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 |
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proofwiki-1200 | Group Action on Sets with k Elements | Let $\struct {G, \circ}$ be a finite group whose identity is $e$.
Let $\Bbb S = \set {S \subseteq G: \card S = k}$, that is, the set of all of subsets of $G$ which have exactly $k$ elements.
Let $G$ act on $\Bbb S$ by the rule:
:$\forall S \in \Bbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
This is a group acti... | First it is necessary to prove that this is a group action.
The action is the same as the one defined in Group Action on Coset Space, but this time we are limiting ourselves to the subsets of $G$ which have the same number of elements.
From the result in Group Action on Coset Space, we only have to prove that:
:$\card ... | Let $\struct {G, \circ}$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\Bbb S = \set {S \subseteq G: \card S = k}$, that is, the [[Definition:Set|set]] of all of [[Definition:Subset|subsets]] of $G$ which have exactly $k$ [[Definition:Element|elements]].
Let... | First it is necessary to prove that this is a [[Definition:Group Action|group action]].
The [[Definition:Group Action|action]] is the same as the one defined in [[Group Action on Coset Space]], but this time we are limiting ourselves to the [[Definition:Subset|subsets]] of $G$ which have the same number of [[Definitio... | Group Action on Sets with k Elements | https://proofwiki.org/wiki/Group_Action_on_Sets_with_k_Elements | https://proofwiki.org/wiki/Group_Action_on_Sets_with_k_Elements | [
"Examples of Group Actions",
"Group Action on Sets with k Elements"
] | [
"Definition:Finite Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Set",
"Definition:Subset",
"Definition:Element",
"Definition:Group Action",
"Definition:Stabilizer"
] | [
"Definition:Group Action",
"Definition:Group Action",
"Action of Group on Coset Space is Group Action",
"Definition:Subset",
"Definition:Element",
"Action of Group on Coset Space is Group Action",
"Order of Subset Product with Singleton",
"Definition:Stabilizer",
"Stabilizer is Subgroup",
"Definit... |
proofwiki-1201 | Group Action on Subgroup of Symmetric Group | Let $S_n$ be the symmetric group of $n$ elements.
Let $H$ be a subgroup of $S_n$.
Let $X$ be any set with $n$ elements.
Then $H$ acts on $X$ as a group of transformations on $X$. | The identity permutation takes each element of $X$ to itself, thus fulfilling {{GroupActionAxiom|2}}.
The group operation in $S_n$ ensures fulfilment of {{GroupActionAxiom|1}}.
{{qed}} | Let $S_n$ be the [[Definition:Symmetric Group on n Letters|symmetric group of $n$ elements]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $S_n$.
Let $X$ be any [[Definition:Set|set]] with $n$ [[Definition:Element|elements]].
Then $H$ [[Definition:Group Action|acts on]] $X$ as a [[Group Action determines Biject... | The [[Definition:Identity Mapping|identity permutation]] takes each element of $X$ to itself, thus fulfilling {{GroupActionAxiom|2}}.
The [[Definition:Group Operation|group operation]] in $S_n$ ensures fulfilment of {{GroupActionAxiom|1}}.
{{qed}} | Group Action on Subgroup of Symmetric Group | https://proofwiki.org/wiki/Group_Action_on_Subgroup_of_Symmetric_Group | https://proofwiki.org/wiki/Group_Action_on_Subgroup_of_Symmetric_Group | [
"Group Actions",
"Symmetric Groups"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Subgroup",
"Definition:Set",
"Definition:Element",
"Definition:Group Action",
"Group Action determines Bijection"
] | [
"Definition:Identity Mapping",
"Definition:Group Product/Group Law"
] |
proofwiki-1202 | Quotient of Transformation Group acts Effectively | Let $G$ be a transformation group (which may or may not be effective) acting on $X$.
Then the quotient group $G / G_0$, where $G_0$ is the kernel, ''does'' act effectively on $X$. | Let $g G_0 \in G / G_0$.
Let $x \in X$.
Let $G / G_0$ act on $X$ by the rule:
:$g G_0 * x := g x$
Then by definition of effective transformation group:
:$G / G_0$ acts effectively on $X$ {{iff}}:
::$g G_0 * x = x \implies g G_0 = G_0$
Suppose $g G_0 * x = g x = x$.
We want to show $g G_0 = G_0$.
By definition of kernel... | Let $G$ be a [[Definition:Transformation Group|transformation group]] (which may or may not be [[Definition:Effective Transformation Group|effective]]) acting on $X$.
Then the [[Definition:Quotient Group|quotient group]] $G / G_0$, where $G_0$ is the [[Definition:Kernel of Group Action|kernel]], ''does'' act [[Definit... | Let $g G_0 \in G / G_0$.
Let $x \in X$.
Let $G / G_0$ act on $X$ by the rule:
:$g G_0 * x := g x$
Then by definition of [[Definition:Effective Transformation Group|effective transformation group]]:
:$G / G_0$ acts [[Definition:Effective Transformation Group|effectively]] on $X$ {{iff}}:
::$g G_0 * x = x \implies g... | Quotient of Transformation Group acts Effectively | https://proofwiki.org/wiki/Quotient_of_Transformation_Group_acts_Effectively | https://proofwiki.org/wiki/Quotient_of_Transformation_Group_acts_Effectively | [
"Group Actions",
"Quotient Groups"
] | [
"Definition:Group Action/Transformation Group",
"Definition:Effective Transformation Group",
"Definition:Quotient Group",
"Definition:Kernel of Group Action",
"Definition:Effective Transformation Group"
] | [
"Definition:Effective Transformation Group",
"Definition:Effective Transformation Group",
"Definition:Kernel of Group Action",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Subgroup"
] |
proofwiki-1203 | Condition for Group to Act Effectively on Left Coset Space | Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Then $G$ acts effectively on the left coset space $G / H$ {{iff}}:
:$\ds \bigcap_{a \mathop \in G} H^a = \set e$
where $H^a$ denotes the conjugate of $H$ by $a$. | $G$ acts effectively on the left coset space $G / H$ {{iff}} $a H \mapsto g a H$ is faithful, {{iff}}:
{{begin-eqn}}
{{eqn | q = \forall g \in G: \forall a H \in G / H
| l = g a H = a H
| o = \implies
| r = g = e
| c = {{Defof|Faithful Group Action}}
}}
{{eqn | ll= \leadstoandfrom
| q = \f... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $G$ [[Definition:Effective Transformation Group|acts effectively]] on the [[Definition:Left Coset Space|left coset space]] $G / H$ {{iff}}:
:$\ds \bigcap_{a \matho... | $G$ [[Definition:Effective Transformation Group|acts effectively]] on the [[Definition:Left Coset Space|left coset space]] $G / H$ {{iff}} $a H \mapsto g a H$ is [[Definition:Faithful Group Action|faithful]], {{iff}}:
{{begin-eqn}}
{{eqn | q = \forall g \in G: \forall a H \in G / H
| l = g a H = a H
| o = ... | Condition for Group to Act Effectively on Left Coset Space | https://proofwiki.org/wiki/Condition_for_Group_to_Act_Effectively_on_Left_Coset_Space | https://proofwiki.org/wiki/Condition_for_Group_to_Act_Effectively_on_Left_Coset_Space | [
"Group Action on Coset Space",
"Cosets"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Effective Transformation Group",
"Definition:Coset Space/Left Coset Space",
"Definition:Conjugate (Group Theory)/Subset"
] | [
"Definition:Effective Transformation Group",
"Definition:Coset Space/Left Coset Space",
"Definition:Faithful Group Action",
"Left Cosets are Equal iff Product with Inverse in Subgroup"
] |
proofwiki-1204 | Conjugacy Action on Identity | Let $G$ be a group whose identity is $e$.
For the conjugacy action:
:$\order {\Orb e} = 1$
and thus:
:$\Stab e = G$ | {{begin-eqn}}
{{eqn | l = g * e
| r = g e g^{-1}
| c =
}}
{{eqn | r = g g^{-1}
| c =
}}
{{eqn | r = e
| c =
}}
{{end-eqn}}
So the only conjugate of $e$ is $e$ itself.
Thus:
:$\Orb e = \set e$
and so:
:$\order {\Orb e} = 1$
From the Orbit-Stabilizer Theorem, it follows immediately that:
:$\Sta... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
For the [[Definition:Conjugacy Action|conjugacy action]]:
:$\order {\Orb e} = 1$
and thus:
:$\Stab e = G$ | {{begin-eqn}}
{{eqn | l = g * e
| r = g e g^{-1}
| c =
}}
{{eqn | r = g g^{-1}
| c =
}}
{{eqn | r = e
| c =
}}
{{end-eqn}}
So the only [[Definition:Conjugate of Group Element|conjugate]] of $e$ is $e$ itself.
Thus:
:$\Orb e = \set e$
and so:
:$\order {\Orb e} = 1$
From the [[Orbit-Stabili... | Conjugacy Action on Identity | https://proofwiki.org/wiki/Conjugacy_Action_on_Identity | https://proofwiki.org/wiki/Conjugacy_Action_on_Identity | [
"Conjugacy Action"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Conjugacy Action"
] | [
"Definition:Conjugate (Group Theory)/Element",
"Orbit-Stabilizer Theorem"
] |
proofwiki-1205 | Cauchy's Lemma (Group Theory) | Let $\struct {G, \circ}$ be a group of finite order whose identity is $e$.
Let $p$ be a prime number which divides the order of $G$.
Then $\struct {G, \circ}$ has an element of order $p$. | Let $\order G = n$ such that $p \divides n$.
Let:
:$X = \set {\tuple {a_1, a_2, \ldots, a_p} \in G^p: a_1 a_2 \cdots a_p = e}$
where $G^p$ is the cartesian product $\underbrace {G \times G \times \cdots \times G}_p$.
The first $p - 1$ coordinates of an element of $X$ can be chosen arbitrarily.
The last coordinate is de... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] of [[Definition:Order of Structure|finite order]] whose [[Definition:Identity Element|identity]] is $e$.
Let $p$ be a [[Definition:Prime Number|prime number]] which [[Definition:Divisor of Integer|divides]] the [[Definition:Order of Structure|order]] of $G$.
T... | Let $\order G = n$ such that $p \divides n$.
Let:
:$X = \set {\tuple {a_1, a_2, \ldots, a_p} \in G^p: a_1 a_2 \cdots a_p = e}$
where $G^p$ is the [[Definition:Cartesian Space|cartesian product]] $\underbrace {G \times G \times \cdots \times G}_p$.
The first $p - 1$ [[Definition:Coordinate of Ordered Tuple|coordinate... | Cauchy's Lemma (Group Theory)/Proof 1 | https://proofwiki.org/wiki/Cauchy's_Lemma_(Group_Theory) | https://proofwiki.org/wiki/Cauchy's_Lemma_(Group_Theory)/Proof_1 | [
"Cauchy's Lemma (Group Theory)",
"Group Theory"
] | [
"Definition:Group",
"Definition:Order of Structure",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure",
"Definition:Element",
"Definition:Order of Group Element"
] | [
"Definition:Cartesian Product/Cartesian Space",
"Definition:Cartesian Product/Coordinate",
"Definition:Element",
"Definition:Cartesian Product/Coordinate",
"Product Rule for Counting",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Cyclic Group/Generator",
"Definition:Elemen... |
proofwiki-1206 | Cauchy's Lemma (Group Theory) | Let $\struct {G, \circ}$ be a group of finite order whose identity is $e$.
Let $p$ be a prime number which divides the order of $G$.
Then $\struct {G, \circ}$ has an element of order $p$. | By {{Corollary|First Sylow Theorem}}, $G$ has subgroups of order $p^r$ for all $r$ such that $p^r \divides \order G$.
Thus $G$ has at least one subgroup $H$ of order $p$.
As a Prime Group is Cyclic, $H$ is a cyclic group.
Thus by definition $H$ has an element of order $p$.
Hence the result.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] of [[Definition:Order of Structure|finite order]] whose [[Definition:Identity Element|identity]] is $e$.
Let $p$ be a [[Definition:Prime Number|prime number]] which [[Definition:Divisor of Integer|divides]] the [[Definition:Order of Structure|order]] of $G$.
T... | By {{Corollary|First Sylow Theorem}}, $G$ has [[Definition:Subgroup|subgroups]] of [[Definition:Order of Group|order]] $p^r$ for all $r$ such that $p^r \divides \order G$.
Thus $G$ has at least one [[Definition:Subgroup|subgroup]] $H$ of [[Definition:Order of Group|order]] $p$.
As a [[Prime Group is Cyclic]], $H$ is ... | Cauchy's Lemma (Group Theory)/Proof 2 | https://proofwiki.org/wiki/Cauchy's_Lemma_(Group_Theory) | https://proofwiki.org/wiki/Cauchy's_Lemma_(Group_Theory)/Proof_2 | [
"Cauchy's Lemma (Group Theory)",
"Group Theory"
] | [
"Definition:Group",
"Definition:Order of Structure",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure",
"Definition:Element",
"Definition:Order of Group Element"
] | [
"Definition:Subgroup",
"Definition:Order of Structure",
"Definition:Subgroup",
"Definition:Order of Structure",
"Prime Group is Cyclic",
"Definition:Cyclic Group",
"Definition:Element",
"Definition:Order of Group Element"
] |
proofwiki-1207 | Transposition is Self-Inverse | All transpositions are self-inverse. | Let $\pi = \begin{bmatrix} k_1 & k_2 \end{bmatrix}$ be a transposition.
Writing $\pi \pi$ in cycle notation gives:
:$\begin{bmatrix} k_1 & k_2 \end{bmatrix} \begin{bmatrix} k_1 & k_2 \end{bmatrix}$
from which we see that $k_1 \to k_2 \to k_1$ and $k_2 \to k_1 \to k_2$.
The result follows from the definition of self-inv... | All [[Definition:Transposition|transpositions]] are [[Definition:Self-Inverse Element|self-inverse]]. | Let $\pi = \begin{bmatrix} k_1 & k_2 \end{bmatrix}$ be a [[Definition:Transposition|transposition]].
Writing $\pi \pi$ in [[Definition:Permutation on n Letters/Cycle Notation|cycle notation]] gives:
:$\begin{bmatrix} k_1 & k_2 \end{bmatrix} \begin{bmatrix} k_1 & k_2 \end{bmatrix}$
from which we see that $k_1 \to k_2 ... | Transposition is Self-Inverse | https://proofwiki.org/wiki/Transposition_is_Self-Inverse | https://proofwiki.org/wiki/Transposition_is_Self-Inverse | [
"Symmetric Groups"
] | [
"Definition:Transposition",
"Definition:Self-Inverse Element"
] | [
"Definition:Transposition",
"Definition:Permutation on n Letters/Cycle Notation",
"Definition:Self-Inverse Element",
"Category:Symmetric Groups"
] |
proofwiki-1208 | Conjugates of Transpositions | Let $k_1, k_2, k_3 \in \set {1, 2, \ldots, n}$.
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = \begin{pmatrix} k_1 & k_2 \end{pmatrix}
| r = \begin{pmatrix} k_3 & k_2 \end{pmatrix} \begin{pmatrix} k_1 & k_3 \end{pmatrix} \begin{pmatrix} k_3 & k_2 \end{pmatrix}
}}
{{eqn | n = 2
| l = \begin{pmatrix} k_1 & k_2... | $(1):$ Calculating the product of $\begin{pmatrix} k_3 & k_2 \end{pmatrix} \begin{pmatrix} k_1 & k_3 \end{pmatrix} \begin{pmatrix} k_3 & k_2 \end{pmatrix}$:
:: $k_1 \to k_3 \to k_2$
:: $k_2 \to k_3 \to k_1$
:: $k_3 \to k_2 \to k_3$
hence the result.
{{qed|lemma}}
$(2):$ Calculating the product of $\begin{pmatrix} k_3 &... | Let $k_1, k_2, k_3 \in \set {1, 2, \ldots, n}$.
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = \begin{pmatrix} k_1 & k_2 \end{pmatrix}
| r = \begin{pmatrix} k_3 & k_2 \end{pmatrix} \begin{pmatrix} k_1 & k_3 \end{pmatrix} \begin{pmatrix} k_3 & k_2 \end{pmatrix}
}}
{{eqn | n = 2
| l = \begin{pmatrix} k_1 & k_... | $(1):$ Calculating the product of $\begin{pmatrix} k_3 & k_2 \end{pmatrix} \begin{pmatrix} k_1 & k_3 \end{pmatrix} \begin{pmatrix} k_3 & k_2 \end{pmatrix}$:
:: $k_1 \to k_3 \to k_2$
:: $k_2 \to k_3 \to k_1$
:: $k_3 \to k_2 \to k_3$
hence the result.
{{qed|lemma}}
$(2):$ Calculating the product of $\begin{pmatrix} k... | Conjugates of Transpositions | https://proofwiki.org/wiki/Conjugates_of_Transpositions | https://proofwiki.org/wiki/Conjugates_of_Transpositions | [
"Symmetric Groups",
"Conjugacy",
"Transpositions"
] | [] | [
"Category:Symmetric Groups",
"Category:Conjugacy",
"Category:Transpositions"
] |
proofwiki-1209 | K-Cycle can be Factored into Transpositions | Every $k$-cycle can be factorised into the product of $k - 1$ transpositions.
This factorisation is not unique. | The cycle $\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix}$ has the factorisation:
:$\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix} = \begin{pmatrix} 1 & k \end{pmatrix} \ldots \begin{pmatrix} 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix}$
Therefore, the general $k$-cycle $\begin{pmatrix} i_1 & i_2 & \ldot... | Every [[Definition:Cyclic Permutation|$k$-cycle]] can be factorised into the product of $k - 1$ [[Definition:Transposition|transpositions]].
This factorisation is not unique. | The cycle $\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix}$ has the factorisation:
:$\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix} = \begin{pmatrix} 1 & k \end{pmatrix} \ldots \begin{pmatrix} 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix}$
Therefore, the general [[Definition:Cyclic Permutation|$k$-cycle... | K-Cycle can be Factored into Transpositions | https://proofwiki.org/wiki/K-Cycle_can_be_Factored_into_Transpositions | https://proofwiki.org/wiki/K-Cycle_can_be_Factored_into_Transpositions | [
"Cyclic Permutations"
] | [
"Definition:Cyclic Permutation",
"Definition:Transposition"
] | [
"Definition:Cyclic Permutation",
"Definition:Cyclic Permutation"
] |
proofwiki-1210 | Sign of Permutation is Plus or Minus Unity | Let $n \in \N$ be a natural number.
Let $\N_n$ denote the set of natural numbers $\set {1, 2, \ldots, n}$.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\sequence {x_k}_{k \mathop \in \N_n}$ be a finite sequence in $\R$.
Let $\pi \in S_n$.
Let $\map {\Delta_n} {x_1, x_2, \ldots, x_n}$ be the product of diff... | If $\exists i, j \in \N_n$ such that $x_i = x_j$, then $\map {\Delta_n} {x_1, x_2, \ldots, x_n} = 0$ and the result follows trivially.
So, suppose all the elements $x_k$ are distinct.
Let us use $\Delta_n$ to denote $\map {\Delta_n} {x_1, x_2, \ldots, x_n}$.
