapping",
"Equivalence of Mappings between Finite Sets of Same Cardinality",
"Definition:Totally Ordered Set",
"Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements",
"Definition:Greatest Element"... |
proofwiki-788 | Isomorphism to Closed Interval | Let $m, n \in \N$ such that $m < n$.
Let $\closedint {m + 1} n$ denote the integer interval from $m + 1$ to $n$.
Let $h: \N_{n - m} \to \closedint {m + 1} n$ be the mapping defined as:
:$\forall x \in \N_{n - m}: \map h x = x + m + 1$
Let the orderings on $\closedint {m + 1} n$ and $\N_{n - m}$ be those induced by the ... | First note that the cardinality of $\closedint {m + 1} n$ is given by:
:$\card {\closedint {m + 1} n} = n - m$
From Unique Isomorphism between Equivalent Finite Totally Ordered Sets, it suffices to show that $h$ is an order isomorphism.
To this end, remark that, for all $x, y \in \N_{n - m}$:
{{begin-eqn}}
{{eqn | l = ... | Let $m, n \in \N$ such that $m < n$.
Let $\closedint {m + 1} n$ denote the [[Definition:Integer Interval|integer interval]] from $m + 1$ to $n$.
Let $h: \N_{n - m} \to \closedint {m + 1} n$ be the [[Definition:Mapping|mapping]] defined as:
:$\forall x \in \N_{n - m}: \map h x = x + m + 1$
Let the [[Definition:Orde... | First note that the [[Definition:Cardinality|cardinality]] of $\closedint {m + 1} n$ is given by:
:$\card {\closedint {m + 1} n} = n - m$
From [[Unique Isomorphism between Equivalent Finite Totally Ordered Sets]], it suffices to show that $h$ is an [[Definition:Order Isomorphism|order isomorphism]].
To this end, re... | Isomorphism to Closed Interval | https://proofwiki.org/wiki/Isomorphism_to_Closed_Interval | https://proofwiki.org/wiki/Isomorphism_to_Closed_Interval | [
"Set Theory",
"Order Theory",
"Mapping Theory"
] | [
"Definition:Closed Interval/Integer Interval",
"Definition:Mapping",
"Definition:Ordering",
"Definition:Restriction/Relation",
"Definition:Ordering",
"Definition:Order Isomorphism"
] | [
"Definition:Cardinality",
"Unique Isomorphism between Equivalent Finite Totally Ordered Sets",
"Definition:Order Isomorphism",
"Natural Number Addition is Cancellable",
"Definition:Injection",
"Equivalence of Mappings between Finite Sets of Same Cardinality",
"Definition:Bijection",
"Ordering on Natur... |
proofwiki-789 | Regular Representation wrt Cancellable Element on Finite Semigroup is Bijection | Let $\left({S, \circ}\right)$ be a finite semigroup.
Let $a \in S$ be cancellable.
Then:
:the left regular representation $\lambda_a$
and:
:the right regular representation $\rho_a$
of $\left({S, \circ}\right)$ with respect to $a$ are both bijections. | By Cancellable iff Regular Representations Injective, $\lambda_a$ and $\rho_a$ are injections.
We have that $S$ is finite.
From Injection from Finite Set to Itself is Surjection, both $\lambda_a$ and $\rho_a$ are surjections.
Thus $\lambda_a$ and $\rho_a$ are injective and surjective, and therefore bijections.
{{qed}}
... | Let $\left({S, \circ}\right)$ be a [[Definition:Finite|finite]] [[Definition:Semigroup|semigroup]].
Let $a \in S$ be [[Definition:Cancellable Element|cancellable]].
Then:
:the [[Definition:Left Regular Representation|left regular representation]] $\lambda_a$
and:
:the [[Definition:Right Regular Representation|right ... | By [[Cancellable iff Regular Representations Injective]], $\lambda_a$ and $\rho_a$ are [[Definition:Injection|injections]].
We have that $S$ is [[Definition:Finite Set|finite]].
From [[Injection from Finite Set to Itself is Surjection]], both $\lambda_a$ and $\rho_a$ are [[Definition:Surjection|surjections]].