Let $1 \le a < b \le n$.
Then $x_a - x_b$ is a divisor of $\D... | Let $n \in \N$ be a [[Definition:Natural Number|natural number]].
Let $\N_n$ denote the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\set {1, 2, \ldots, n}$.
Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $\sequence {x_k}_{k \mathop \... | If $\exists i, j \in \N_n$ such that $x_i = x_j$, then $\map {\Delta_n} {x_1, x_2, \ldots, x_n} = 0$ and the result follows trivially.
So, suppose all the elements $x_k$ are [[Definition:Distinct|distinct]].
Let us use $\Delta_n$ to denote $\map {\Delta_n} {x_1, x_2, \ldots, x_n}$.
Let $1 \le a < b \le n$.
Then $x... | Sign of Permutation is Plus or Minus Unity | https://proofwiki.org/wiki/Sign_of_Permutation_is_Plus_or_Minus_Unity | https://proofwiki.org/wiki/Sign_of_Permutation_is_Plus_or_Minus_Unity | [
"Symmetric Groups",
"Sign of Permutation"
] | [
"Definition:Natural Numbers",
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Symmetric Group/n Letters",
"Definition:Finite Sequence",
"Definition:Real Sequence",
"Definition:Product of Differences",
"Definition:Sign of Permutation on n Letters"
] | [
"Definition:Distinct",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-1211 | Parity Function is Homomorphism | Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$.
Let $\map \sgn \pi$ be the sign of $\pi$.
Let the parity function of $\pi$ be defined as:
:Parity of $\pi = \begin {cases} \mathrm {Even} & : \map \sgn \pi = 1 \\ \mathrm {Odd} & : \map \sgn \pi = -1 \end {cases}$
The mapping $\sgn: S_n \to C_2$, w... | We need to show that:
:$\forall \pi, \rho \in S_n: \map \sgn \pi \, \map \sgn \rho = \map \sgn {\pi \rho}$
Let $\Delta_n$ be an arbitrary product of differences.
{{begin-eqn}}
{{eqn | l = \map \sgn {\pi \rho} \Delta_n
| r = \pi \rho \cdot \Delta_n
| c = {{Defof|Sign of Permutation}}
}}
{{eqn | r = \pi \cdot... | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $\pi \in S_n$.
Let $\map \sgn \pi$ be the [[Definition:Sign of Permutation on n Letters|sign of $\pi$]].
Let the [[Definition:Parity of Permutation|parity function]] of $\pi$ be defined as:
:Parity of $\pi = \begin ... | We need to show that:
:$\forall \pi, \rho \in S_n: \map \sgn \pi \, \map \sgn \rho = \map \sgn {\pi \rho}$
Let $\Delta_n$ be an arbitrary [[Definition:Product of Differences|product of differences]].
{{begin-eqn}}
{{eqn | l = \map \sgn {\pi \rho} \Delta_n
| r = \pi \rho \cdot \Delta_n
| c = {{Defof|Sign o... | Parity Function is Homomorphism | https://proofwiki.org/wiki/Parity_Function_is_Homomorphism | https://proofwiki.org/wiki/Parity_Function_is_Homomorphism | [
"Symmetric Groups"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Sign of Permutation on n Letters",
"Definition:Parity of Permutation",
"Definition:Mapping",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Group Homomorphism"
] | [
"Definition:Product of Differences",
"Permutation on Polynomial is Group Action",
"Permutation on Polynomial is Group Action",
"Definition:Parity Group"
] |
proofwiki-1212 | Permutation on Polynomial is Group Action | Let $n \in \Z: n > 0$.
Let $F_n$ be the set of all polynomials in $n$ variables $x_1, x_2, \ldots, x_n$:
:$F = \set {\map f {x_1, x_2, \ldots, x_n}: f \text{ is a polynomial in $n$ variables} }$
Let $S_n$ denote the symmetric group on $n$ letters.
Let $*: S_n \times F \to F$ be the mapping defined as:
:$\forall \pi \in... | Let $\pi, \rho \in S_n$.
Let $\pi * f$ be the permutation on the polynomial $f$ by $\pi$.
Let $e \in S_n$ be the identity of $S_n$.
From Symmetric Group is Group:
:$e * f = f$
thus fulfilling {{GroupActionAxiom|1}}.
Then we have that:
{{begin-eqn}}
{{eqn | l = \paren {\pi \circ \rho} * f
| r = \map \pi {\rho * f}... | Let $n \in \Z: n > 0$.
Let $F_n$ be the set of all [[Definition:Polynomial|polynomials]] in $n$ variables $x_1, x_2, \ldots, x_n$:
:$F = \set {\map f {x_1, x_2, \ldots, x_n}: f \text{ is a polynomial in $n$ variables} }$
Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
... | Let $\pi, \rho \in S_n$.
Let $\pi * f$ be the [[Definition:Permutation on Polynomial|permutation on the polynomial]] $f$ by $\pi$.
Let $e \in S_n$ be the [[Definition:Identity Element|identity]] of $S_n$.
From [[Symmetric Group is Group]]:
:$e * f = f$
thus fulfilling {{GroupActionAxiom|1}}.
Then we have that:
{... | Permutation on Polynomial is Group Action | https://proofwiki.org/wiki/Permutation_on_Polynomial_is_Group_Action | https://proofwiki.org/wiki/Permutation_on_Polynomial_is_Group_Action | [
"Group Actions",
"Permutations"
] | [
"Definition:Polynomial",
"Definition:Symmetric Group/n Letters",
"Definition:Mapping",
"Definition:Group Action"
] | [
"Definition:Permutation on Polynomial",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Symmetric Group is Group"
] |
proofwiki-1213 | Parity of Inverse of Permutation | Let $S_n$ denote the symmetric group on $n$ letters.
Then:
:$\forall \pi \in S_n: \map \sgn \pi = \map \sgn {\pi^{-1} }$ | From Parity Function is Homomorphism:
:$\map \sgn {I_{S_n} } = 1$
Thus:
:$\pi \pi^{-1} = I_{S_n} \implies \map \sgn \pi \, \map \sgn {\pi^{-1} } = 1$
The result follows immediately.
{{qed}} | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Then:
:$\forall \pi \in S_n: \map \sgn \pi = \map \sgn {\pi^{-1} }$ | From [[Parity Function is Homomorphism]]:
:$\map \sgn {I_{S_n} } = 1$
Thus:
:$\pi \pi^{-1} = I_{S_n} \implies \map \sgn \pi \, \map \sgn {\pi^{-1} } = 1$
The result follows immediately.
{{qed}} | Parity of Inverse of Permutation | https://proofwiki.org/wiki/Parity_of_Inverse_of_Permutation | https://proofwiki.org/wiki/Parity_of_Inverse_of_Permutation | [
"Symmetric Groups"
] | [
"Definition:Symmetric Group/n Letters"
] | [
"Parity Function is Homomorphism"
] |
proofwiki-1214 | Parity of Conjugate of Permutation | Let $S_n$ denote the symmetric group on $n$ letters.
Then:
:$\forall \pi, \rho \in S_n: \map \sgn {\pi \rho \pi^{-1} } = \map \sgn \rho$
where $\map \sgn \pi$ is the sign of $\pi$. | As $\map \sgn \pi = \pm 1$ for any $\pi \in S_n$, we can apply the laws of commutativity and associativity:
{{begin-eqn}}
{{eqn | l = \map \sgn \pi \, \map \sgn \rho
| r = \map \sgn \rho \, \map \sgn \pi
| c =
}}
{{eqn | ll= \leadsto
| l = \map \sgn \pi \, \map \sgn \rho \, \map \sgn {\pi^{-1} }
... | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Then:
:$\forall \pi, \rho \in S_n: \map \sgn {\pi \rho \pi^{-1} } = \map \sgn \rho$
where $\map \sgn \pi$ is the [[Definition:Sign of Permutation|sign of $\pi$]]. | As $\map \sgn \pi = \pm 1$ for any $\pi \in S_n$, we can apply the laws of [[Definition:Commutative Operation|commutativity]] and [[Definition:Associative Operation|associativity]]:
{{begin-eqn}}
{{eqn | l = \map \sgn \pi \, \map \sgn \rho
| r = \map \sgn \rho \, \map \sgn \pi
| c =
}}
{{eqn | ll= \leadst... | Parity of Conjugate of Permutation | https://proofwiki.org/wiki/Parity_of_Conjugate_of_Permutation | https://proofwiki.org/wiki/Parity_of_Conjugate_of_Permutation | [
"Symmetric Groups"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Sign of Permutation"
] | [
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Parity Function is Homomorphism"
] |
proofwiki-1215 | Parity of K-Cycle | Let $\pi$ be a $k$-cycle.
Then:
:$\map \sgn \pi = \begin{cases} 1 & : k \ \text {odd} \\ -1 & : k \ \text {even} \end{cases}$
Thus:
:$\map \sgn \pi = \paren {-1}^{k - 1}$
or equivalently:
:$\map \sgn \pi = \paren {-1}^{k + 1}$ | From Transposition is of Odd Parity, any transposition is of odd parity.
From K-Cycle can be Factored into Transpositions, we see that a $k$-cycle is the product of $k - 1$ transpositions.
Thus $\pi$ is even {{iff}} $k$ is odd.
{{qed}} | Let $\pi$ be a [[Definition:Cyclic Permutation|$k$-cycle]].
Then:
:$\map \sgn \pi = \begin{cases} 1 & : k \ \text {odd} \\ -1 & : k \ \text {even} \end{cases}$
Thus:
:$\map \sgn \pi = \paren {-1}^{k - 1}$
or equivalently:
:$\map \sgn \pi = \paren {-1}^{k + 1}$ | From [[Transposition is of Odd Parity]], any [[Definition:Transposition|transposition]] is of [[Definition:Parity of Permutation|odd parity]].
From [[K-Cycle can be Factored into Transpositions]], we see that a [[Definition:Cyclic Permutation|$k$-cycle]] is the product of $k - 1$ [[Definition:Transposition|transpositi... | Parity of K-Cycle | https://proofwiki.org/wiki/Parity_of_K-Cycle | https://proofwiki.org/wiki/Parity_of_K-Cycle | [
"Cyclic Permutations"
] | [
"Definition:Cyclic Permutation"
] | [
"Transposition is of Odd Parity",
"Definition:Transposition",
"Definition:Parity of Permutation",
"K-Cycle can be Factored into Transpositions",
"Definition:Cyclic Permutation",
"Definition:Transposition",
"Definition:Odd Integer"
] |
proofwiki-1216 | Alternating Group is Normal Subgroup of Symmetric Group | Let $n \ge 2$ be a natural number.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $A_n$ be the alternating group on $n$ letters.
Then $A_n$ is a normal subgroup of $S_n$ whose index is $2$. | We have that $\map \sgn {S_n}$ is onto $C_2$.
Thus from the First Isomorphism Theorem, $A_n$ consists of the set of even permutations of $S_n$.
The result follows from Subgroup of Index 2 is Normal.
{{Qed}} | Let $n \ge 2$ be a [[Definition:Natural Number|natural number]].
Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $A_n$ be the [[Definition:Alternating Group|alternating group on $n$ letters]].
Then $A_n$ is a [[Definition:Normal Subgroup|normal subgroup]] of $S_n... | We have that $\map \sgn {S_n}$ is [[Definition:Surjection|onto]] $C_2$.
Thus from the [[First Isomorphism Theorem for Groups|First Isomorphism Theorem]], $A_n$ consists of the set of [[Definition:Even Permutation|even permutations]] of $S_n$.
The result follows from [[Subgroup of Index 2 is Normal]].
{{Qed}} | Alternating Group is Normal Subgroup of Symmetric Group | https://proofwiki.org/wiki/Alternating_Group_is_Normal_Subgroup_of_Symmetric_Group | https://proofwiki.org/wiki/Alternating_Group_is_Normal_Subgroup_of_Symmetric_Group | [
"Symmetric Groups",
"Alternating Groups"
] | [
"Definition:Natural Numbers",
"Definition:Symmetric Group/n Letters",
"Definition:Alternating Group",
"Definition:Normal Subgroup",
"Definition:Index of Subgroup"
] | [
"Definition:Surjection",
"First Isomorphism Theorem/Groups",
"Definition:Even Permutation",
"Subgroup of Index 2 is Normal"
] |
proofwiki-1217 | Group of Permutations either All or Half Even | Let $G$ be a group of permutations.
Then either ''exactly half'' of the permutations in $G$ are even, or they are ''all'' even. | From Parity Function is Homomorphism, the mapping:
:$\sgn : G \to \Z_2$
is a homomorphism.
Then:
:$G / \map \ker \sgn \cong \Img \sgn$
from the First Isomorphism Theorem.
The only possibilities for $\Img \sgn$ are $\set 0$ or $\Z_2$.
So either:
:$\order {G / \map \ker \sgn} = 1$
in which case all the permutations of $G... | Let $G$ be a [[Definition:Group|group]] of [[Definition:Permutation|permutations]].
Then either ''exactly half'' of the permutations in $G$ are [[Definition:Even Permutation|even]], or they are ''all'' [[Definition:Even Permutation|even]]. | From [[Parity Function is Homomorphism]], the [[Definition:Mapping|mapping]]:
:$\sgn : G \to \Z_2$
is a [[Definition:Group Homomorphism|homomorphism]].
Then:
:$G / \map \ker \sgn \cong \Img \sgn$
from the [[First Isomorphism Theorem for Groups|First Isomorphism Theorem]].
The only possibilities for $\Img \sgn$ are $\... | Group of Permutations either All or Half Even | https://proofwiki.org/wiki/Group_of_Permutations_either_All_or_Half_Even | https://proofwiki.org/wiki/Group_of_Permutations_either_All_or_Half_Even | [
"Symmetric Groups"
] | [
"Definition:Group",
"Definition:Permutation",
"Definition:Even Permutation",
"Definition:Even Permutation"
] | [
"Parity Function is Homomorphism",
"Definition:Mapping",
"Definition:Group Homomorphism",
"First Isomorphism Theorem/Groups",
"Definition:Permutation",
"Definition:Even Permutation",
"Category:Symmetric Groups"
] |
proofwiki-1218 | Cycle Decomposition of Conjugate | Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi, \rho \in S_n$.
The cycle decomposition of the permutation $\pi \rho \pi^{-1}$ can be obtained from that of $\rho$ by replacing each $i$ in the cycle decomposition of $\rho$ with $\map \pi i$. | Consider the effect of $\pi \rho \pi^{-1}$ on $\map \pi i$:
:$\map {\pi \rho \pi^{-1} } {\map \pi i} = \map \pi {\map \rho i}$
That is:
:$\pi \rho \pi^{-1}$ maps $\map \pi i$ to $\map \pi {\map \rho i}$
In the cycle decomposition of $\pi \rho \pi^{-1}$, $\map \pi i$ lies to the left of $\map \pi {\map \rho i}$, whereas... | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $\pi, \rho \in S_n$.
The [[Definition:Cycle Decomposition|cycle decomposition]] of the [[Definition:Permutation|permutation]] $\pi \rho \pi^{-1}$ can be obtained from that of $\rho$ by replacing each $i$ in the [[Def... | Consider the effect of $\pi \rho \pi^{-1}$ on $\map \pi i$:
:$\map {\pi \rho \pi^{-1} } {\map \pi i} = \map \pi {\map \rho i}$
That is:
:$\pi \rho \pi^{-1}$ maps $\map \pi i$ to $\map \pi {\map \rho i}$
In the [[Definition:Cycle Decomposition|cycle decomposition]] of $\pi \rho \pi^{-1}$, $\map \pi i$ lies to the le... | Cycle Decomposition of Conjugate | https://proofwiki.org/wiki/Cycle_Decomposition_of_Conjugate | https://proofwiki.org/wiki/Cycle_Decomposition_of_Conjugate | [
"Symmetric Groups",
"Conjugacy",
"Permutations",
"Cycle Decompositions"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Cycle Decomposition",
"Definition:Permutation",
"Definition:Cycle Decomposition"
] | [
"Definition:Cycle Decomposition",
"Definition:Cycle Decomposition"
] |
proofwiki-1219 | Conjugate Permutations have Same Cycle Type | Let $n \ge 1$ be a natural number.
Let $G$ be a subgroup of the symmetric group on $n$ letters $S_n$.
Let $\sigma, \rho \in G$.
Then $\sigma$ and $\rho$ are conjugate {{iff}} they have the same cycle type. | Let $\sigma \in G$ have cycle type $\tuple {k_1, k_2, \ldots, k_n}$.
From Existence and Uniqueness of Cycle Decomposition, $\sigma$ can be expressed uniquely as the product of disjoint cycles:
:$\sigma = \alpha_1 \alpha_2 \dotsm \alpha_l$
where $\alpha_i$ is a $k_i$-cycle for some $i\in \set {1, 2, \ldots, l}$.
Let $\t... | Let $n \ge 1$ be a [[Definition:Natural Number|natural number]].
Let $G$ be a [[Definition:Subgroup|subgroup]] of the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]] $S_n$.
Let $\sigma, \rho \in G$.
Then $\sigma$ and $\rho$ are [[Definition:Conjugate of Group Element|conjugate]] {{iff}} t... | Let $\sigma \in G$ have [[Definition:Cycle Type|cycle type]] $\tuple {k_1, k_2, \ldots, k_n}$.
From [[Existence and Uniqueness of Cycle Decomposition]], $\sigma$ can be expressed uniquely as the [[Definition:Product Element|product]] of [[Definition:Disjoint Permutations|disjoint]] [[Definition:Cyclic Permutation|cycl... | Conjugate Permutations have Same Cycle Type | https://proofwiki.org/wiki/Conjugate_Permutations_have_Same_Cycle_Type | https://proofwiki.org/wiki/Conjugate_Permutations_have_Same_Cycle_Type | [
"Conjugacy",
"Permutations"
] | [
"Definition:Natural Numbers",
"Definition:Subgroup",
"Definition:Symmetric Group/n Letters",
"Definition:Conjugate (Group Theory)/Element",
"Definition:Cycle Type"
] | [
"Definition:Cycle Type",
"Existence and Uniqueness of Cycle Decomposition",
"Definition:Group Product/Product Element",
"Definition:Disjoint Permutations",
"Definition:Cyclic Permutation",
"Definition:Cyclic Permutation",
"Product of Conjugates equals Conjugate of Products",
"Cycle Decomposition of Co... |
proofwiki-1220 | Transpositions of Adjacent Elements generate Symmetric Group | Let $n \in \Z: n > 1$.
Let $S_n$ denote the symmetric group on $n$ letters.
Then the transpositions $a_k = \begin{pmatrix} k & k + 1 \end{pmatrix}$ for $1 \le k < n$ are a set of generators for $S_n$.
They satisfy the relations:
{{begin-eqn}}
{{eqn | l = a_k^2
| r = e
| c = (for $1 \le k < n$)
}}
{{eqn | l ... | First, we show that each $\begin{pmatrix} i & j \end{pmatrix}$ where $i < j$ is in the subgroup $\gen {a_1, a_2, \ldots, a_{n - 1} }$.
From Cycle Decomposition of Conjugate, we can conjugate $a_i$ by $a_{i + 1}$ to give:
:$\begin{pmatrix} i & i + 2 \end{pmatrix}$
Conjugating $a_i$ by the product $a_{j - 1} a_{j - 2} \l... | Let $n \in \Z: n > 1$.
Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Then the [[Definition:Transposition|transpositions]] $a_k = \begin{pmatrix} k & k + 1 \end{pmatrix}$ for $1 \le k < n$ are a [[Definition:Generator of Group|set of generators]] for $S_n$.
They sat... | First, we show that each $\begin{pmatrix} i & j \end{pmatrix}$ where $i < j$ is in the [[Definition:Subgroup|subgroup]] $\gen {a_1, a_2, \ldots, a_{n - 1} }$.
From [[Cycle Decomposition of Conjugate]], we can [[Definition:Conjugate of Group Element|conjugate]] $a_i$ by $a_{i + 1}$ to give:
:$\begin{pmatrix} i & i + 2... | Transpositions of Adjacent Elements generate Symmetric Group | https://proofwiki.org/wiki/Transpositions_of_Adjacent_Elements_generate_Symmetric_Group | https://proofwiki.org/wiki/Transpositions_of_Adjacent_Elements_generate_Symmetric_Group | [
"Symmetric Groups",
"Examples of Generators of Groups"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Transposition",
"Definition:Generator of Group"
] | [
"Definition:Subgroup",
"Cycle Decomposition of Conjugate",
"Definition:Conjugate (Group Theory)/Element",
"Definition:Conjugate (Group Theory)/Element",
"Definition:Group Product/Product Element",
"K-Cycle can be Factored into Transpositions",
"Definition:Cyclic Permutation",
"Definition:Group Product... |
proofwiki-1221 | Hermitian Matrix has Real Eigenvalues | Every Hermitian matrix has eigenvalues which are all real numbers. | Let $\mathbf A$ be a Hermitian matrix.
Then, by definition:
:$\mathbf A = \mathbf A^\dagger$
where $\mathbf A^\dagger$ denotes the Hermitian conjugate of $\mathbf A$.
Let $\lambda$ be an eigenvalue of $\mathbf A$.
Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$.
Let $\mathbf v$ be represente... | Every [[Definition:Hermitian Matrix|Hermitian matrix]] has [[Definition:Eigenvalue of Square Matrix|eigenvalues]] which are all [[Definition:Real Number|real numbers]]. | Let $\mathbf A$ be a [[Definition:Hermitian Matrix|Hermitian matrix]].
Then, by definition:
:$\mathbf A = \mathbf A^\dagger$
where $\mathbf A^\dagger$ denotes the [[Definition:Hermitian Conjugate|Hermitian conjugate]] of $\mathbf A$.
Let $\lambda$ be an [[Definition:Eigenvalue of Square Matrix|eigenvalue]] of $\math... | Hermitian Matrix has Real Eigenvalues/Proof 1 | https://proofwiki.org/wiki/Hermitian_Matrix_has_Real_Eigenvalues | https://proofwiki.org/wiki/Hermitian_Matrix_has_Real_Eigenvalues/Proof_1 | [
"Hermitian Matrix has Real Eigenvalues",
"Hermitian Matrices",
"Eigenvalues of Square Matrices"
] | [
"Definition:Hermitian Matrix",
"Definition:Eigenvalue/Square Matrix",
"Definition:Real Number"
] | [
"Definition:Hermitian Matrix",
"Definition:Hermitian Conjugate",
"Definition:Eigenvalue/Square Matrix",
"Definition:Eigenvector/Square Matrix",
"Definition:Eigenvalue/Square Matrix",
"Definition:Column Matrix",
"Definition:Eigenvector/Square Matrix",
"Definition:Matrix Product (Conventional)",
"Defi... |
proofwiki-1222 | Hermitian Matrix has Real Eigenvalues | Every Hermitian matrix has eigenvalues which are all real numbers. | The result immediately follows from Hermitian Matrix is Hermitian Operator and Eigenvalues of Hermitian Operator are Real.
{{qed}} | Every [[Definition:Hermitian Matrix|Hermitian matrix]] has [[Definition:Eigenvalue of Square Matrix|eigenvalues]] which are all [[Definition:Real Number|real numbers]]. | The result immediately follows from [[Hermitian Matrix is Hermitian Operator]] and [[Eigenvalues of Hermitian Operator are Real]].
{{qed}} | Hermitian Matrix has Real Eigenvalues/Proof 2 | https://proofwiki.org/wiki/Hermitian_Matrix_has_Real_Eigenvalues | https://proofwiki.org/wiki/Hermitian_Matrix_has_Real_Eigenvalues/Proof_2 | [
"Hermitian Matrix has Real Eigenvalues",
"Hermitian Matrices",
"Eigenvalues of Square Matrices"
] | [
"Definition:Hermitian Matrix",
"Definition:Eigenvalue/Square Matrix",
"Definition:Real Number"
] | [
"Hermitian Matrix is Hermitian Operator",
"Eigenvalues of Hermitian Operator are Real"
] |
proofwiki-1223 | Hermitian Matrix has Real Eigenvalues | Every Hermitian matrix has eigenvalues which are all real numbers. | Let $\mathbf A$ be a Hermitian matrix.
Then, by definition, $\mathbf A = \mathbf A^\dagger$, where $^\dagger$ designates its Hermitian conjugate.
Let $\lambda$ be an eigenvalue of $\mathbf A$.
Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$ of $\mathbf A$.
Denote with $\innerprod \cdot \cdot... | Every [[Definition:Hermitian Matrix|Hermitian matrix]] has [[Definition:Eigenvalue of Square Matrix|eigenvalues]] which are all [[Definition:Real Number|real numbers]]. | Let $\mathbf A$ be a [[Definition:Hermitian Matrix|Hermitian matrix]].
Then, by definition, $\mathbf A = \mathbf A^\dagger$, where $^\dagger$ designates its [[Definition:Hermitian Conjugate|Hermitian conjugate]].
Let $\lambda$ be an [[Definition:Eigenvalue of Linear Operator|eigenvalue]] of $\mathbf A$.
Let $\mathb... | Hermitian Matrix has Real Eigenvalues/Proof 3 | https://proofwiki.org/wiki/Hermitian_Matrix_has_Real_Eigenvalues | https://proofwiki.org/wiki/Hermitian_Matrix_has_Real_Eigenvalues/Proof_3 | [
"Hermitian Matrix has Real Eigenvalues",
"Hermitian Matrices",
"Eigenvalues of Square Matrices"
] | [
"Definition:Hermitian Matrix",
"Definition:Eigenvalue/Square Matrix",
"Definition:Real Number"
] | [
"Definition:Hermitian Matrix",
"Definition:Hermitian Conjugate",
"Definition:Eigenvalue/Linear Operator",
"Definition:Eigenvector/Linear Operator",
"Definition:Eigenvalue/Linear Operator",
"Definition:Inner Product",
"Axiom:Complex Inner Product Axioms",
"Hermitian Matrix is Hermitian Operator",
"Ax... |
proofwiki-1224 | Permutation of Cosets | Let $G$ be a group and let $H \le G$.
Let $\mathbb S$ be the set of all distinct left cosets of $H$ in $G$.
Then:
:$(1): \quad$ For any $g \in G$, the mapping $\theta_g: \mathbb S \to \mathbb S$ defined by $\map {\theta_g} {x H} = g x H$ is a permutation of $\mathbb S$.
:$(2): \quad$ The mapping $\theta$ defined by $\m... | First we need to show that $\theta_g$ is well-defined and injective.
{{begin-eqn}}
{{eqn | l = x H
| r = y H
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y^{-1} x
| o = \in
| r = H
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {g y}^{-1} g x
| r = y^{-1} x \in H
... | Let $G$ be a [[Definition:Group|group]] and let $H \le G$.
Let $\mathbb S$ be the set of all distinct [[Definition:Left Coset|left cosets]] of $H$ in $G$.
Then:
:$(1): \quad$ For any $g \in G$, the [[Definition:Mapping|mapping]] $\theta_g: \mathbb S \to \mathbb S$ defined by $\map {\theta_g} {x H} = g x H$ is a [[Def... | First we need to show that $\theta_g$ is [[Definition:Well-Defined Mapping|well-defined]] and [[Definition:Injection|injective]].
{{begin-eqn}}
{{eqn | l = x H
| r = y H
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y^{-1} x
| o = \in
| r = H
| c =
}}
{{eqn | ll= \leadstoandfrom
... | Permutation of Cosets | https://proofwiki.org/wiki/Permutation_of_Cosets | https://proofwiki.org/wiki/Permutation_of_Cosets | [
"Symmetric Groups",
"Cosets",
"Permutations"
] | [
"Definition:Group",
"Definition:Coset/Left Coset",
"Definition:Mapping",
"Definition:Permutation",
"Definition:Group Homomorphism",
"Definition:Symmetric Group",
"Definition:Kernel of Group Homomorphism",
"Definition:Subgroup"
] | [
"Definition:Well-Defined/Mapping",
"Definition:Injection",
"Definition:Well-Defined/Mapping",
"Definition:Injection",
"Definition:Surjection",
"Definition:Well-Defined/Mapping",
"Definition:Bijection",
"Definition:Permutation",
"Definition:Group Homomorphism"
] |
proofwiki-1225 | Cayley's Representation Theorem | Let $S_n$ denote the symmetric group on $n$ letters.
Every finite group is isomorphic to a subgroup of $S_n$ for some $n \in \Z$. | Let $G$ be a group and let $a \in G$.
Consider the left regular representation $\lambda_a: G \to G$ defined as:
:$\map {\lambda_a} x = a \cdot x$
From Regular Representations in Group are Permutations we have that $\lambda_a$ is a permutation.
Now let $b \in G$ and consider $\lambda_b: G \to G$ defined as:
:$\map {\lam... | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Every finite [[Definition:Group|group]] is [[Definition:Group Isomorphism|isomorphic]] to a [[Definition:Subgroup|subgroup]] of $S_n$ for some $n \in \Z$. | Let $G$ be a [[Definition:Group|group]] and let $a \in G$.
Consider the [[Definition:Left Regular Representation|left regular representation]] $\lambda_a: G \to G$ defined as:
:$\map {\lambda_a} x = a \cdot x$
From [[Regular Representations in Group are Permutations]] we have that $\lambda_a$ is a [[Definition:Permu... | Cayley's Representation Theorem/General Case/Proof 1 | https://proofwiki.org/wiki/Cayley's_Representation_Theorem | https://proofwiki.org/wiki/Cayley's_Representation_Theorem/General_Case/Proof_1 | [
"Cayley's Representation Theorem",
"Symmetric Groups",
"Group Theory",
"Subgroups",
"Representation Theorems",
"Cayley's Theorems"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Subgroup"
] | [
"Definition:Group",
"Definition:Regular Representations/Left Regular Representation",
"Regular Representations in Group are Permutations",
"Definition:Permutation",
"Cancellation Laws",
"Definition:Bijection",
"Composition of Regular Representations",
"Definition:Composition of Mappings",
"Definitio... |
proofwiki-1226 | Cayley's Representation Theorem | Let $S_n$ denote the symmetric group on $n$ letters.
Every finite group is isomorphic to a subgroup of $S_n$ for some $n \in \Z$. | We interpret $\struct {G,\cdot}$ as an group category:
{{DefineCategory
| ob = Only one, say $*$
| mor = $a: * \to *$ for all $a \in G$
| comp = $a \cdot b: * \to *$
| id = $e: * \to *$
}}
where $e$ is the identity element of $G$.
In particular, $G$ is a small category.
By Cayley's Theorem, $G$ is isomorphic to a sub... | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Every finite [[Definition:Group|group]] is [[Definition:Group Isomorphism|isomorphic]] to a [[Definition:Subgroup|subgroup]] of $S_n$ for some $n \in \Z$. | We interpret $\struct {G,\cdot}$ as an [[Definition:Group Category|group category]]:
{{DefineCategory
| ob = Only one, say $*$
| mor = $a: * \to *$ for all $a \in G$
| comp = $a \cdot b: * \to *$
| id = $e: * \to *$
}}
where $e$ is the [[Definition:Identity Element|identity element]] of $G$.
In particular, $G$ is a ... | Cayley's Representation Theorem/General Case/Proof 2 | https://proofwiki.org/wiki/Cayley's_Representation_Theorem | https://proofwiki.org/wiki/Cayley's_Representation_Theorem/General_Case/Proof_2 | [
"Cayley's Representation Theorem",
"Symmetric Groups",
"Group Theory",
"Subgroups",
"Representation Theorems",
"Cayley's Theorems"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Subgroup"
] | [
"Definition:Group Category",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Small Category",
"Cayley's Theorem (Category Theory)",
"Definition:Isomorphism of Categories/Isomorphic Categories",
"Definition:Subcategory",
"Definition:Functor/Covariant",
"Definition:Identity Func... |
proofwiki-1227 | Cayley's Representation Theorem | Let $S_n$ denote the symmetric group on $n$ letters.
Every finite group is isomorphic to a subgroup of $S_n$ for some $n \in \Z$. | Let $H = \set e$.
We can apply Permutation of Cosets to $H$ so that:
:$\mathbb S = G$
and:
:$\map \ker \theta = \set e$
The result follows by the First Isomorphism Theorem for Groups.
{{Qed}} | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Every finite [[Definition:Group|group]] is [[Definition:Group Isomorphism|isomorphic]] to a [[Definition:Subgroup|subgroup]] of $S_n$ for some $n \in \Z$. | Let $H = \set e$.
We can apply [[Permutation of Cosets]] to $H$ so that:
:$\mathbb S = G$
and:
:$\map \ker \theta = \set e$
The result follows by the [[First Isomorphism Theorem for Groups]].
{{Qed}} | Cayley's Representation Theorem/Proof 1 | https://proofwiki.org/wiki/Cayley's_Representation_Theorem | https://proofwiki.org/wiki/Cayley's_Representation_Theorem/Proof_1 | [
"Cayley's Representation Theorem",
"Symmetric Groups",
"Group Theory",
"Subgroups",
"Representation Theorems",
"Cayley's Theorems"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Subgroup"
] | [
"Permutation of Cosets",
"First Isomorphism Theorem/Groups"
] |
proofwiki-1228 | Cayley's Representation Theorem | Let $S_n$ denote the symmetric group on $n$ letters.
Every finite group is isomorphic to a subgroup of $S_n$ for some $n \in \Z$. | Let $G$ be any arbitrary finite group whose identity is $e_G$.
Let $S$ be the symmetric group on the elements of $G$, where $e_S$ is the identity of $S$.
For any $x \in G$, let $\lambda_x$ be the left regular representation of $G$ with respect to $x$.
From Regular Representations in Group are Permutations, $\forall x \... | Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Every finite [[Definition:Group|group]] is [[Definition:Group Isomorphism|isomorphic]] to a [[Definition:Subgroup|subgroup]] of $S_n$ for some $n \in \Z$. | Let $G$ be any arbitrary [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e_G$.
Let $S$ be the [[Definition:Symmetric Group|symmetric group]] on the elements of $G$, where $e_S$ is the [[Definition:Identity Element|identity]] of $S$.
For any $x \in G$, let $\lambda_x$ be the... | Cayley's Representation Theorem/Proof 2 | https://proofwiki.org/wiki/Cayley's_Representation_Theorem | https://proofwiki.org/wiki/Cayley's_Representation_Theorem/Proof_2 | [
"Cayley's Representation Theorem",
"Symmetric Groups",
"Group Theory",
"Subgroups",
"Representation Theorems",
"Cayley's Theorems"
] | [
"Definition:Symmetric Group/n Letters",
"Definition:Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Subgroup"
] | [
"Definition:Finite Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Symmetric Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Regular Representations/Left Regular Representation",
"Regular Representations in Group are Permutations",
"Defin... |
proofwiki-1229 | Epimorphism Preserves Modules | Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $\struct {H, +_H, \circ}_R$ be an $R$-algebraic structure.
Let $\phi: G \to H$ be an epimorphism.
Then $H$ is an $R$-module. | If $\struct {G, +_G, \circ}_R$ is an $R$-module, then:
$\forall x, y, \in G, \forall \lambda, \mu \in R$:
:$(1): \quad \lambda \circ \paren {x +_G y} = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$
:$(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} +_G \paren {\mu \circ x}$
:$(3): \quad \... | Let $\struct {G, +_G, \circ}_R$ be an [[Definition:Module over Ring|$R$-module]].
Let $\struct {H, +_H, \circ}_R$ be an [[Definition:R-Algebraic Structure|$R$-algebraic structure]].
Let $\phi: G \to H$ be an [[Definition:R-Algebraic Structure Epimorphism|epimorphism]].
Then $H$ is an [[Definition:Module over Ring|$... | If $\struct {G, +_G, \circ}_R$ is an [[Definition:Module over Ring|$R$-module]], then:
$\forall x, y, \in G, \forall \lambda, \mu \in R$:
:$(1): \quad \lambda \circ \paren {x +_G y} = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$
:$(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} +_G \p... | Epimorphism Preserves Modules | https://proofwiki.org/wiki/Epimorphism_Preserves_Modules | https://proofwiki.org/wiki/Epimorphism_Preserves_Modules | [
"Epimorphism Preserves Modules",
"Epimorphisms (Abstract Algebra)",
"Module Theory"
] | [
"Definition:Module over Ring",
"Definition:R-Algebraic Structure",
"Definition:R-Algebraic Structure Epimorphism",
"Definition:Module over Ring"
] | [
"Definition:Module over Ring",
"Definition:R-Algebraic Structure Epimorphism",
"Definition:R-Algebraic Structure Epimorphism",
"Definition:R-Algebraic Structure Epimorphism",
"Definition:Surjection",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:R-Algebraic Structure Epimorphism",
"Axiom... |
proofwiki-1230 | Condition for Linear Transformation | Let $G$ be a unitary $R$-module.
Let $H$ be an $R$-module.
Let $\phi: G \to H$ be a mapping.
Then $\phi$ is a linear transformation {{iff}}:
:$\forall x, y \in G: \forall \lambda, \mu \in R: \map \phi {\lambda x + \mu y} = \lambda \map \phi x + \mu \map \phi y$ | === Sufficient Condition ===
{{begin-eqn}}
{{eqn | q = \forall x, y \in G: \forall \lambda, \mu \in R
| l = \map \phi {\lambda x + \mu y}
| r = \map \phi {\lambda x} + \map \phi {\mu y}
| c = {{Defof|Linear Transformation}}: condition $(1)$
}}
{{eqn | r = \lambda \map \phi x + \mu \map \phi y
| ... | Let $G$ be a [[Definition:Unitary Module over Ring|unitary $R$-module]].
Let $H$ be an [[Definition:Module over Ring|$R$-module]].
Let $\phi: G \to H$ be a [[Definition:Mapping|mapping]].