Thus $... | Regular Representation wrt Cancellable Element on Finite Semigroup is Bijection | https://proofwiki.org/wiki/Regular_Representation_wrt_Cancellable_Element_on_Finite_Semigroup_is_Bijection | https://proofwiki.org/wiki/Regular_Representation_wrt_Cancellable_Element_on_Finite_Semigroup_is_Bijection | [
"Semigroups",
"Regular Representations",
"Cancellability"
] | [
"Definition:Finite",
"Definition:Semigroup",
"Definition:Cancellable Element",
"Definition:Regular Representations/Left Regular Representation",
"Definition:Regular Representations/Right Regular Representation",
"Definition:Bijection"
] | [
"Cancellable iff Regular Representations Injective",
"Definition:Injection",
"Definition:Finite Set",
"Injection from Finite Set to Itself is Surjection",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijection",
"Category:Semigroups",
"Category:Regular Repr... |
proofwiki-790 | Power Set of Natural Numbers is Uncountable | The power set $\powerset \N$ of the natural numbers $\N$ is uncountable. | There is no bijection from a set to its power set.
From Injection from Set to Power Set, we have that there exists an injection $f: \N \to \powerset \N$.
From the Cantor-Bernstein-Schröder Theorem, there can be no injection $g: \powerset \N \to \N$.
So, by definition, $\powerset \N$ is uncountable.
{{qed}} | The [[Definition:Power Set|power set]] $\powerset \N$ of the [[Definition:Natural Numbers|natural numbers]] $\N$ is [[Definition:Uncountable Set|uncountable]]. | There is [[No Bijection from Set to its Power Set|no bijection from a set to its power set]].
From [[Injection from Set to Power Set]], we have that there exists an [[Definition:Injection|injection]] $f: \N \to \powerset \N$.
From the [[Cantor-Bernstein-Schröder Theorem]], there can be no [[Definition:Injection|injec... | Power Set of Natural Numbers is Uncountable | https://proofwiki.org/wiki/Power_Set_of_Natural_Numbers_is_Uncountable | https://proofwiki.org/wiki/Power_Set_of_Natural_Numbers_is_Uncountable | [
"Power Set",
"Natural Numbers",
"Uncountable Sets"
] | [
"Definition:Power Set",
"Definition:Natural Numbers",
"Definition:Uncountable/Set"
] | [
"No Bijection from Set to its Power Set",
"Injection from Set to Power Set",
"Definition:Injection",
"Cantor-Bernstein-Schröder Theorem",
"Definition:Injection",
"Definition:Uncountable/Set"
] |
proofwiki-791 | Subset of Countably Infinite Set is Countable | Every subset of a countably infinite set is countable. | Let $S = \set {a_0, a_1, a_2, \ldots}$ be countably infinite.
Let $T \subseteq S = \set {a_{n_0}, a_{n_1}, a_{n_2}, \ldots}$, where $a_{n_0}, a_{n_1}, a_{n_2}, \ldots$ are the elements of $S$ also in $T$.
If the set of numbers $\set {n_0, n_1, n_2, \ldots}$ has a largest number, then $T$ is finite.
Otherwise, consider ... | Every [[Definition:Subset|subset]] of a [[Definition:Countably Infinite Set|countably infinite set]] is [[Definition:Countable Set|countable]]. | Let $S = \set {a_0, a_1, a_2, \ldots}$ be [[Definition:Countably Infinite Set|countably infinite]].
Let $T \subseteq S = \set {a_{n_0}, a_{n_1}, a_{n_2}, \ldots}$, where $a_{n_0}, a_{n_1}, a_{n_2}, \ldots$ are the [[Definition:Element|elements]] of $S$ also in $T$.
If the set of numbers $\set {n_0, n_1, n_2, \ldots}$... | Subset of Countably Infinite Set is Countable | https://proofwiki.org/wiki/Subset_of_Countably_Infinite_Set_is_Countable | https://proofwiki.org/wiki/Subset_of_Countably_Infinite_Set_is_Countable | [
"Countable Sets",
"Subsets"
] | [
"Definition:Subset",
"Definition:Countably Infinite/Set",
"Definition:Countable Set"
] | [
"Definition:Countably Infinite/Set",
"Definition:Element",
"Definition:Finite Set",
"Definition:Bijection",
"Definition:Bijection",
"Definition:Bijection",
"Definition:Bijection",
"Definition:Element",
"Definition:Countable Set"
] |
proofwiki-792 | Infinite Set has Countably Infinite Subset | Every infinite set has a countably infinite subset. | Let $S$ be an infinite set.
We use Between Two Sets Exists Injection or Surjection.
Suppose that there exists an injection $\psi: \N \to S$.
Let $T$ be the image of $\psi$.
From Injection to Image is Bijection, it follows that $\psi^{-1}: T \to \N$ is a bijection.
Hence, $T$ is a countably infinite subset of $S$.
Now, ... | Every [[Definition:Infinite Set|infinite set]] has a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Subset|subset]]. | Let $S$ be an [[Definition:Infinite Set|infinite set]].