Then $\phi$ is a [[Definition:Linear Transformation|linear transformation]] {{iff}}:
:$\forall x, y \in G: \forall \lambda, \mu ... | === Sufficient Condition ===
{{begin-eqn}}
{{eqn | q = \forall x, y \in G: \forall \lambda, \mu \in R
| l = \map \phi {\lambda x + \mu y}
| r = \map \phi {\lambda x} + \map \phi {\mu y}
| c = {{Defof|Linear Transformation}}: condition $(1)$
}}
{{eqn | r = \lambda \map \phi x + \mu \map \phi y
|... | Condition for Linear Transformation | https://proofwiki.org/wiki/Condition_for_Linear_Transformation | https://proofwiki.org/wiki/Condition_for_Linear_Transformation | [
"Linear Transformations"
] | [
"Definition:Unitary Module over Ring",
"Definition:Module over Ring",
"Definition:Mapping",
"Definition:Linear Transformation"
] | [] |
proofwiki-1231 | Module of All Mappings is Module | Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $S$ be a set.
Let $\struct {G^S, +_G', \circ}_R$ be the module of all mappings from $S$ to $G$.
Then $\struct {G^S, +_G', \circ}_R$ is an $R$-module. | To show that $\struct {G^S, +_G', \circ}_R$ is an $R$-module, we verify the following:
$\forall f, g, \in G^S, \forall \lambda, \mu \in R$:
:$(1): \quad \lambda \circ \paren {f +_G' g} = \paren {\lambda \circ f} +_G' \paren {\lambda \circ g}$
:$(2): \quad \paren {\lambda +_R \mu} \circ f = \paren {\lambda \circ f} +_G ... | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G, +_G, \circ}_R$ be an [[Definition:Module over Ring|$R$-module]].
Let $S$ be a [[Definition:Set|set]].
Let $\struct {G^S, +_G', \circ}_R$ be the [[Definition:Module of All Mappings|module of all mappings]] from $S$ to ... | To show that $\struct {G^S, +_G', \circ}_R$ is an [[Definition:Module over Ring|$R$-module]], we verify the following:
$\forall f, g, \in G^S, \forall \lambda, \mu \in R$:
:$(1): \quad \lambda \circ \paren {f +_G' g} = \paren {\lambda \circ f} +_G' \paren {\lambda \circ g}$
:$(2): \quad \paren {\lambda +_R \mu} \cir... | Module of All Mappings is Module | https://proofwiki.org/wiki/Module_of_All_Mappings_is_Module | https://proofwiki.org/wiki/Module_of_All_Mappings_is_Module | [
"Examples of Modules"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Module over Ring",
"Definition:Set",
"Definition:Module of All Mappings",
"Definition:Module over Ring"
] | [
"Definition:Module over Ring"
] |
proofwiki-1232 | Kuratowski's Theorem | The following conditions on a graph $\Gamma$ are equivalent:
:$(1): \quad \Gamma$ is planar
:$(2): \quad \Gamma$ contains no subdivision of either the complete graph $K_5$ or the complete bipartite graph $K_{3, 3}$. | Let $\Gamma$ be a graph with $V$ vertices, $E$ edges and $F$ faces.
The proof proceeds in two parts: $(1) \implies (2)$ and $(2) \implies (1)$. | The following conditions on a [[Definition:Graph (Graph Theory)|graph]] $\Gamma$ are equivalent:
:$(1): \quad \Gamma$ is [[Definition:Planar Graph|planar]]
:$(2): \quad \Gamma$ contains no [[Definition:Graph Subdivision|subdivision]] of either the [[Definition:Complete Graph|complete graph $K_5$]] or the [[Definition... | Let $\Gamma$ be a [[Definition:Graph (Graph Theory)|graph]] with $V$ [[Definition:Vertex of Graph|vertices]], $E$ [[Definition:Edge of Graph|edges]] and $F$ [[Definition:Face of Graph|faces]].
The proof proceeds in two parts: $(1) \implies (2)$ and $(2) \implies (1)$. | Kuratowski's Theorem | https://proofwiki.org/wiki/Kuratowski's_Theorem | https://proofwiki.org/wiki/Kuratowski's_Theorem | [
"Graph Theory"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Planar Graph",
"Definition:Subdivision (Graph Theory)/Graph",
"Definition:Complete Graph",
"Definition:Complete Bipartite Graph"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Planar Graph/Face",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Planar Graph/Face",
"Definition:Planar Graph/Face",
"Defi... |
proofwiki-1233 | Fundamental Theorem of Finite Abelian Groups | Every finite abelian group is an internal group direct product of cyclic groups whose orders are prime powers.
The number of terms in the product and the orders of the cyclic groups are uniquely determined by the group. | Let $G$ be a finite abelian group.
By means of Abelian Group is Product of Prime-power Order Groups, we factor it uniquely into groups of prime-power order.
Then Abelian Group of Prime-power Order is Product of Cyclic Groups applies to each of these factors.
Hence we conclude $G$ factors into prime-power order cyclic g... | Every [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]] is an [[Definition:Internal Group Direct Product|internal group direct product]] of [[Definition:Cyclic Group|cyclic groups]] whose [[Definition:Order of Group|orders]] are [[Definition:Prime Power|prime powers]].
The number of terms i... | Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
By means of [[Abelian Group is Product of Prime-power Order Groups]], we factor it uniquely into groups of [[Definition:Prime Power|prime-power]] [[Definition:Order of Group|order]].
Then [[Abelian Group of Prime-power Order i... | Fundamental Theorem of Finite Abelian Groups | https://proofwiki.org/wiki/Fundamental_Theorem_of_Finite_Abelian_Groups | https://proofwiki.org/wiki/Fundamental_Theorem_of_Finite_Abelian_Groups | [
"Fundamental Theorem of Finite Abelian Groups",
"Fundamental Theorems",
"Abelian Groups",
"Proofs by Induction"
] | [
"Definition:Finite Group",
"Definition:Abelian Group",
"Definition:Internal Group Direct Product",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Prime Power",
"Definition:Internal Group Direct Product",
"Definition:Order of Structure",
"Definition:Cyclic Group",
"Definiti... | [
"Definition:Finite Group",
"Definition:Abelian Group",
"Abelian Group Factored by Prime/Corollary",
"Definition:Prime Power",
"Definition:Order of Structure",
"Abelian Group of Prime-power Order is Product of Cyclic Groups",
"Definition:Prime Power",
"Definition:Order of Structure",
"Definition:Cycl... |
proofwiki-1234 | Finite Direct Product of Modules is Module | Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be $R$-modules.
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their direct product.
Then $G$ is a module. | This is a special case of Direct Product of Modules is Module. | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be [[Definition:Module over Ring|$R$-modules]].
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their [[Definition:Module Direct P... | This is a special case of [[Direct Product of Modules is Module]]. | Finite Direct Product of Modules is Module/Proof 1 | https://proofwiki.org/wiki/Finite_Direct_Product_of_Modules_is_Module | https://proofwiki.org/wiki/Finite_Direct_Product_of_Modules_is_Module/Proof_1 | [
"Finite Direct Product of Modules is Module",
"Module Theory",
"Direct Products"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Module over Ring",
"Definition:Module Direct Product",
"Definition:Module over Ring"
] | [
"Direct Product of Modules is Module"
] |
proofwiki-1235 | Finite Direct Product of Modules is Module | Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be $R$-modules.
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their direct product.
Then $G$ is a module. | === {{Module-axiom|1|nolink}} ===
Let $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in G$.
{{begin-eqn}}
{{eqn | l = \lambda \circ \paren {x + y}
| r = \lambda \circ \paren {\tuple {x_1, x_2, \ldots, x_n} + \tuple {y_1, y_2, \ldots, y_n} }
| c =
}}
{{eqn | r = \lambda \circ \tupl... | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be [[Definition:Module over Ring|$R$-modules]].
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their [[Definition:Module Direct P... | === {{Module-axiom|1|nolink}} ===
Let $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in G$.
{{begin-eqn}}
{{eqn | l = \lambda \circ \paren {x + y}
| r = \lambda \circ \paren {\tuple {x_1, x_2, \ldots, x_n} + \tuple {y_1, y_2, \ldots, y_n} }
| c =
}}
{{eqn | r = \lambda \circ \tu... | Finite Direct Product of Modules is Module/Proof 2 | https://proofwiki.org/wiki/Finite_Direct_Product_of_Modules_is_Module | https://proofwiki.org/wiki/Finite_Direct_Product_of_Modules_is_Module/Proof_2 | [
"Finite Direct Product of Modules is Module",
"Module Theory",
"Direct Products"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Module over Ring",
"Definition:Module Direct Product",
"Definition:Module over Ring"
] | [] |
proofwiki-1236 | Subring Module is Module | Let $\struct {R, +, \times}$ be a ring.
Let $\struct {S, +_S, \times_S}$ be a subring of $R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $\circ_S$ be the restriction of $\circ$ to $S \times G$.
Let $\struct {G, +_G, \circ_S}_S$ be subring module induced by $S$.
Then $\struct {G, +_G, \circ_S}_S$ is an $S$-mo... | We have that:
:$\forall a, b \in S: a +_S b = a + b$
:$\forall a, b \in S: a \times_S b = a \times b$
:$\forall a \in S: \forall x \in G = a \circ_S x = a \circ x$
as $+_S$, $\times_S$ and $\circ_S$ are restrictions.
Let us verify the module axioms. | Let $\struct {R, +, \times}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {S, +_S, \times_S}$ be a [[Definition:Subring|subring]] of $R$.
Let $\struct {G, +_G, \circ}_R$ be an [[Definition:Module over Ring|$R$-module]].
Let $\circ_S$ be the [[Definition:Restriction of Operation|restriction]] of $\c... | We have that:
:$\forall a, b \in S: a +_S b = a + b$
:$\forall a, b \in S: a \times_S b = a \times b$
:$\forall a \in S: \forall x \in G = a \circ_S x = a \circ x$
as $+_S$, $\times_S$ and $\circ_S$ are [[Definition:Restriction of Operation|restrictions]].
Let us verify the [[Axiom:Module Axioms|module axioms]]. | Subring Module is Module | https://proofwiki.org/wiki/Subring_Module_is_Module | https://proofwiki.org/wiki/Subring_Module_is_Module | [
"Subring Module is Module",
"Subrings",
"Module Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Subring",
"Definition:Module over Ring",
"Definition:Restriction/Operation",
"Definition:Subring Module",
"Definition:Module over Ring"
] | [
"Definition:Restriction/Operation",
"Axiom:Left Module Axioms"
] |
proofwiki-1237 | Z-Module Associated with Abelian Group is Unitary Z-Module | Let $\struct {G, *}$ be an abelian group with identity $e$.
Let $\struct {G, *, \circ}_\Z$ be the $Z$-module associated with $G$.
Then $\struct {G, *, \circ}_\Z$ is a unitary $Z$-module. | The notation $*^n x$ can be written as $x^n$.
Let us verify that $\struct {G, *, \circ}_\Z$ is a unitary $\Z$-module by verifying the axioms in turn. | Let $\struct {G, *}$ be an [[Definition:Abelian Group|abelian group]] with [[Definition:Identity Element|identity]] $e$.
Let $\struct {G, *, \circ}_\Z$ be the [[Definition:Z-Module Associated with Abelian Group|$Z$-module associated with $G$]].
Then $\struct {G, *, \circ}_\Z$ is a [[Definition:Unitary Module|unitary... | The notation $*^n x$ can be written as $x^n$.
Let us verify that $\struct {G, *, \circ}_\Z$ is a [[Definition:Unitary Module|unitary $\Z$-module]] by verifying the axioms in turn. | Z-Module Associated with Abelian Group is Unitary Z-Module | https://proofwiki.org/wiki/Z-Module_Associated_with_Abelian_Group_is_Unitary_Z-Module | https://proofwiki.org/wiki/Z-Module_Associated_with_Abelian_Group_is_Unitary_Z-Module | [
"Examples of Unitary Modules",
"Z-Module Associated with Abelian Group"
] | [
"Definition:Abelian Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Z-Module Associated with Abelian Group",
"Definition:Unitary Module over Ring"
] | [
"Definition:Unitary Module over Ring",
"Definition:Unitary Module over Ring"
] |
proofwiki-1238 | Projection on Cartesian Product of Modules | Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +, \circ}_R$ be the cartesian product of a sequence $\sequence {\struct {G_n, +_n, \circ_n}_R}$ of $R$-modules.
Then for each $j \in \closedint 1 n$, the projection $\pr_j$ on the $j$th co-ordinate is an epimorphism from $\struct {G, +, \circ}_R$ onto $\struc... | To demonstrate that $\pr_j$ is an epimorphism, we need to show that:
:$(1): \quad \pr_j$ is a surjection
:$(2): \quad \forall x, y \in G: \map {\pr_j} {x + y} = \map {\pr_j} x +_j \map {\pr_j} y$
:$(3): \quad \forall x \in G: \forall \lambda \in R: \map {\pr_j} {\lambda \circ x} = \lambda \circ_j \map {\pr_j} x$
Criter... | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G, +, \circ}_R$ be the [[Definition:Cartesian Product|cartesian product]] of a [[Definition:Sequence|sequence]] $\sequence {\struct {G_n, +_n, \circ_n}_R}$ of [[Definition:Module over Ring|$R$-modules]].
Then for each $j... | To demonstrate that $\pr_j$ is an [[Definition:R-Algebraic Structure Epimorphism|epimorphism]], we need to show that:
:$(1): \quad \pr_j$ is a [[Definition:Surjection|surjection]]
:$(2): \quad \forall x, y \in G: \map {\pr_j} {x + y} = \map {\pr_j} x +_j \map {\pr_j} y$
:$(3): \quad \forall x \in G: \forall \lambda \i... | Projection on Cartesian Product of Modules | https://proofwiki.org/wiki/Projection_on_Cartesian_Product_of_Modules | https://proofwiki.org/wiki/Projection_on_Cartesian_Product_of_Modules | [
"Cartesian Product",
"Projections",
"Epimorphisms (Abstract Algebra)",
"Module Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Cartesian Product",
"Definition:Sequence",
"Definition:Module over Ring",
"Definition:Projection (Mapping Theory)",
"Definition:R-Algebraic Structure Epimorphism"
] | [
"Definition:R-Algebraic Structure Epimorphism",
"Definition:Surjection",
"Projection is Epimorphism/General Result"
] |
proofwiki-1239 | Submodule Test | Let $\struct {R, +_R, \times_R}$ be a ring with unity $1_R$.
Let $\struct {G, +, \circ}_R$ be a unitary $R$-module.
Let $H$ be a non-empty subset of $G$.
Then $\struct {H, +, \circ}_R$ is a submodule of $G$ {{iff}}:
{{begin-eqn}}
{{eqn | n = SM1
| q = \forall x, y \in H
| l = x + y
| o = \in
| r... | === Necessary Condition ===
Let $\struct {H, +, \circ}_R$ fulfil the conditions $\text {SM} 1$ and $\text {SM} 2$.
We have {{hypothesis}} that $H \subseteq G$ such that $H \ne \O$.
We have from $\text {SM} 1$:
:$\forall x, y \in H: x + y \in H$
We have {{hypothesis}} that $\struct {R, +_R, \times_R}$ is a ring with uni... | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring with Unity|ring with unity $1_R$]].
Let $\struct {G, +, \circ}_R$ be a [[Definition:Unitary Module over Ring|unitary $R$-module]].
Let $H$ be a non-[[Definition:Empty Set|empty]] [[Definition:Subset|subset]] of $G$.
Then $\struct {H, +, \circ}_R$ is a [[Defini... | === Necessary Condition ===
Let $\struct {H, +, \circ}_R$ fulfil the conditions $\text {SM} 1$ and $\text {SM} 2$.
We have {{hypothesis}} that $H \subseteq G$ such that $H \ne \O$.
We have from $\text {SM} 1$:
:$\forall x, y \in H: x + y \in H$
We have {{hypothesis}} that $\struct {R, +_R, \times_R}$ is a [[Defini... | Submodule Test | https://proofwiki.org/wiki/Submodule_Test | https://proofwiki.org/wiki/Submodule_Test | [
"Submodules"
] | [
"Definition:Ring with Unity",
"Definition:Unitary Module over Ring",
"Definition:Empty Set",
"Definition:Subset",
"Definition:Submodule"
] | [
"Definition:Ring with Unity",
"Definition:Unity (Abstract Algebra)/Ring",
"Two-Step Subgroup Test",
"Definition:Subgroup",
"Axiom:Left Module Axioms",
"Definition:Scalar Multiplication/Module",
"Definition:Distributive Operation/Left",
"Definition:Group Product/Group Law",
"Definition:Scalar Multipl... |
proofwiki-1240 | Module is Submodule of Itself | Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Then $\struct {G, +_G, \circ}_R$ is a submodule of itself. | Follows directly from Group is Subgroup of Itself.
{{qed}} | Let $\struct {G, +_G, \circ}_R$ be an [[Definition:Module over Ring|$R$-module]].
Then $\struct {G, +_G, \circ}_R$ is a [[Definition:Submodule|submodule]] of itself. | Follows directly from [[Group is Subgroup of Itself]].
{{qed}} | Module is Submodule of Itself | https://proofwiki.org/wiki/Module_is_Submodule_of_Itself | https://proofwiki.org/wiki/Module_is_Submodule_of_Itself | [
"Submodules"
] | [
"Definition:Module over Ring",
"Definition:Submodule"
] | [
"Group is Subgroup of Itself"
] |
proofwiki-1241 | Null Module Submodule of All | Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Then the null module:
:$\struct {\set {e_G}, +_G, \circ}_R$
is a submodule of $\struct {G, +_G, \circ}_R$. | From Trivial Subgroup is Subgroup, the trivial subgroup is a subgroup of the group $\struct {G, +_G}$.
The result follows from the Submodule Test.
{{qed}} | Let $\struct {G, +_G, \circ}_R$ be an [[Definition:Module over Ring|$R$-module]].
Then the [[Definition:Null Module|null module]]:
:$\struct {\set {e_G}, +_G, \circ}_R$
is a [[Definition:Submodule|submodule]] of $\struct {G, +_G, \circ}_R$. | From [[Trivial Subgroup is Subgroup]], the [[Definition:Trivial Subgroup|trivial subgroup]] is a [[Definition:Subgroup|subgroup]] of the [[Definition:Group|group]] $\struct {G, +_G}$.
The result follows from the [[Submodule Test]].
{{qed}} | Null Module Submodule of All | https://proofwiki.org/wiki/Null_Module_Submodule_of_All | https://proofwiki.org/wiki/Null_Module_Submodule_of_All | [
"Submodules",
"Module Theory"
] | [
"Definition:Module over Ring",
"Definition:Null Module",
"Definition:Submodule"
] | [
"Trivial Subgroup is Subgroup",
"Definition:Trivial Subgroup",
"Definition:Subgroup",
"Definition:Group",
"Submodule Test"
] |
proofwiki-1242 | First Sylow Theorem | Let $p$ be a prime number.
Let $G$ be a group such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the order of $G$
:$p$ is not a divisor of $k$.
Then $G$ has at least one Sylow $p$-subgroup. | Let $\order G = k p^r$ where $p \nmid k$.
From the First Sylow Theorem, $G$ has a subgroup $S$ of order $p^r$.
From (need to find it), $S$ itself has subgroups of order $p^n$ for all $n \in \set {1, 2, \ldots, r}$.
{{qed}} | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $G$ be a [[Definition:Group|group]] such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the [[Definition:Order of Structure|order]] of $G$
:$p$ is not a [[Definition:Divisor of Integer|divisor]] of $k$.
Then $G$ has at least one [[Definition:Sylow p-Su... | Let $\order G = k p^r$ where $p \nmid k$.
From the [[First Sylow Theorem]], $G$ has a [[Definition:Subgroup|subgroup]] $S$ of [[Definition:Order of Group|order]] $p^r$.
From (need to find it), $S$ itself has [[Definition:Subgroup|subgroups]] of [[Definition:Order of Group|order]] $p^n$ for all $n \in \set {1, 2, \ldo... | First Sylow Theorem/Corollary/Proof 1 | https://proofwiki.org/wiki/First_Sylow_Theorem | https://proofwiki.org/wiki/First_Sylow_Theorem/Corollary/Proof_1 | [
"Sylow Theorems",
"First Sylow Theorem"
] | [
"Definition:Prime Number",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Divisor (Algebra)/Integer",
"Definition:Sylow p-Subgroup"
] | [
"First Sylow Theorem",
"Definition:Subgroup",
"Definition:Order of Structure",
"Definition:Subgroup",
"Definition:Order of Structure"
] |
proofwiki-1243 | First Sylow Theorem | Let $p$ be a prime number.
Let $G$ be a group such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the order of $G$
:$p$ is not a divisor of $k$.
Then $G$ has at least one Sylow $p$-subgroup. | Let $\order G = m p^n$.
Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.
Let $N = \order {\mathbb S}$.
Now $N$ is the number of ways $p^n$ elements can be chosen from a set containing $p^n k$ elements.
From Cardinality of Set of Subsets... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $G$ be a [[Definition:Group|group]] such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the [[Definition:Order of Structure|order]] of $G$
:$p$ is not a [[Definition:Divisor of Integer|divisor]] of $k$.
Then $G$ has at least one [[Definition:Sylow p-Su... | Let $\order G = m p^n$.
Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of [[Definition:Subset|subsets]] of $G$ which have exactly $p^n$ [[Definition:Element|elements]].
Let $N = \order {\mathbb S}$.
Now $N$ is the number of ways $p^n$ [[Definition:Element|elements]] can be chosen fr... | First Sylow Theorem/Corollary/Proof 2 | https://proofwiki.org/wiki/First_Sylow_Theorem | https://proofwiki.org/wiki/First_Sylow_Theorem/Corollary/Proof_2 | [
"Sylow Theorems",
"First Sylow Theorem"
] | [
"Definition:Prime Number",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Divisor (Algebra)/Integer",
"Definition:Sylow p-Subgroup"
] | [
"Definition:Subset",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Cardinality of Set of Subsets",
"Definition:Power (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Power (Algebra)/Integer",
"Definition:Fraction/Numerator",
"Definition:Fraction/Denomina... |
proofwiki-1244 | First Sylow Theorem | Let $p$ be a prime number.
Let $G$ be a group such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the order of $G$
:$p$ is not a divisor of $k$.
Then $G$ has at least one Sylow $p$-subgroup. | Let $\order G = k p^n$ such that $p \nmid k$.
Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.
Let $N = \order {\mathbb S}$.
Now $N$ is the number of ways $p^n$ elements can be chosen from a set containing $p^n k$ elements.
From Cardina... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $G$ be a [[Definition:Group|group]] such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the [[Definition:Order of Structure|order]] of $G$
:$p$ is not a [[Definition:Divisor of Integer|divisor]] of $k$.
Then $G$ has at least one [[Definition:Sylow p-Su... | Let $\order G = k p^n$ such that $p \nmid k$.
Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of [[Definition:Subset|subsets]] of $G$ which have exactly $p^n$ [[Definition:Element|elements]].
Let $N = \order {\mathbb S}$.
Now $N$ is the number of ways $p^n$ [[Definition:Element|eleme... | First Sylow Theorem/Proof 1 | https://proofwiki.org/wiki/First_Sylow_Theorem | https://proofwiki.org/wiki/First_Sylow_Theorem/Proof_1 | [
"Sylow Theorems",
"First Sylow Theorem"
] | [
"Definition:Prime Number",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Divisor (Algebra)/Integer",
"Definition:Sylow p-Subgroup"
] | [
"Definition:Subset",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Cardinality of Set of Subsets",
"Binomial Coefficient involving Power of Prime",
"Definition:Group Action",
"Definition:Coset/Left Coset",
"Group Action on Sets with k Elements",
"Definition:Group Action",
"D... |
proofwiki-1245 | First Sylow Theorem | Let $p$ be a prime number.
Let $G$ be a group such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the order of $G$
:$p$ is not a divisor of $k$.
Then $G$ has at least one Sylow $p$-subgroup. | Let $\Bbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.
Now let $G$ act on $\Bbb S$ by the rule:
:$\forall S \in \Bbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
From Set of Orbits forms Partition, the orbits partition $\mathbb S$.
Let the... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $G$ be a [[Definition:Group|group]] such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the [[Definition:Order of Structure|order]] of $G$
:$p$ is not a [[Definition:Divisor of Integer|divisor]] of $k$.
Then $G$ has at least one [[Definition:Sylow p-Su... | Let $\Bbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of [[Definition:Subset|subsets]] of $G$ which have exactly $p^n$ [[Definition:Element|elements]].
Now let $G$ [[Definition:Group Action|act]] on $\Bbb S$ by the rule:
:$\forall S \in \Bbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
Fr... | First Sylow Theorem/Proof 2 | https://proofwiki.org/wiki/First_Sylow_Theorem | https://proofwiki.org/wiki/First_Sylow_Theorem/Proof_2 | [
"Sylow Theorems",
"First Sylow Theorem"
] | [
"Definition:Prime Number",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Divisor (Algebra)/Integer",
"Definition:Sylow p-Subgroup"
] | [
"Definition:Subset",
"Definition:Element",
"Definition:Group Action",
"Set of Orbits forms Partition",
"Definition:Orbit (Group Theory)",
"Definition:Set Partition",
"Definition:Orbit (Group Theory)",
"Definition:Orbit (Group Theory)",
"Definition:Order of Structure",
"Orbit-Stabilizer Theorem",
... |
proofwiki-1246 | H-Cobordism Theorem | Let $X^n, Y^n$ be two simply connected manifolds.
Let $n \in \N: n \ge 5$ and $\exists W$ such that $W$ is an h-cobordism between $X$ and $Y$.
Then $\exists \psi: W \to X \times \closedint 0 1$ such that $\psi$ is a diffeomorphism.
In particular, $X$ and $Y$ are diffeomorphic. | {{ProofWanted|In progress}}
Category:Cobordisms
a0li2st5wmkmtvb6ao38mv19ri10w03 | Let $X^n, Y^n$ be two [[Definition:Simply Connected|simply connected]] [[Definition:Topological Manifold|manifolds]].
Let $n \in \N: n \ge 5$ and $\exists W$ such that $W$ is an [[Definition:Cobordism|h-cobordism]] between $X$ and $Y$.
Then $\exists \psi: W \to X \times \closedint 0 1$ such that $\psi$ is a [[Defini... | {{ProofWanted|In progress}}
[[Category:Cobordisms]]
a0li2st5wmkmtvb6ao38mv19ri10w03 | H-Cobordism Theorem | https://proofwiki.org/wiki/H-Cobordism_Theorem | https://proofwiki.org/wiki/H-Cobordism_Theorem | [
"Cobordisms"
] | [
"Definition:Simply Connected",
"Definition:Topological Manifold",
"Definition:Cobordism",
"Definition:Diffeomorphism",
"Definition:Diffeomorphism"
] | [
"Category:Cobordisms"
] |
proofwiki-1247 | Group has Subgroups of All Prime Power Factors | Let $p$ be a prime.
Let $G$ be a finite group of order $n$.
If $p^k \divides n$ then $G$ has at least one subgroup of order $p^k$. | From Composition Series of Group of Prime Power Order, a $p$-group has subgroups corresponding to every divisor of its order.
Thus, taken with the First Sylow Theorem, a finite group has a subgroup corresponding to every prime power divisor of its order.
{{qed}} | Let $p$ be a [[Definition:Prime Number|prime]].
Let $G$ be a [[Definition:Finite Group|finite group]] of [[Definition:Order of Structure|order]] $n$.
If $p^k \divides n$ then $G$ has at least one [[Definition:Subgroup|subgroup]] of [[Definition:Order of Structure|order]] $p^k$. | From [[Composition Series of Group of Prime Power Order]], a [[Definition:P-Group|$p$-group]] has [[Definition:Subgroup|subgroups]] corresponding to every [[Definition:Divisor of Integer|divisor]] of its [[Definition:Order of Structure|order]].
Thus, taken with the [[First Sylow Theorem]], a [[Definition:Finite Group|... | Group has Subgroups of All Prime Power Factors | https://proofwiki.org/wiki/Group_has_Subgroups_of_All_Prime_Power_Factors | https://proofwiki.org/wiki/Group_has_Subgroups_of_All_Prime_Power_Factors | [
"Order of Groups",
"P-Groups",
"Subgroups"
] | [
"Definition:Prime Number",
"Definition:Finite Group",
"Definition:Order of Structure",
"Definition:Subgroup",
"Definition:Order of Structure"
] | [
"Composition Series of Group of Prime Power Order",
"Definition:P-Group",
"Definition:Subgroup",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure",
"First Sylow Theorem",
"Definition:Finite Group",
"Definition:Subgroup",
"Definition:Prime Power",
"Definition:Divisor (Algebra)... |
proofwiki-1248 | Normalizer of Sylow p-Subgroup | Let $P$ be a Sylow $p$-subgroup of a finite group $G$.
Let $\map {N_G} P$ be the normalizer of $P$.
Then any $p$-subgroup of $\map {N_G} P$ is contained in $P$.
In particular, $P$ is the unique Sylow $p$-subgroup of $\map {N_G} P$. | Let $Q$ be a $p$-subgroup of $N = \map {N_G} P$.
Let $\order Q = p^m, \order P = p^n$.
By Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:
:$P \lhd \map {N_G} P$
thus by Subset Product with Normal Subgroup as Generator:
:$\gen {P, Q} = P Q$
Thus by Order of Subgroup Product:
:$P Q... | Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of a [[Definition:Finite Group|finite group]] $G$.
Let $\map {N_G} P$ be the [[Definition:Normalizer|normalizer]] of $P$.
Then any [[Definition:P-Subgroup|$p$-subgroup]] of $\map {N_G} P$ is contained in $P$.
In particular, $P$ is the unique [[Definit... | Let $Q$ be a [[Definition:P-Subgroup|$p$-subgroup]] of $N = \map {N_G} P$.
Let $\order Q = p^m, \order P = p^n$.
By [[Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup]]:
:$P \lhd \map {N_G} P$
thus by [[Subset Product with Normal Subgroup as Generator]]:
:$\gen {P, Q} = P Q$
Thu... | Normalizer of Sylow p-Subgroup | https://proofwiki.org/wiki/Normalizer_of_Sylow_p-Subgroup | https://proofwiki.org/wiki/Normalizer_of_Sylow_p-Subgroup | [
"P-Groups",
"Subgroups",
"Normalizers",
"Sylow p-Subgroups"
] | [
"Definition:Sylow p-Subgroup",
"Definition:Finite Group",
"Definition:Normalizer",
"Definition:P-Subgroup",
"Definition:Sylow p-Subgroup"
] | [
"Definition:P-Subgroup",
"Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup",
"Subset Product with Normal Subgroup as Generator",
"Order of Subgroup Product",
"Intersection with Subset is Subset",
"Definition:Sylow p-Subgroup"
] |
proofwiki-1249 | Second Sylow Theorem | Let $P$ be a Sylow $p$-subgroup of the finite group $G$.
Let $Q$ be any $p$-subgroup of $G$.
Then $Q$ is a subset of a conjugate of $P$. | Let $P$ be a Sylow $p$-subgroup of $G$.
Let $\mathbb S$ be the set of all distinct $G$-conjugates of $P$:
:$\mathbb S = \set {g P g^{-1}: g \in G}$
Let $h * S$ be the conjugacy action:
:$\forall h \in P, S \in \mathbb S: h * S = h S h^{-1}$
From Conjugacy Action on Subgroups is Group Action, this is a group action for ... | Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of the [[Definition:Finite Group|finite group]] $G$.
Let $Q$ be any [[Definition:P-Subgroup|$p$-subgroup]] of $G$.
Then $Q$ is a [[Definition:Subset|subset]] of a [[Definition:Conjugate of Group Subset|conjugate]] of $P$. | Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
Let $\mathbb S$ be the [[Definition:Set|set]] of all [[Definition:Distinct|distinct]] [[Definition:Conjugate of Group Subset|$G$-conjugates]] of $P$:
:$\mathbb S = \set {g P g^{-1}: g \in G}$
Let $h * S$ be the [[Definition:Conjugacy Action|conj... | Second Sylow Theorem | https://proofwiki.org/wiki/Second_Sylow_Theorem | https://proofwiki.org/wiki/Second_Sylow_Theorem | [
"Sylow Theorems"
] | [
"Definition:Sylow p-Subgroup",
"Definition:Finite Group",
"Definition:P-Subgroup",
"Definition:Subset",
"Definition:Conjugate (Group Theory)/Subset"
] | [
"Definition:Sylow p-Subgroup",
"Definition:Set",
"Definition:Distinct",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Conjugacy Action",
"Conjugacy Action on Subgroups is Group Action",
"Definition:Group Action",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Or... |
proofwiki-1250 | Third Sylow Theorem | All the Sylow $p$-subgroups of a finite group are conjugate. | Suppose $P$ and $Q$ are Sylow $p$-subgroups of $G$.
By the Second Sylow Theorem, $Q$ is a subset of a conjugate of $P$.
But since $\order P = \order Q$, it follows that $Q$ must equal a conjugate of $P$.
{{qed}} | All the [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of a [[Definition:Finite Group|finite group]] are [[Definition:Conjugate of Group Subset|conjugate]]. | Suppose $P$ and $Q$ are [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$.
By the [[Second Sylow Theorem]], $Q$ is a [[Definition:Subset|subset]] of a [[Definition:Conjugate of Group Subset|conjugate]] of $P$.
But since $\order P = \order Q$, it follows that $Q$ must equal a [[Definition:Conjugate of Group S... | Third Sylow Theorem/Proof 1 | https://proofwiki.org/wiki/Third_Sylow_Theorem | https://proofwiki.org/wiki/Third_Sylow_Theorem/Proof_1 | [
"Sylow Theorems",
"Third Sylow Theorem"
] | [
"Definition:Sylow p-Subgroup",
"Definition:Finite Group",
"Definition:Conjugate (Group Theory)/Subset"
] | [
"Definition:Sylow p-Subgroup",
"Second Sylow Theorem",
"Definition:Subset",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Conjugate (Group Theory)/Subset"
] |
proofwiki-1251 | Third Sylow Theorem | All the Sylow $p$-subgroups of a finite group are conjugate. | Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$.
Let $H$ be a Sylow $p$-subgroup of $G$.
We have that:
:$\order H = p^n$
:$\index G H = m$
Let $S_1, S_2, \ldots, S_m$ denote the left cosets of $G \pmod H$.
We have that $G$ acts on $G / H$ by the rule:
:$g * S_i = g S_i$.
Let $H_i$ denote the s... | All the [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of a [[Definition:Finite Group|finite group]] are [[Definition:Conjugate of Group Subset|conjugate]]. | Let $G$ be a [[Definition:Finite Group|finite group]] of [[Definition:Order of Group|order]] $p^n m$, where $p \nmid m$ and $n > 0$.
Let $H$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
We have that:
:$\order H = p^n$
:$\index G H = m$
Let $S_1, S_2, \ldots, S_m$ denote the [[Definition:Left Cose... | Third Sylow Theorem/Proof 2 | https://proofwiki.org/wiki/Third_Sylow_Theorem | https://proofwiki.org/wiki/Third_Sylow_Theorem/Proof_2 | [
"Sylow Theorems",
"Third Sylow Theorem"
] | [
"Definition:Sylow p-Subgroup",
"Definition:Finite Group",
"Definition:Conjugate (Group Theory)/Subset"
] | [
"Definition:Finite Group",
"Definition:Order of Structure",
"Definition:Sylow p-Subgroup",
"Definition:Coset/Left Coset",
"Definition:Group Action",
"Definition:Stabilizer",
"Orbit-Stabilizer Theorem",
"Definition:Sylow p-Subgroup",
"Definition:Orbit (Group Theory)",
"Definition:Cardinality",
"D... |
proofwiki-1252 | Fourth Sylow Theorem | The number of Sylow $p$-subgroups of a finite group is congruent to $1 \pmod p$. | Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$.
Let $r$ be the number of Sylow $p$-subgroups of $G$.
Let $H$ be a Sylow $p$-subgroup of $G$.
We have that:
:$\order H = p^n$
:$\index G H = m$
Let $S_1, S_2, \ldots, S_m$ denote the elements of the left coset space of $G / H$.
We have that $H$ a... | The number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of a [[Definition:Finite Group|finite group]] is [[Definition:Congruence (Number Theory)|congruent]] to $1 \pmod p$. | Let $G$ be a [[Definition:Finite Group|finite group]] of [[Definition:Order of Group|order]] $p^n m$, where $p \nmid m$ and $n > 0$.
Let $r$ be the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$.
Let $H$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
We have that:
:$\order H =... | Fourth Sylow Theorem/Proof 2 | https://proofwiki.org/wiki/Fourth_Sylow_Theorem | https://proofwiki.org/wiki/Fourth_Sylow_Theorem/Proof_2 | [
"Sylow Theorems",
"Fourth Sylow Theorem"
] | [
"Definition:Sylow p-Subgroup",
"Definition:Finite Group",
"Definition:Congruence (Number Theory)"
] | [
"Definition:Finite Group",
"Definition:Order of Structure",
"Definition:Sylow p-Subgroup",
"Definition:Sylow p-Subgroup",
"Definition:Element",
"Definition:Coset Space/Left Coset Space",
"Definition:Group Action",
"Definition:Orbit (Group Theory)",
"Orbits of Group Action on Sets with Power of Prime... |
proofwiki-1253 | Sylow p-Subgroup is Unique iff Normal | A group $G$ has exactly one Sylow $p$-subgroup $P$ {{iff}} $P$ is normal. | If $G$ has precisely one Sylow $p$-subgroup, it must be normal from Unique Subgroup of a Given Order is Normal.
Suppose a Sylow $p$-subgroup $P$ is normal.
Then it equals its conjugates.
Thus, by the Third Sylow Theorem, there can be only one such Sylow $p$-subgroup.
{{qed}} | A [[Definition:Group|group]] $G$ has exactly one [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] $P$ {{iff}} $P$ is [[Definition:Normal Subgroup|normal]]. | If $G$ has precisely one [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]], it must be [[Definition:Normal Subgroup|normal]] from [[Unique Subgroup of a Given Order is Normal]].
Suppose a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] $P$ is [[Definition:Normal Subgroup|normal]].