We use [[Between Two Sets Exists Injection or Surjection]].
Suppose that there exists an [[Definition:Injection|injection]] $\psi: \N \to S$.
Let $T$ be the [[Definition:Image of Mapping|image]] of $\psi$.
From [[Injection to Image is Bijection]], it follows ... | Infinite Set has Countably Infinite Subset/Proof 1 | https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset | https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset/Proof_1 | [
"Countable Sets",
"Infinite Sets",
"Subsets",
"Infinite Set has Countably Infinite Subset"
] | [
"Definition:Infinite Set",
"Definition:Countably Infinite/Set",
"Definition:Subset"
] | [
"Definition:Infinite Set",
"Between Two Sets Exists Injection or Surjection",
"Definition:Injection",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Injection to Image is Bijection",
"Definition:Bijection",
"Definition:Countably Infinite/Set",
"Definition:Subset",
"Definition:Surjection",
"Surj... |
proofwiki-793 | Infinite Set has Countably Infinite Subset | Every infinite set has a countably infinite subset. | Let $S$ be an infinite set.
First an injection $f: \N \to S$ is constructed.
Let $g$ be a choice function on $\powerset S \setminus \set \O$.
Then define $f: \N \to S$ as follows:
:$\forall n \in \N: \map f n = \begin {cases} \map g S & : n = 0 \\ \map g {S \setminus \set {\map f 0, \ldots, \map f {n - 1} } } & : n > 0... | Every [[Definition:Infinite Set|infinite set]] has a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Subset|subset]]. | Let $S$ be an [[Definition:Infinite Set|infinite set]].
First an [[Definition:Injection|injection]] $f: \N \to S$ is constructed.
Let $g$ be a [[Definition:Choice Function|choice function]] on $\powerset S \setminus \set \O$.
Then define $f: \N \to S$ as follows:
:$\forall n \in \N: \map f n = \begin {cases} \map g... | Infinite Set has Countably Infinite Subset/Proof 2 | https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset | https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset/Proof_2 | [
"Countable Sets",
"Infinite Sets",
"Subsets",
"Infinite Set has Countably Infinite Subset"
] | [
"Definition:Infinite Set",
"Definition:Countably Infinite/Set",
"Definition:Subset"
] | [
"Definition:Infinite Set",
"Definition:Injection",
"Definition:Choice Function",
"Definition:Infinite Set",
"Definition:Non-Empty Set",
"Definition:Infinite Set",
"Definition:Injection",
"Injection to Image is Bijection",
"Definition:Bijection",
"Definition:Countable Set"
] |
proofwiki-794 | Infinite Set has Countably Infinite Subset | Every infinite set has a countably infinite subset. | Let $S$ be an infinite set.
First an injection $f: \N \to S$ is constructed.
Let $f$ be a choice function on $\powerset S \setminus \set \O$.
That is:
:$\forall A \in \powerset S \setminus \set \O: \map f A \in A$
This is justified only if the Axiom of Choice is accepted.
Let $\CC$ be the set of all finite subsets of $... | Every [[Definition:Infinite Set|infinite set]] has a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Subset|subset]]. | Let $S$ be an [[Definition:Infinite Set|infinite set]].
First an [[Definition:Injection|injection]] $f: \N \to S$ is constructed.
Let $f$ be a [[Definition:Choice Function|choice function]] on $\powerset S \setminus \set \O$.
That is:
:$\forall A \in \powerset S \setminus \set \O: \map f A \in A$
This is justified ... | Infinite Set has Countably Infinite Subset/Proof 3 | https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset | https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset/Proof_3 | [
"Countable Sets",
"Infinite Sets",
"Subsets",
"Infinite Set has Countably Infinite Subset"
] | [
"Definition:Infinite Set",
"Definition:Countably Infinite/Set",
"Definition:Subset"
] | [
"Definition:Infinite Set",
"Definition:Injection",
"Definition:Choice Function",
"Axiom:Axiom of Choice",
"Definition:Set of Sets",
"Definition:Finite Set",
"Definition:Infinite Set",
"Definition:Mapping",
"Principle of Recursive Definition",
"Definition:Mapping",
"Definition:Set",
"Definition... |
proofwiki-795 | Infinite Set has Countably Infinite Subset | Every infinite set has a countably infinite subset. | Let $S$ be an infinite set.
For all $n \in \N$, let:
:$\FF_n = \set {T \subseteq S: \size T = n}$
where $\size T$ denotes the cardinality of $T$.
From Set is Infinite iff exist Subsets of all Finite Cardinalities:
:$\FF_n$ is non-empty.