Then it equals its [[Definition... | Sylow p-Subgroup is Unique iff Normal | https://proofwiki.org/wiki/Sylow_p-Subgroup_is_Unique_iff_Normal | https://proofwiki.org/wiki/Sylow_p-Subgroup_is_Unique_iff_Normal | [
"Normal Subgroups",
"Sylow p-Subgroups"
] | [
"Definition:Group",
"Definition:Sylow p-Subgroup",
"Definition:Normal Subgroup"
] | [
"Definition:Sylow p-Subgroup",
"Definition:Normal Subgroup",
"Unique Subgroup of a Given Order is Normal",
"Definition:Sylow p-Subgroup",
"Definition:Normal Subgroup",
"Definition:Conjugate (Group Theory)/Subset",
"Third Sylow Theorem",
"Definition:Sylow p-Subgroup"
] |
proofwiki-1254 | Intersection of Normal Subgroup with Sylow P-Subgroup | Let $P$ be a Sylow $p$-subgroup of a finite group $G$.
Let $N$ be a normal subgroup of $G$.
Then $P \cap N$ is a Sylow $p$-subgroup of $N$. | Since $N \lhd G$, we see that:
:$\gen {P, N} = P N$
from Subset Product with Normal Subgroup as Generator.
Since $P \cap N \le P$, it follows that:
:$\order {P \cap N} = p^k$
where $k > 0$.
By Order of Subgroup Product:
:$\order {P N} \order {P \cap N} = \order P \order N$
Hence from Lagrange's Theorem:
:$\index N {P \... | Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of a [[Definition:Finite Group|finite group]] $G$.
Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Then $P \cap N$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $N$. | Since $N \lhd G$, we see that:
:$\gen {P, N} = P N$
from [[Subset Product with Normal Subgroup as Generator]].
Since $P \cap N \le P$, it follows that:
:$\order {P \cap N} = p^k$
where $k > 0$.
By [[Order of Subgroup Product]]:
:$\order {P N} \order {P \cap N} = \order P \order N$
Hence from [[Lagrange's Theorem (Gr... | Intersection of Normal Subgroup with Sylow P-Subgroup | https://proofwiki.org/wiki/Intersection_of_Normal_Subgroup_with_Sylow_P-Subgroup | https://proofwiki.org/wiki/Intersection_of_Normal_Subgroup_with_Sylow_P-Subgroup | [
"Normal Subgroups",
"Set Intersection",
"Sylow p-Subgroups"
] | [
"Definition:Sylow p-Subgroup",
"Definition:Finite Group",
"Definition:Normal Subgroup",
"Definition:Sylow p-Subgroup"
] | [
"Subset Product with Normal Subgroup as Generator",
"Order of Subgroup Product",
"Lagrange's Theorem (Group Theory)",
"Tower Law for Subgroups",
"Definition:Divisor (Algebra)/Integer",
"Definition:Sylow p-Subgroup"
] |
proofwiki-1255 | Quotient of Sylow P-Subgroup | Let $P$ be a Sylow $p$-subgroup of a finite group $G$.
Let $N$ be a normal subgroup of $G$.
Then $P N / N$ is a Sylow $p$-subgroup of $G / N$. | We have that $P \le G$ and $ N \lhd G$.
So by the Second Isomorphism Theorem for Groups:
:$P N / N \cong P / \paren {P \cap N}$
We have that:
:$P N / N = \set {p N : p \in P}$
and so every element of $P N / N$ has order a power of $p$.
Hence $P N / N$ is a $p$-subgroup of $G / N$.
From Intersection of Normal Subgroup w... | Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of a [[Definition:Finite Group|finite group]] $G$.
Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Then $P N / N$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G / N$. | We have that $P \le G$ and $ N \lhd G$.
So by the [[Second Isomorphism Theorem for Groups]]:
:$P N / N \cong P / \paren {P \cap N}$
We have that:
:$P N / N = \set {p N : p \in P}$
and so every [[Definition:Element|element]] of $P N / N$ has [[Definition:Order of Group Element|order]] a [[Definition:Power of Group Ele... | Quotient of Sylow P-Subgroup | https://proofwiki.org/wiki/Quotient_of_Sylow_P-Subgroup | https://proofwiki.org/wiki/Quotient_of_Sylow_P-Subgroup | [
"Quotient Groups",
"Sylow p-Subgroups"
] | [
"Definition:Sylow p-Subgroup",
"Definition:Finite Group",
"Definition:Normal Subgroup",
"Definition:Sylow p-Subgroup"
] | [
"Second Isomorphism Theorem/Groups",
"Definition:Element",
"Definition:Order of Group Element",
"Definition:Power of Element/Group",
"Definition:P-Subgroup",
"Intersection of Normal Subgroup with Sylow P-Subgroup",
"Definition:Sylow p-Subgroup"
] |
proofwiki-1256 | Fifth Sylow Theorem | The number of Sylow $p$-subgroups of a finite group is a divisor of their common index. | Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$.
Let $r$ be the number of Sylow $p$-subgroups of $G$.
It is to be shown that $r \divides m$.
Let $H$ be a Sylow $p$-subgroup of $G$.
We have that:
:$\order H = p^n$
:$\index G H = m$
Let $S_1, S_2, \ldots, S_m$ denote the elements of the left cos... | The number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of a [[Definition:Finite Group|finite group]] is a [[Definition:Divisor of Integer|divisor]] of their common [[Definition:Index of Subgroup|index]]. | Let $G$ be a [[Definition:Finite Group|finite group]] of [[Definition:Order of Group|order]] $p^n m$, where $p \nmid m$ and $n > 0$.
Let $r$ be the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$.
It is to be shown that $r \divides m$.
Let $H$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgro... | Fifth Sylow Theorem/Proof 2 | https://proofwiki.org/wiki/Fifth_Sylow_Theorem | https://proofwiki.org/wiki/Fifth_Sylow_Theorem/Proof_2 | [
"Sylow Theorems",
"Fifth Sylow Theorem"
] | [
"Definition:Sylow p-Subgroup",
"Definition:Finite Group",
"Definition:Divisor (Algebra)/Integer",
"Definition:Index of Subgroup"
] | [
"Definition:Finite Group",
"Definition:Order of Structure",
"Definition:Sylow p-Subgroup",
"Definition:Sylow p-Subgroup",
"Definition:Element",
"Definition:Coset Space/Left Coset Space",
"Definition:Group Action",
"Definition:Orbit (Group Theory)",
"Orbits of Group Action on Sets with Power of Prime... |
proofwiki-1257 | Finite Submodule of Function Space | Let $\struct {G, +}$ be a group whose identity is $e$.
Let $R$ be a ring.
Let $\struct {G, +, \circ}_R$ be an $R$-module.
Let $S$ be a set.
Let $G^S$ the set of all mappings $f: S \to G$.
Let $G^{\paren S}$ be the set of all mappings $f: S \to G$ such that $\map f x = e$ for all but finitely many elements $x$ of $S$.
T... | Let $\struct {G, +, \circ}_R$ be an $R$-module and $S$ be a set.
We need to show that $\struct {G^{\paren S}, +'}$ is a group.
Let $f, g \in G^{\paren S}$.
Let:
:$F = \set {x \in S: \map f x \ne e}$
:$G = \set {x \in S: \map g x \ne e}$
From the definition of $f$ and $g$, both $F$ and $G$ are finite.
Since $e + e = e$ ... | Let $\struct {G, +}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G, +, \circ}_R$ be an [[Definition:Module over Ring|$R$-module]].
Let $S$ be a set.
Let $G^S$ the set of all [[Definition:Mapping|mappi... | Let $\struct {G, +, \circ}_R$ be an [[Definition:Module over Ring|$R$-module]] and $S$ be a [[Definition:Set|set]].
We need to show that $\struct {G^{\paren S}, +'}$ is a [[Definition:Group|group]].
Let $f, g \in G^{\paren S}$.
Let:
:$F = \set {x \in S: \map f x \ne e}$
:$G = \set {x \in S: \map g x \ne e}$
From ... | Finite Submodule of Function Space | https://proofwiki.org/wiki/Finite_Submodule_of_Function_Space | https://proofwiki.org/wiki/Finite_Submodule_of_Function_Space | [
"Submodules",
"Function Spaces"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Ring (Abstract Algebra)",
"Definition:Module over Ring",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Finite",
"Definition:Submodule",
"Definition:Pointwise Operation/Induced Structure"
] | [
"Definition:Module over Ring",
"Definition:Set",
"Definition:Group",
"Definition:Finite",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Pointwise Inverse",
"Two... |
proofwiki-1258 | Elements of Module with Equal Images under Linear Transformations form Submodule | Let $G$ and $H$ be $R$-modules.
Let $\phi$ and $\psi$ be linear transformations from $G$ into $H$.
Then the set $S = \set {x \in G: \map \phi x = \map \psi x}$ is a submodule of $G$. | Let $x, y \in S$.
Let $\lambda \in R$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {x + y}
| r = \map \phi x + \map \phi y
| c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \map \psi x + \map \psi y
| c = $x, y \in S$
}}
{{eqn | r = \map \psi {x + y}
| c = {{Defof|Linear Transformation}}
}}
{{e... | Let $G$ and $H$ be [[Definition:Module over Ring|$R$-modules]].
Let $\phi$ and $\psi$ be [[Definition:Linear Transformation|linear transformations]] from $G$ into $H$.
Then the set $S = \set {x \in G: \map \phi x = \map \psi x}$ is a [[Definition:Submodule|submodule]] of $G$. | Let $x, y \in S$.
Let $\lambda \in R$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {x + y}
| r = \map \phi x + \map \phi y
| c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \map \psi x + \map \psi y
| c = $x, y \in S$
}}
{{eqn | r = \map \psi {x + y}
| c = {{Defof|Linear Transformation}}
}}... | Elements of Module with Equal Images under Linear Transformations form Submodule | https://proofwiki.org/wiki/Elements_of_Module_with_Equal_Images_under_Linear_Transformations_form_Submodule | https://proofwiki.org/wiki/Elements_of_Module_with_Equal_Images_under_Linear_Transformations_form_Submodule | [
"Linear Transformations"
] | [
"Definition:Module over Ring",
"Definition:Linear Transformation",
"Definition:Submodule"
] | [
"Submodule Test",
"Definition:Submodule"
] |
proofwiki-1259 | Poincaré Conjecture | Let $\Sigma^m$ be a smooth $m$-manifold.
Let $\Sigma^m$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_m \struct {\Sigma; \Z} = \Z$
Then $\Sigma^m$ is homeomorphic to the $m$-sphere $\Bbb S^m$. | The proof proceeds on several dimensional-cases. Note that the case $m = 3$ is incredibly intricate, and that a full proof would be impractical to produce here. An outline of the $m = 3$ case will be given instead. | Let $\Sigma^m$ be a [[Definition:Smooth Manifold|smooth $m$-manifold]].
Let $\Sigma^m$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_m \struct {\Sigma; \Z} = \Z$
Then $\Sigma^m$ is homeomorphic to the $m$-sphere $\Bbb S^m$. | The proof proceeds on several dimensional-cases. Note that the case $m = 3$ is incredibly intricate, and that a full proof would be impractical to produce here. An outline of the $m = 3$ case will be given instead. | Poincaré Conjecture | https://proofwiki.org/wiki/Poincaré_Conjecture | https://proofwiki.org/wiki/Poincaré_Conjecture | [
"Poincaré Conjecture",
"Topological Manifolds",
"Millennium Problems"
] | [
"Definition:Topological Manifold/Smooth Manifold"
] | [] |
proofwiki-1260 | Generated Submodule is Linear Combinations | Let $G$ be a unitary $R$-module.
Let $S \subseteq G$.
Then the submodule $H$ generated by $S$ is the set of all linear combinations of $S$. | First the extreme case:
The smallest submodule of $G$ containing $\O$ is $\set {e_G}$.
By definition of linear combination of empty set, $\set {e_G}$ is the set of all linear combinations of $\O$.
Now the general case:
Let $\O \subset S \subseteq G$.
Let $L$ be the set of all linear combinations of $S$.
Since $G$ is a ... | Let $G$ be a [[Definition:Unitary Module|unitary $R$-module]].
Let $S \subseteq G$.
Then the [[Definition:Generator of Module|submodule $H$ generated by $S$]] is the [[Definition:Set|set]] of all [[Definition:Linear Combination of Subset|linear combinations]] of $S$. | First the extreme case:
The smallest submodule of $G$ containing $\O$ is $\set {e_G}$.
By definition of [[Definition:Linear Combination of Empty Set|linear combination of empty set]], $\set {e_G}$ is the [[Definition:Set|set]] of all [[Definition:Linear Combination of Subset|linear combinations]] of $\O$.
Now the g... | Generated Submodule is Linear Combinations | https://proofwiki.org/wiki/Generated_Submodule_is_Linear_Combinations | https://proofwiki.org/wiki/Generated_Submodule_is_Linear_Combinations | [
"Unitary Modules",
"Linear Algebra"
] | [
"Definition:Unitary Module over Ring",
"Definition:Generator of Module",
"Definition:Set",
"Definition:Linear Combination/Subset"
] | [
"Definition:Linear Combination/Empty Set",
"Definition:Set",
"Definition:Linear Combination/Subset",
"Definition:Linear Combination/Subset",
"Definition:Unitary Module over Ring",
"Definition:Element",
"Definition:Linear Combination/Subset",
"Submodule Test",
"Definition:Linear Combination/Subset",
... |
proofwiki-1261 | Empty Set is Linearly Independent | The empty set is a linearly independent set. | There are no sequences at all of $n$ terms of the empty set for any $n > 0$.
Hence the result holds vacuously.
{{qed}} | The [[Definition:Empty Set|empty set]] is a [[Definition:Linearly Independent Set|linearly independent set]]. | There are no [[Definition:Sequence|sequences]] at all of $n$ terms of the [[Definition:Empty Set|empty set]] for any $n > 0$.
Hence the result holds [[Definition:Vacuous Truth|vacuously]].
{{qed}} | Empty Set is Linearly Independent | https://proofwiki.org/wiki/Empty_Set_is_Linearly_Independent | https://proofwiki.org/wiki/Empty_Set_is_Linearly_Independent | [
"Empty Set",
"Linear Independence"
] | [
"Definition:Empty Set",
"Definition:Linearly Independent/Set"
] | [
"Definition:Sequence",
"Definition:Empty Set",
"Definition:Vacuous Truth"
] |
proofwiki-1262 | Subset of Module Containing Identity is Linearly Dependent | Let $G$ be a group whose identity is $e$.
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $H \subseteq G$ such that $e \in H$.
Then $H$ is a linearly dependent set. | From Scalar Product with Identity, $\forall \lambda: \lambda \circ e = e$.
Let $H \subseteq G$ such that $e \in H$.
Consider any sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ in $H$ which includes $e$.
So, let $a_j = e$ for some $j \in \closedint 1 n$.
Let $c \in R \ne 0_R$.
Consider the sequence $\sequenc... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $\struct {G, +_G, \circ}_R$ be an [[Definition:Module over Ring|$R$-module]].
Let $H \subseteq ... | From [[Scalar Product with Identity]], $\forall \lambda: \lambda \circ e = e$.
Let $H \subseteq G$ such that $e \in H$.
Consider any [[Definition:Sequence|sequence]] $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ in $H$ which includes $e$.
So, let $a_j = e$ for some $j \in \closedint 1 n$.
Let $c \in R \ne 0_R$.... | Subset of Module Containing Identity is Linearly Dependent | https://proofwiki.org/wiki/Subset_of_Module_Containing_Identity_is_Linearly_Dependent | https://proofwiki.org/wiki/Subset_of_Module_Containing_Identity_is_Linearly_Dependent | [
"Module Theory",
"Linear Dependence"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Module over Ring",
"Definition:Linearly Dependent/Set"
] | [
"Scalar Product with Identity",
"Definition:Sequence",
"Definition:Sequence",
"Definition:Element",
"Definition:Sequence"
] |
proofwiki-1263 | Subset of Linearly Independent Set is Linearly Independent | A subset of a linearly independent set is also linearly independent. | Let $G$ be an unitary $R$-module.
Then $\sequence {a_n}$ is a linearly independent sequence {{iff}} $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$.
So suppose that $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$.
Then clearly $\sequence {a_n}$ is a linearly independent sequence ... | A [[Definition:Subset|subset]] of a [[Definition:Linearly Independent Set|linearly independent set]] is also [[Definition:Linearly Independent Set|linearly independent]]. | Let $G$ be an [[Definition:Unitary Module|unitary $R$-module]].
Then $\sequence {a_n}$ is a [[Definition:Linearly Independent Sequence|linearly independent sequence]] {{iff}} $\set {a_1, a_2, \ldots, a_n}$ is a [[Definition:Linearly Independent Set|linearly independent set]] of $G$.
So suppose that $\set {a_1, a_2, ... | Subset of Linearly Independent Set is Linearly Independent | https://proofwiki.org/wiki/Subset_of_Linearly_Independent_Set_is_Linearly_Independent | https://proofwiki.org/wiki/Subset_of_Linearly_Independent_Set_is_Linearly_Independent | [
"Unitary Modules",
"Linear Independence"
] | [
"Definition:Subset",
"Definition:Linearly Independent/Set",
"Definition:Linearly Independent/Set"
] | [
"Definition:Unitary Module over Ring",
"Definition:Linearly Independent/Sequence",
"Definition:Linearly Independent/Set",
"Definition:Linearly Independent/Set",
"Definition:Linearly Independent/Sequence",
"Definition:Linearly Independent/Sequence",
"Definition:Sequence of Distinct Terms",
"Definition:... |
proofwiki-1264 | Superset of Linearly Dependent Set is Linearly Dependent | Let $S$ be a set of elements of a unitary module.
Let $S$ contain a subset $T$ which is linearly dependent.
Then $S$ is also linearly dependent. | Let $S$ be a linearly independent set.
Let $T$ be a subset of $S$.
By Subset of Linearly Independent Set is Linearly Independent, $T$ is also linearly independent.
Thus by Proof by Contraposition, if $T$ is linearly dependent, then so must $S$ be.
{{qed}} | Let $S$ be a [[Definition:Set|set]] of [[Definition:Element|elements]] of a [[Definition:Unitary Module|unitary module]].
Let $S$ contain a [[Definition:Subset|subset]] $T$ which is [[Definition:Linearly Dependent Set|linearly dependent]].
Then $S$ is also [[Definition:Linearly Dependent Set|linearly dependent]]. | Let $S$ be a [[Definition:Linearly Independent Set|linearly independent set]].
Let $T$ be a [[Definition:Subset|subset]] of $S$.
By [[Subset of Linearly Independent Set is Linearly Independent]], $T$ is also [[Definition:Linearly Independent Set|linearly independent]].
Thus by [[Proof by Contraposition]], if $T$ is ... | Superset of Linearly Dependent Set is Linearly Dependent | https://proofwiki.org/wiki/Superset_of_Linearly_Dependent_Set_is_Linearly_Dependent | https://proofwiki.org/wiki/Superset_of_Linearly_Dependent_Set_is_Linearly_Dependent | [
"Unitary Modules",
"Linear Dependence",
"Proofs by Contraposition"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Unitary Module over Ring",
"Definition:Subset",
"Definition:Linearly Dependent/Set",
"Definition:Linearly Dependent/Set"
] | [
"Definition:Linearly Independent/Set",
"Definition:Subset",
"Subset of Linearly Independent Set is Linearly Independent",
"Definition:Linearly Independent/Set",
"Proof by Contraposition",
"Definition:Linearly Dependent/Set"
] |
proofwiki-1265 | Number of Ordered Bases from a Basis | Each basis of $n$ elements determines $n!$ ordered bases. | Follows directly from Cardinality of Set of Bijections.
{{qed}} | Each [[Definition:Basis (Linear Algebra)|basis]] of $n$ elements determines $n!$ [[Definition:Ordered Basis|ordered bases]]. | Follows directly from [[Cardinality of Set of Bijections]].
{{qed}} | Number of Ordered Bases from a Basis | https://proofwiki.org/wiki/Number_of_Ordered_Bases_from_a_Basis | https://proofwiki.org/wiki/Number_of_Ordered_Bases_from_a_Basis | [
"Module Theory"
] | [
"Definition:Basis (Linear Algebra)",
"Definition:Ordered Basis"
] | [
"Cardinality of Set of Bijections"
] |
proofwiki-1266 | Classification of Compact Two-Manifolds | A smooth, compact, path-connected manifold of dimension $2$ is diffeomorphic to one of the following:
:the sphere $\mathbb S^2$
:a connected sum of tori $\mathbb T^2$
:a connected sum of projective spaces $\mathbb{RP}^2$.
Such a $2$-manifold with boundary is diffeomorphic to one of the following:
:the sphere $\mathbb S... | It is known that the connected sum of $g$ tori, $\mathbb T_1^2 \# \mathbb T_2^2 \# \ldots \# \mathbb T_g^2$, which we denote $g \mathbb T^2$, is orientable and has Euler characteristic $2 - 2 g - b$, where $b$ is the number of boundary curves.
It is also known that the connected sum of $p$ projective spaces $p \mathbb{... | A [[Definition:Smooth Manifold|smooth]], [[Definition:Compact Topological Space|compact]], [[Definition:Path-Connected|path-connected]] [[Definition:Topological Manifold|manifold]] of [[Definition:Dimension (Topology)|dimension $2$]] is [[Definition:Diffeomorphism|diffeomorphic]] to one of the following:
:the [[Definit... | It is known that the [[Definition:Connected Sum|connected sum]] of $g$ [[Definition:Torus (Topology)|tori]], $\mathbb T_1^2 \# \mathbb T_2^2 \# \ldots \# \mathbb T_g^2$, which we denote $g \mathbb T^2$, is [[Definition:Orientable|orientable]] and has [[Definition:Euler Characteristic|Euler characteristic]] $2 - 2 g - b... | Classification of Compact Two-Manifolds | https://proofwiki.org/wiki/Classification_of_Compact_Two-Manifolds | https://proofwiki.org/wiki/Classification_of_Compact_Two-Manifolds | [
"Classification of Compact Two-Manifolds",
"Smooth Manifolds",
"Algebraic Topology"
] | [
"Definition:Topological Manifold/Smooth Manifold",
"Definition:Compact Topological Space",
"Definition:Path-Connected",
"Definition:Topological Manifold",
"Definition:Dimension (Topology)",
"Definition:Diffeomorphism",
"Definition:Sphere/Topology",
"Definition:Connected Sum",
"Definition:Torus (Topo... | [
"Definition:Connected Sum",
"Definition:Torus (Topology)",
"Definition:Orientable",
"Definition:Euler Characteristic",
"Definition:Boundary Curve",
"Definition:Connected Sum",
"Definition:Projective Space",
"Definition:Non-Orientable",
"Definition:Euler Characteristic",
"Definition:Euler Character... |
proofwiki-1267 | Standard Ordered Basis is Basis | Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $n$ be a positive integer.
For each $j \in \closedint 1 n$, let $e_j$ be the ordered $n$-tuple of elements of $R$ whose $j$th entry is $1_R$ and all of whose other entries is $0_R$.
Then $\sequence {e_n}$ is an ordered ba... | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \lambda_k e_k
| r = \lambda_1 \tuple {1_R, 0_R, 0_R, \ldots, 0_R}
| c =
}}
{{eqn | o = +
| r = \lambda_2 \tuple {0_R, 1_R, 0_R, \ldots, 0_R}
| c =
}}
{{eqn | o = +
| r = \ldots
| c =
}}
{{eqn | o = +
| r = \lambda_n \tuple {0_... | 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 $n$ be a [[Definition:Positive Integer|positive integer]].
For each $j \in \closedint 1 n$, let $e_j$ be the [[Definition:Ordered Tupl... | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \lambda_k e_k
| r = \lambda_1 \tuple {1_R, 0_R, 0_R, \ldots, 0_R}
| c =
}}
{{eqn | o = +
| r = \lambda_2 \tuple {0_R, 1_R, 0_R, \ldots, 0_R}
| c =
}}
{{eqn | o = +
| r = \ldots
| c =
}}
{{eqn | o = +
| r = \lambda_n \tuple {0_... | Standard Ordered Basis is Basis | https://proofwiki.org/wiki/Standard_Ordered_Basis_is_Basis | https://proofwiki.org/wiki/Standard_Ordered_Basis_is_Basis | [
"Standard Ordered Bases"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Positive/Integer",
"Definition:Ordered Tuple",
"Definition:Element",
"Definition:Ordered Basis",
"Definition:Module on Cartesian Product",
"Definition:Ordered Basis",
"Definition:Standard ... | [] |
proofwiki-1268 | Basis for Finite Submodule of Function Space | Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $A$ be a set.
For each $a \in A$, let $f_a: A \to R$ be defined as:
:$\forall x \in A: \map {f_a} x = \begin{cases}
1 & : x = a \\
0 & : x \ne a
\end{cases}$
Then $B = \set {f_a: a \in A}$ is a basis of the Finite Submodu... | Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of distinct terms of $A$.
Let $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of scalars.
Then: $\ds \sum_{k \mathop = 1}^n \lambda_k f_{a_k}$ is the mapping whose value at $a_k$ is $\lambda_k$ and whose value at any $x$ not in $\... | 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 $A$ be a [[Definition:Set|set]].
For each $a \in A$, let $f_a: A \to R$ be defined as:
:$\forall x \in A: \map {f_a} x = \begin{cases... | Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Sequence of Distinct Terms|sequence of distinct terms]] of $A$.
Let $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Scalar (Module)|scalars]].
Then: $\ds \sum_{k \mathop = 1}^n \lambda... | Basis for Finite Submodule of Function Space | https://proofwiki.org/wiki/Basis_for_Finite_Submodule_of_Function_Space | https://proofwiki.org/wiki/Basis_for_Finite_Submodule_of_Function_Space | [
"Submodules",
"Function Spaces",
"Linear Algebra",
"Bases of Modules"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Set",
"Definition:Basis (Linear Algebra)",
"Finite Submodule of Function Space"
] | [
"Definition:Sequence of Distinct Terms",
"Definition:Sequence",
"Definition:Scalar/Module",
"Definition:Mapping",
"Definition:Generator of Module",
"Definition:Linearly Independent/Set",
"Definition:Basis (Linear Algebra)",
"Definition:Standard Basis"
] |
proofwiki-1269 | Unique Representation by Ordered Basis | Let $G$ be a unitary $R$-module.
Then $\sequence {a_n}$ is an ordered basis of $G$ {{iff}}:
:For every $x \in G$ there exists one and only one sequence $\sequence {\lambda_n}$ of scalars such that $\ds x = \sum_{k \mathop = 1}^n \lambda_k a_k$. | === Necessary Condition ===
Let $\sequence {a_n}$ be an ordered basis of $G$.
Then every element of $G$ is a linear combination of $\set {a_1, \ldots, a_n}$, which is a generator of $G$, by Generated Submodule is Linear Combinations.
Thus there exists at least one such sequence of scalar.
Now suppose there were two suc... | Let $G$ be a [[Definition:Unitary Module|unitary $R$-module]].
Then $\sequence {a_n}$ is an [[Definition:Ordered Basis|ordered basis]] of $G$ {{iff}}:
:For every $x \in G$ there exists [[Definition:Exactly One|one and only one]] [[Definition:Sequence|sequence]] $\sequence {\lambda_n}$ of [[Definition:Scalar (Module)|... | === Necessary Condition ===
Let $\sequence {a_n}$ be an [[Definition:Ordered Basis|ordered basis]] of $G$.
Then every element of $G$ is a [[Definition:Linear Combination|linear combination]] of $\set {a_1, \ldots, a_n}$, which is a [[Definition:Generator of Module|generator]] of $G$, by [[Generated Submodule is Linea... | Unique Representation by Ordered Basis | https://proofwiki.org/wiki/Unique_Representation_by_Ordered_Basis | https://proofwiki.org/wiki/Unique_Representation_by_Ordered_Basis | [
"Unitary Modules",
"Linear Algebra",
"Bases of Modules"
] | [
"Definition:Unitary Module over Ring",
"Definition:Ordered Basis",
"Definition:Unique",
"Definition:Sequence",
"Definition:Scalar/Module"
] | [
"Definition:Ordered Basis",
"Definition:Linear Combination",
"Definition:Generator of Module",
"Generated Submodule is Linear Combinations",
"Definition:Sequence",
"Definition:Scalar/Module",
"Definition:Sequence",
"Definition:Scalar/Module",
"Definition:Linearly Independent/Sequence",
"Definition... |
proofwiki-1270 | Isomorphism from R^n via n-Term Sequence | Let $G$ be a unitary $R$-module.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an ordered basis of $G$.
Let $R^n$ be the $R$-module $R^n$.
Let $\psi: R^n \to G$ be defined as:
:$\ds \map \psi {\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n} } = \sum_{k \mathop = 1}^n \lambda_k a_k$
Then $\psi$ is an is... | By Unique Representation by Ordered Basis, $\psi$ is a bijection.
We have:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \lambda_k a_k + \sum_{k \mathop = 1}^n \mu_k a_k
| r = \sum_{k \mathop = 1}^n \paren {\lambda_k a_k + \mu_k a_k}
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {\lambda_k + \mu_k}... | Let $G$ be a [[Definition:Unitary Module|unitary $R$-module]].
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an [[Definition:Ordered Basis|ordered basis]] of $G$.
Let $R^n$ be the [[Definition:Module on Cartesian Product|$R$-module $R^n$]].
Let $\psi: R^n \to G$ be defined as:
:$\ds \map \psi {\sequence ... | By [[Unique Representation by Ordered Basis]], $\psi$ is a [[Definition:Bijection|bijection]].
We have:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \lambda_k a_k + \sum_{k \mathop = 1}^n \mu_k a_k
| r = \sum_{k \mathop = 1}^n \paren {\lambda_k a_k + \mu_k a_k}
| c =
}}
{{eqn | r = \sum_{k \mathop =... | Isomorphism from R^n via n-Term Sequence | https://proofwiki.org/wiki/Isomorphism_from_R^n_via_n-Term_Sequence | https://proofwiki.org/wiki/Isomorphism_from_R^n_via_n-Term_Sequence | [
"Module Theory",
"Unitary Modules",
"R-Algebraic Structure Isomorphisms"
] | [
"Definition:Unitary Module over Ring",
"Definition:Ordered Basis",
"Definition:Module on Cartesian Product",
"Definition:Isomorphism (Abstract Algebra)/R-Algebraic Structure Isomorphism"
] | [
"Unique Representation by Ordered Basis",
"Definition:Bijection",
"Definition:R-Algebraic Structure Homomorphism"
] |
proofwiki-1271 | Unitary R-Modules with n-Element Bases are Isomorphic | Let $G$ and $H$ be unitary $R$-modules for some ring with unity $R$.
Let $G$ and $G$ both have bases with $n$ elements.
Then $G$ and $H$ are isomorphic. | From Isomorphism from $R^n$ via $n$-Term Sequence, both $G$ and $H$ are isomorphic to the $R$-module $R^n$.
{{qed}} | Let $G$ and $H$ be [[Definition:Unitary Module|unitary $R$-modules]] for some [[Definition:Ring with Unity|ring with unity]] $R$.
Let $G$ and $G$ both have [[Definition:Basis (Linear Algebra)|bases]] with $n$ [[Definition:Element|elements]].
Then $G$ and $H$ are [[Definition:R-Algebraic Structure Isomorphism|isomorph... | From [[Isomorphism from R^n via n-Term Sequence|Isomorphism from $R^n$ via $n$-Term Sequence]], both $G$ and $H$ are [[Definition:R-Algebraic Structure Isomorphism|isomorphic]] to the [[Definition:Module on Cartesian Product|$R$-module $R^n$]].
{{qed}} | Unitary R-Modules with n-Element Bases are Isomorphic | https://proofwiki.org/wiki/Unitary_R-Modules_with_n-Element_Bases_are_Isomorphic | https://proofwiki.org/wiki/Unitary_R-Modules_with_n-Element_Bases_are_Isomorphic | [
"Unitary Modules"
] | [
"Definition:Unitary Module over Ring",
"Definition:Ring with Unity",
"Definition:Basis (Linear Algebra)",
"Definition:Element",
"Definition:Isomorphism (Abstract Algebra)/R-Algebraic Structure Isomorphism"
] | [
"Isomorphism from R^n via n-Term Sequence",
"Definition:Isomorphism (Abstract Algebra)/R-Algebraic Structure Isomorphism",
"Definition:Module on Cartesian Product"
] |
proofwiki-1272 | R-Module R^n is n-Dimensional | The $R$-module $R^n$ is $n$-dimensional. | Follows directly from Standard Ordered Basis is Basis.
{{qed}}
Category:Module on Cartesian Product
4e2u0vlwjpxp8kx7zj4kipknso5ie65 | The [[Definition:Module on Cartesian Product|$R$-module $R^n$]] is [[Definition:Dimension (Linear Algebra)|$n$-dimensional]]. | Follows directly from [[Standard Ordered Basis is Basis]].
{{qed}}
[[Category:Module on Cartesian Product]]
4e2u0vlwjpxp8kx7zj4kipknso5ie65 | R-Module R^n is n-Dimensional | https://proofwiki.org/wiki/R-Module_R^n_is_n-Dimensional | https://proofwiki.org/wiki/R-Module_R^n_is_n-Dimensional | [
"Module on Cartesian Product"
] | [
"Definition:Module on Cartesian Product",
"Definition:Dimension (Linear Algebra)"
] | [
"Standard Ordered Basis is Basis",
"Category:Module on Cartesian Product"
] |
proofwiki-1273 | Unique Linear Transformation Between Modules | Let $\struct {G, +_G, \times_G}_R$ and $\struct {H, +_H, \times_H}_R$ be unitary $R$-modules.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an ordered basis of $G$.
Let $\sequence {b_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of $n$ elements of $H$.
Then there exists a unique linear transformation $\p... | By Unique Representation by Ordered Basis, the mapping $\phi: G \to H$ defined as:
:$(1): \quad \ds \map \phi {\sum_{j \mathop = 1}^n \lambda_j \times_G a_j} = \sum_{j \mathop = 1}^n \lambda_j \times_H b_j$
is well-defined.
{{begin-eqn}}
{{eqn | q = \forall k \in \closedint 1 n
| l = \map \phi {\sum_{j \mathop = ... | Let $\struct {G, +_G, \times_G}_R$ and $\struct {H, +_H, \times_H}_R$ be [[Definition:Unitary Module|unitary $R$-modules]].
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an [[Definition:Ordered Basis|ordered basis]] of $G$.
Let $\sequence {b_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Finite Sequ... | By [[Unique Representation by Ordered Basis]], the [[Definition:Mapping|mapping]] $\phi: G \to H$ defined as:
:$(1): \quad \ds \map \phi {\sum_{j \mathop = 1}^n \lambda_j \times_G a_j} = \sum_{j \mathop = 1}^n \lambda_j \times_H b_j$
is [[Definition:Well-Defined Mapping|well-defined]].
{{begin-eqn}}
{{eqn | q = \foral... | Unique Linear Transformation Between Modules | https://proofwiki.org/wiki/Unique_Linear_Transformation_Between_Modules | https://proofwiki.org/wiki/Unique_Linear_Transformation_Between_Modules | [
"Linear Transformations"
] | [
"Definition:Unitary Module over Ring",
"Definition:Ordered Basis",
"Definition:Finite Sequence",
"Definition:Element",
"Definition:Unique",
"Definition:Linear Transformation"
] | [
"Unique Representation by Ordered Basis",
"Definition:Mapping",
"Definition:Well-Defined/Mapping",
"Definition:Kronecker Delta",
"Definition:Linear Transformation",
"Definition:Element",
"Definition:Element",
"Linear Transformation of Generated Module",
"Definition:Linear Transformation"
] |
proofwiki-1274 | Linear Transformation from Ordered Basis less Kernel | Let $G$ and $H$ be unitary $R$-modules.
Let $\phi: G \to H$ be a non-zero linear transformation.
Let $G$ be $n$-dimensional.
Let $\sequence {a_n}$ be any ordered basis of $G$ such that $\set {a_k: r + 1 \le k \le n}$ is the basis of the kernel of $\phi$.
Then $\sequence {\map \phi {a_r} }$ is an ordered basis of the im... | Suppose:
:$\ds \sum_{k \mathop = 1}^r \lambda_k \map \phi {a_k} = 0$
Then:
:$\ds \map \phi {\sum_{k \mathop = 1}^r \lambda_k a_k} = 0$
So $\ds \sum_{k \mathop = 1}^r \lambda_k \map \phi {a_k}$ belongs to the kernel of $\phi$ and hence is also a linear combination of $\set {a_k: r + 1 \le k \le n}$.
Thus $\forall k \in ... | Let $G$ and $H$ be [[Definition:Unitary Module|unitary $R$-modules]].
Let $\phi: G \to H$ be a non-zero [[Definition:Linear Transformation|linear transformation]].
Let $G$ be [[Definition:Dimension of Module|$n$-dimensional]].
Let $\sequence {a_n}$ be any [[Definition:Ordered Basis|ordered basis]] of $G$ such that $... | Suppose:
:$\ds \sum_{k \mathop = 1}^r \lambda_k \map \phi {a_k} = 0$
Then:
:$\ds \map \phi {\sum_{k \mathop = 1}^r \lambda_k a_k} = 0$
So $\ds \sum_{k \mathop = 1}^r \lambda_k \map \phi {a_k}$ belongs to the [[Definition:Kernel of Linear Transformation|kernel]] of $\phi$ and hence is also a [[Definition:Linear Combin... | Linear Transformation from Ordered Basis less Kernel | https://proofwiki.org/wiki/Linear_Transformation_from_Ordered_Basis_less_Kernel | https://proofwiki.org/wiki/Linear_Transformation_from_Ordered_Basis_less_Kernel | [
"Linear Transformations"
] | [
"Definition:Unitary Module over Ring",
"Definition:Linear Transformation",
"Definition:Dimension of Module",
"Definition:Ordered Basis",
"Definition:Basis (Linear Algebra)",
"Definition:Kernel of Linear Transformation",
"Definition:Ordered Basis",
"Definition:Image (Set Theory)/Mapping/Mapping"
] | [
"Definition:Kernel of Linear Transformation",
"Definition:Linear Combination",
"Definition:Linearly Independent/Sequence",
"Definition:Linearly Independent/Sequence",
"Definition:Ordered Basis",
"Definition:Image (Set Theory)/Mapping/Mapping"
] |
proofwiki-1275 | Addition of Linear Transformations | Let $\struct {G, +_G}$ and $\struct {H, +_H}$ be abelian groups.