Using the axiom of countable choice, we can obtain a sequence $\sequence {S_n}_{n \... | Every [[Definition:Infinite Set|infinite set]] has a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Subset|subset]]. | Let $S$ be an [[Definition:Infinite Set|infinite set]].
For all $n \in \N$, let:
:$\FF_n = \set {T \subseteq S: \size T = n}$
where $\size T$ denotes the [[Definition:Cardinality|cardinality]] of $T$.
From [[Set is Infinite iff exist Subsets of all Finite Cardinalities]]:
:$\FF_n$ is [[Definition:Non-Empty Set|non-e... | Infinite Set has Countably Infinite Subset/Proof 4 | https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset | https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset/Proof_4 | [
"Countable Sets",
"Infinite Sets",
"Subsets",
"Infinite Set has Countably Infinite Subset"
] | [
"Definition:Infinite Set",
"Definition:Countably Infinite/Set",
"Definition:Subset"
] | [
"Definition:Infinite Set",
"Definition:Cardinality",
"Set is Infinite iff exist Subsets of all Finite Cardinalities",
"Definition:Non-Empty Set",
"Axiom:Axiom of Countable Choice",
"Definition:Sequence",
"Definition:Subset",
"Definition:Cardinality",
"Set is Infinite iff exist Subsets of all Finite ... |
proofwiki-796 | No Bijection between Finite Set and Proper Subset | A finite set can not be in one-to-one correspondence with one of its proper subsets.
That is, a finite set is not Dedekind-infinite. | Let $S$ be a finite set.
Let $T$ be a proper subset of $S$.
Let $f: T \to S$ be an injection.
By Cardinality of Image of Injection and Cardinality of Subset of Finite Set:
:$\card {\Img f} = \card T < \card S$
where $\Img f$ denotes the image of $f$.
Thus $\Img f \ne S$, and so $f$ is not a bijection.
{{qed}} | A [[Definition:Finite Set|finite set]] can not be in [[Definition:Bijection|one-to-one correspondence]] with one of its [[Definition:Proper Subset|proper subsets]].
That is, a [[Definition:Finite Set|finite set]] is not [[Definition:Dedekind-Infinite|Dedekind-infinite]]. | Let $S$ be a [[Definition:Finite Set|finite set]].
Let $T$ be a [[Definition:Proper Subset|proper subset]] of $S$.
Let $f: T \to S$ be an [[Definition:Injection|injection]].
By [[Cardinality of Image of Injection]] and [[Cardinality of Subset of Finite Set]]:
:$\card {\Img f} = \card T < \card S$
where $\Img f$ den... | No Bijection between Finite Set and Proper Subset/Proof 1 | https://proofwiki.org/wiki/No_Bijection_between_Finite_Set_and_Proper_Subset | https://proofwiki.org/wiki/No_Bijection_between_Finite_Set_and_Proper_Subset/Proof_1 | [
"No Bijection between Finite Set and Proper Subset",
"Proper Subsets",
"Finite Sets"
] | [
"Definition:Finite Set",
"Definition:Bijection",
"Definition:Proper Subset",
"Definition:Finite Set",
"Definition:Dedekind-Infinite"
] | [
"Definition:Finite Set",
"Definition:Proper Subset",
"Definition:Injection",
"Cardinality of Image of Injection",
"Cardinality of Subset of Finite Set",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Bijection"
] |
proofwiki-797 | No Bijection between Finite Set and Proper Subset | A finite set can not be in one-to-one correspondence with one of its proper subsets.
That is, a finite set is not Dedekind-infinite. | Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:The set $\N_n = \set {0, 1, \ldots, n - 1}$ of natural numbers less than $n$ is equivalent to none of its proper subsets.
Here we use the definition of set equivalence to mean that $S$ is equivalent to $T$ if there exists a bijection betwe... | A [[Definition:Finite Set|finite set]] can not be in [[Definition:Bijection|one-to-one correspondence]] with one of its [[Definition:Proper Subset|proper subsets]].
That is, a [[Definition:Finite Set|finite set]] is not [[Definition:Dedekind-Infinite|Dedekind-infinite]]. | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:The set $\N_n = \set {0, 1, \ldots, n - 1}$ of [[Definition:Initial Segment of Natural Numbers|natural numbers less than $n$]] is [[Definition:Set Equivalence|equivalent]]... | No Bijection between Finite Set and Proper Subset/Proof 2 | https://proofwiki.org/wiki/No_Bijection_between_Finite_Set_and_Proper_Subset | https://proofwiki.org/wiki/No_Bijection_between_Finite_Set_and_Proper_Subset/Proof_2 | [
"No Bijection between Finite Set and Proper Subset",
"Proper Subsets",
"Finite Sets"
] | [
"Definition:Finite Set",
"Definition:Bijection",
"Definition:Proper Subset",
"Definition:Finite Set",
"Definition:Dedekind-Infinite"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Initial Segment of Natural Numbers",
"Definition:Set Equivalence",
"Definition:Proper Subset",
"Definition:Set Equivalence",
"Definition:Set Equivalence",
"Definition:Bijection",
"Definition:Set Equivalence",
"Definition:... |
proofwiki-798 | Infinite Set is Equivalent to Proper Subset | A set is infinite {{iff}} it is equivalent to one of its proper subsets. | Let $T$ be an infinite set.
By Infinite Set has Countably Infinite Subset, it is possible to construct a countably infinite subset of $T$.
Let $S = \set {a_1, a_2, a_3, \ldots}$ be such a countably infinite subset of $T$.
Create a Partition of $S$ into:
:$S_1 = \set {a_1, a_3, a_5, \ldots}, S_2 = \set {a_2, a_4, a_6, \... | A [[Definition:Set|set]] is [[Definition:Infinite Set|infinite]] {{iff}} it is [[Definition:Set Equivalence|equivalent]] to one of its [[Definition:Proper Subset|proper subsets]]. | Let $T$ be an [[Definition:Infinite Set|infinite set]].
By [[Infinite Set has Countably Infinite Subset]], it is possible to construct a [[Definition:Countable|countably infinite]] [[Definition:Subset|subset]] of $T$.
Let $S = \set {a_1, a_2, a_3, \ldots}$ be such a [[Definition:Countable|countably infinite]] [[Defin... | Infinite Set is Equivalent to Proper Subset/Proof 1 | https://proofwiki.org/wiki/Infinite_Set_is_Equivalent_to_Proper_Subset | https://proofwiki.org/wiki/Infinite_Set_is_Equivalent_to_Proper_Subset/Proof_1 | [
"Infinite Set is Equivalent to Proper Subset",
"Set Equivalence",
"Infinite Sets",
"Proper Subsets"
] | [
"Definition:Set",
"Definition:Infinite Set",
"Definition:Set Equivalence",
"Definition:Proper Subset"
] | [
"Definition:Infinite Set",
"Infinite Set has Countably Infinite Subset",
"Definition:Countable Set",
"Definition:Subset",
"Definition:Countable Set",
"Definition:Subset",
"Definition:Set Partition",
"Definition:Bijection",
"Definition:Bijection",
"Definition:Element",
"Definition:Bijection",
"... |
proofwiki-799 | Infinite Set is Equivalent to Proper Subset | A set is infinite {{iff}} it is equivalent to one of its proper subsets. | Let $S$ be a set.
Suppose $S$ is finite.
From No Bijection between Finite Set and Proper Subset we have that $S$ can not be equivalent to one of its proper subsets.
Suppose $S$ is infinite.
From Infinite Set has Countably Infinite Subset, we can construct $v: \N \to S$ such that $v$ is an injection.
We now construct th... | A [[Definition:Set|set]] is [[Definition:Infinite Set|infinite]] {{iff}} it is [[Definition:Set Equivalence|equivalent]] to one of its [[Definition:Proper Subset|proper subsets]]. | Let $S$ be a [[Definition:Set|set]].
Suppose $S$ is [[Definition:Finite Set|finite]].
From [[No Bijection between Finite Set and Proper Subset]] we have that $S$ can not be [[Definition:Set Equivalence|equivalent]] to one of its [[Definition:Proper Subset|proper subsets]].
Suppose $S$ is [[Definition:Infinite Set|... | Infinite Set is Equivalent to Proper Subset/Proof 2 | https://proofwiki.org/wiki/Infinite_Set_is_Equivalent_to_Proper_Subset | https://proofwiki.org/wiki/Infinite_Set_is_Equivalent_to_Proper_Subset/Proof_2 | [
"Infinite Set is Equivalent to Proper Subset",
"Set Equivalence",
"Infinite Sets",
"Proper Subsets"
] | [
"Definition:Set",
"Definition:Infinite Set",
"Definition:Set Equivalence",
"Definition:Proper Subset"
] | [
"Definition:Set",
"Definition:Finite Set",
"No Bijection between Finite Set and Proper Subset",
"Definition:Set Equivalence",
"Definition:Proper Subset",
"Definition:Infinite Set",
"Infinite Set has Countably Infinite Subset",
"Definition:Injection",
"Definition:Mapping",
"Definition:Injection",
... |
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