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ_G}$ and $\struct {H, +_H, \circ_H}$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $\oplus_H$ be defined as pointwise addition... | From the definition of a module, the group $\struct {H, +_H}$ is abelian.
Hence $\struct {H, +_H}$ is {{afortiori}} a commutative semigroup.
Therefore we can apply Homomorphism on Induced Structure to Commutative Semigroup to show that $\phi +_H \psi: G \to H$ is a homomorphism.
Let $\lambda \in R, x \in G$.
Then:
{{be... | Let $\struct {G, +_G}$ and $\struct {H, +_H}$ be [[Definition:Abelian Group|abelian groups]].
Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G, +_G, \circ_G}$ and $\struct {H, +_H, \circ_H}$ be [[Definition:Module over Ring|$R$-modules]].
Let $\map {\LL_R} {G, H}$ be... | From the definition of a [[Definition:Module over Ring|module]], the [[Definition:Group|group]] $\struct {H, +_H}$ is [[Definition:Abelian Group|abelian]].
Hence $\struct {H, +_H}$ is {{afortiori}} a [[Definition:Commutative Semigroup|commutative semigroup]].
Therefore we can apply [[Homomorphism on Induced Structure... | Addition of Linear Transformations | https://proofwiki.org/wiki/Addition_of_Linear_Transformations | https://proofwiki.org/wiki/Addition_of_Linear_Transformations | [
"Linear Transformations"
] | [
"Definition:Abelian Group",
"Definition:Ring (Abstract Algebra)",
"Definition:Module over Ring",
"Definition:Set of All Linear Transformations",
"Definition:Pointwise Addition of Linear Transformations",
"Definition:Linear Transformation"
] | [
"Definition:Module over Ring",
"Definition:Group",
"Definition:Abelian Group",
"Definition:Commutative Semigroup",
"Homomorphism on Induced Structure to Commutative Semigroup",
"Definition:Homomorphism (Abstract Algebra)"
] |
proofwiki-1276 | Negative Linear Transformation | Let $\struct {G, +_G}$ and $\struct {H, +_H}$ be abelian groups.
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ_G}$ and $\struct {H, +_H, \circ_H}$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $\phi: G \to H$ be a linear transformation.... | From the definition of a module, the group $\struct {H, +_H}$ is abelian.
Therefore we can apply Inverse Mapping in Induced Structure of Homomorphism to Abelian Group to show that $-\phi: G \to H$ is a homomorphism.
Let $\lambda \in R, x \in G$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {-\phi} } {\lambda \circ_G x}... | Let $\struct {G, +_G}$ and $\struct {H, +_H}$ be [[Definition:Abelian Group|abelian groups]].
Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G, +_G, \circ_G}$ and $\struct {H, +_H, \circ_H}$ be [[Definition:Module over Ring|$R$-modules]].
Let $\map {\LL_R} {G, H}$ be... | From the definition of a [[Definition:Module over Ring|module]], the [[Definition:Group|group]] $\struct {H, +_H}$ is [[Definition:Abelian Group|abelian]].
Therefore we can apply [[Inverse Mapping in Induced Structure of Homomorphism to Abelian Group]] to show that $-\phi: G \to H$ is a [[Definition:R-Algebraic Struct... | Negative Linear Transformation | https://proofwiki.org/wiki/Negative_Linear_Transformation | https://proofwiki.org/wiki/Negative_Linear_Transformation | [
"Linear Transformations"
] | [
"Definition:Abelian Group",
"Definition:Ring (Abstract Algebra)",
"Definition:Module over Ring",
"Definition:Set of All Linear Transformations",
"Definition:Linear Transformation",
"Definition:Pointwise Negative of Linear Transformation",
"Definition:Linear Transformation"
] | [
"Definition:Module over Ring",
"Definition:Group",
"Definition:Abelian Group",
"Inverse Mapping in Induced Structure of Homomorphism to Abelian Group",
"Definition:R-Algebraic Structure Homomorphism"
] |
proofwiki-1277 | Set of Linear Transformations under Pointwise Addition forms Abelian Group | Let $\struct {G, +_G}$ and $\struct {H, +_H}$ be abelian groups.
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $\oplus_H$ be defined as pointwise addition... | From Structure Induced by Group Operation is Group, $\struct {H^G, \oplus_H}$ is a group
Let $\phi, \psi \in \map {\LL_R} {G, H}$.
From Addition of Linear Transformations:
:$\phi \oplus_H \psi \in \map {\LL_R} {G, H}$
From Negative Linear Transformation:
:$-\phi \in \map {\LL_R} {G, H}$
Thus, from the Two-Step Subgroup... | Let $\struct {G, +_G}$ and $\struct {H, +_H}$ be [[Definition:Abelian Group|abelian groups]].
Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be [[Definition:Module over Ring|$R$-modules]].
Let $\map {\LL_R} {G, H}$ b... | From [[Structure Induced by Group Operation is Group]], $\struct {H^G, \oplus_H}$ is a [[Definition:Group|group]]
Let $\phi, \psi \in \map {\LL_R} {G, H}$.
From [[Addition of Linear Transformations]]:
:$\phi \oplus_H \psi \in \map {\LL_R} {G, H}$
From [[Negative Linear Transformation]]:
:$-\phi \in \map {\LL_R} {G, ... | Set of Linear Transformations under Pointwise Addition forms Abelian Group | https://proofwiki.org/wiki/Set_of_Linear_Transformations_under_Pointwise_Addition_forms_Abelian_Group | https://proofwiki.org/wiki/Set_of_Linear_Transformations_under_Pointwise_Addition_forms_Abelian_Group | [
"Linear Transformations",
"Abelian Groups"
] | [
"Definition:Abelian Group",
"Definition:Ring (Abstract Algebra)",
"Definition:Module over Ring",
"Definition:Set of All Linear Transformations",
"Definition:Pointwise Addition of Linear Transformations",
"Definition:Abelian Group"
] | [
"Structure Induced by Group Operation is Group",
"Definition:Group",
"Addition of Linear Transformations",
"Negative Linear Transformation",
"Two-Step Subgroup Test",
"Definition:Subgroup",
"Definition:Abelian Group",
"Definition:Element",
"Definition:Abelian Group"
] |
proofwiki-1278 | Morse-Sard Theorem | Let $f: X \to Y$ be any smooth map of manifolds.
Then almost every point in $Y$ is a regular value of $f$. | {{ProofWanted|In progress}} | Let $f: X \to Y$ be any [[Definition:Smooth Mapping|smooth map]] of [[Definition:Topological Manifold|manifolds]].
Then [[Definition:Almost Every|almost every]] point in $Y$ is a [[Definition:Regular Value|regular value]] of $f$. | {{ProofWanted|In progress}} | Morse-Sard Theorem | https://proofwiki.org/wiki/Morse-Sard_Theorem | https://proofwiki.org/wiki/Morse-Sard_Theorem | [
"Topological Manifolds",
"Smooth Mappings"
] | [
"Definition:Smooth Mapping",
"Definition:Topological Manifold",
"Definition:Almost Everywhere",
"Definition:Regular Value"
] | [] |
proofwiki-1279 | Linear Transformation from Center of Scalar Ring | Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.
Let $\phi: G \to H$ be a linear transformation.
Let $\map Z R$ be the center of the scalar ring $R$.
Let $\lambda \in \map Z R$.
Then $\lambda \circ \phi$ is a linear transformation. | By definition of linear transformation, we need to show that:
:$(1): \quad \forall x, y \in G: \map {\paren {\lambda \circ \phi} } {x +_G y} = \lambda \circ \map \phi x +_H \lambda \circ \map \phi y$
:$(2): \quad \forall x \in G: \forall \mu \in R: \map {\paren {\lambda \circ \phi} } {\mu \circ x} = \mu \circ \map {\pa... | Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be [[Definition:Module over Ring|$R$-modules]].
Let $\phi: G \to H$ be a [[Definition:Linear Transformation|linear transformation]].
Let $\map Z R$ be the [[Definition:Center of Ring|center]] of the [[Definition:Scalar Ring of Module|scalar ring]] $R$.
... | By definition of [[Definition:Linear Transformation|linear transformation]], we need to show that:
:$(1): \quad \forall x, y \in G: \map {\paren {\lambda \circ \phi} } {x +_G y} = \lambda \circ \map \phi x +_H \lambda \circ \map \phi y$
:$(2): \quad \forall x \in G: \forall \mu \in R: \map {\paren {\lambda \circ \phi} ... | Linear Transformation from Center of Scalar Ring | https://proofwiki.org/wiki/Linear_Transformation_from_Center_of_Scalar_Ring | https://proofwiki.org/wiki/Linear_Transformation_from_Center_of_Scalar_Ring | [
"Linear Transformations"
] | [
"Definition:Module over Ring",
"Definition:Linear Transformation",
"Definition:Center (Abstract Algebra)/Ring",
"Definition:Scalar Ring/Module",
"Definition:Linear Transformation"
] | [
"Definition:Linear Transformation",
"Definition:Commutative/Elements"
] |
proofwiki-1280 | Set of Linear Transformations over Commutative Ring forms Submodule of Module of All Mappings | Let $R$ be a commutative ring.
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Then $\map {\LL_R} {G, H}$ is a submodule of the $R$-module $H^G$. | By definition of abelian group, the center of a commutative ring $R$ is $R$ itself.
So, we need to show that every $\phi \in \map {\LL_R} {G, H}$ fulfils the conditions to be a linear transformation:
:$(1): \quad \forall x, y \in G: \map {\paren {\lambda \circ \phi} } {x +_G y} = \lambda \circ \map \phi x +_H \lambda \... | Let $R$ be a [[Definition:Commutative Ring|commutative ring]].
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be [[Definition:Module over Ring|$R$-modules]].
Let $\map {\LL_R} {G, H}$ be [[Definition:Set of All Linear Transformations|the set of all linear transformations]] from $G$ to $H$.
Then $\m... | By definition of [[Definition:Abelian Group|abelian group]], the [[Definition:Center of Ring|center]] of a [[Definition:Commutative Ring|commutative ring]] $R$ is $R$ itself.
So, we need to show that every $\phi \in \map {\LL_R} {G, H}$ fulfils the conditions to be a [[Definition:Linear Transformation|linear transform... | Set of Linear Transformations over Commutative Ring forms Submodule of Module of All Mappings | https://proofwiki.org/wiki/Set_of_Linear_Transformations_over_Commutative_Ring_forms_Submodule_of_Module_of_All_Mappings | https://proofwiki.org/wiki/Set_of_Linear_Transformations_over_Commutative_Ring_forms_Submodule_of_Module_of_All_Mappings | [
"Linear Transformations"
] | [
"Definition:Commutative Ring",
"Definition:Module over Ring",
"Definition:Set of All Linear Transformations",
"Definition:Submodule",
"Definition:Module of All Mappings"
] | [
"Definition:Abelian Group",
"Definition:Center (Abstract Algebra)/Ring",
"Definition:Commutative Ring",
"Definition:Linear Transformation",
"Linear Transformation from Center of Scalar Ring"
] |
proofwiki-1281 | Dimension of Set of Linear Transformations | Let $R$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module such that $\map \dim G = n$.
Let $\struct {H, +_H, \circ}_R$ be a unitary $R$-module such that $\map \dim H = m$.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformati... | Consider the set:
:$B = \set {\phi_{i j}: i \in \closedint 1 n, j \in \closedint 1 m}$
From Basis for Set of Linear Transformations, $B$ is a basis for $\map {\LL_R} {G, H}$.
It is seen that by construction, $B$ has $n m$ elements.
The result follows by definition of dimension.
{{Qed}} | Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a [[Definition:Unitary Module|unitary $R$-module]] such that $\map \dim G = n$.
Let $\struct {H, +... | Consider the [[Definition:Set|set]]:
:$B = \set {\phi_{i j}: i \in \closedint 1 n, j \in \closedint 1 m}$
From [[Basis for Set of Linear Transformations]], $B$ is a [[Definition:Basis (Linear Algebra)|basis]] for $\map {\LL_R} {G, H}$.
It is seen that by construction, $B$ has $n m$ [[Definition:Element|elements]].
... | Dimension of Set of Linear Transformations | https://proofwiki.org/wiki/Dimension_of_Set_of_Linear_Transformations | https://proofwiki.org/wiki/Dimension_of_Set_of_Linear_Transformations | [
"Linear Transformations"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Unitary Module over Ring",
"Definition:Unitary Module over Ring",
"Definition:Set of All Linear Transformations",
"Definition:Dimension (Linear Algebra)"
] | [
"Definition:Set",
"Basis for Set of Linear Transformations",
"Definition:Basis (Linear Algebra)",
"Definition:Element",
"Definition:Dimension (Linear Algebra)"
] |
proofwiki-1282 | Dimension of Algebraic Dual | Let $R$ be a commutative ring with unity.
Let $G$ be an $n$-dimensional $R$-module.
Let $G^*$ be the algebraic dual of $G$.
Then $G^*$ is also $n$-dimensional. | Let $\struct {R, +_R, \circ}_R$ denote the $R$-module $R$.
From Dimension of $R$-Module $R$ is $1$ we have that the dimension of $\struct {R, +_R, \circ}_R$ is $1$.
By definition, the algebraic dual of $G$ is the $R$-module $\map {\LL_R} {G, R}$ of all linear forms on $G$.
From Dimension of Set of Linear Transformation... | Let $R$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $G$ be an [[Definition:Dimension (Linear Algebra)|$n$-dimensional]] [[Definition:Module over Ring|$R$-module]].
Let $G^*$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G$.
Then $G^*$ is also [[Definition:Dimension (Li... | Let $\struct {R, +_R, \circ}_R$ denote the [[Definition:R-Module R|$R$-module $R$]].
From [[Dimension of R-Module R is 1|Dimension of $R$-Module $R$ is $1$]] we have that the [[Definition:Dimension (Linear Algebra)|dimension]] of $\struct {R, +_R, \circ}_R$ is $1$.
By definition, the [[Definition:Algebraic Dual|algeb... | Dimension of Algebraic Dual | https://proofwiki.org/wiki/Dimension_of_Algebraic_Dual | https://proofwiki.org/wiki/Dimension_of_Algebraic_Dual | [
"Algebraic Duals"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Dimension (Linear Algebra)",
"Definition:Module over Ring",
"Definition:Algebraic Dual",
"Definition:Dimension (Linear Algebra)"
] | [
"Definition:Module on Cartesian Product/Special Case",
"Dimension of R-Module R is 1",
"Definition:Dimension (Linear Algebra)",
"Definition:Algebraic Dual",
"Definition:Module over Ring",
"Definition:Linear Form (Linear Algebra)",
"Dimension of Set of Linear Transformations",
"Definition:Dimension (Li... |
proofwiki-1283 | Annihilator is Submodule of Algebraic Dual | Let $R$ be a commutative ring with unity.
Let $G$ be a module over $R$.
Let $M$ be a submodule of $G$.
Let $G^*$ be the algebraic dual of $G$.
Then the annihilator $M^\circ$ of $M$ is a submodule of $G^*$. | By definition, $M^\circ$ is a subset of $G^*$.
Recall that by definition of algebraic dual, the elements of $G^*$ are linear transformations from $G$ to the $R$-module $R$.
By Submodule Test, it remains to be shown that:
{{begin-eqn}}
{{eqn | n = SM1
| q = \forall u, v \in M^\circ
| l = u + v
| o = \i... | Let $R$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $G$ be a [[Definition:Module over Ring|module]] over $R$.
Let $M$ be a [[Definition:Submodule|submodule]] of $G$.
Let $G^*$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G$.
Then the [[Definition:Annihilator on Algeb... | By definition, $M^\circ$ is a [[Definition:Subset|subset]] of $G^*$.
Recall that by definition of [[Definition:Algebraic Dual|algebraic dual]], the [[Definition:Element|elements]] of $G^*$ are [[Definition:Linear Transformation|linear transformations]] from $G$ to the [[Definition:R-Module R|$R$-module $R$]].
By [[S... | Annihilator is Submodule of Algebraic Dual | https://proofwiki.org/wiki/Annihilator_is_Submodule_of_Algebraic_Dual | https://proofwiki.org/wiki/Annihilator_is_Submodule_of_Algebraic_Dual | [
"Annihilator is Submodule of Algebraic Dual",
"Submodules",
"Annihilators",
"Algebraic Duals"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Module over Ring",
"Definition:Submodule",
"Definition:Algebraic Dual",
"Definition:Annihilator on Algebraic Dual",
"Definition:Submodule"
] | [
"Definition:Subset",
"Definition:Algebraic Dual",
"Definition:Element",
"Definition:Linear Transformation",
"Definition:Module on Cartesian Product/Special Case",
"Submodule Test",
"Definition:Indeed"
] |
proofwiki-1284 | Inverse Evaluation Isomorphism of Annihilator | Let $R$ be a commutative ring with unity.
Let $G$ be a module over $R$ whose dimension is finite.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the algebraic dual of $G^*$.
Let $N$ be a submodule of $G^*$.
Let $J$ be the evaluation isomorphism from $G$ onto $G^{**}$.
Let $N^\circ$ be the annihilator of $N$.
T... | We are given {{hypothesis}} that $G$ is finite-dimensional.
Hence by Evaluation Isomorphism is Isomorphism, $J: G \to G^{**}$ is an isomorphism, and therefore a surjection.
Thus:
:$N^\circ = \set {x^\wedge \in G^{**}: \forall t \in N: \map {x^\wedge} t = 0}$
where $x^\wedge$ is as defined in the definition of the evalu... | Let $R$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $G$ be a [[Definition:Module over Ring|module over $R$]] whose [[Definition:Dimension of Module|dimension]] is [[Definition:Finite|finite]].
Let $G^*$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G$.
Let $G^{**}$ be t... | We are given {{hypothesis}} that $G$ is [[Definition:Finite|finite]]-[[Definition:Dimension of Module|dimensional]].
Hence by [[Evaluation Isomorphism is Isomorphism]], $J: G \to G^{**}$ is an [[Definition:Module Isomorphism|isomorphism]], and therefore a [[Definition:Surjection|surjection]].
Thus:
:$N^\circ = \set {... | Inverse Evaluation Isomorphism of Annihilator | https://proofwiki.org/wiki/Inverse_Evaluation_Isomorphism_of_Annihilator | https://proofwiki.org/wiki/Inverse_Evaluation_Isomorphism_of_Annihilator | [
"Linear Transformations"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Module over Ring",
"Definition:Dimension of Module",
"Definition:Finite",
"Definition:Algebraic Dual",
"Definition:Algebraic Dual",
"Definition:Submodule",
"Definition:Evaluation Isomorphism",
"Definition:Annihilator on Algebraic Dual",
"Defin... | [
"Definition:Finite",
"Definition:Dimension of Module",
"Evaluation Isomorphism is Isomorphism",
"Definition:Isomorphism (Abstract Algebra)/R-Algebraic Structure Isomorphism/Module Isomorphism",
"Definition:Surjection",
"Definition:Evaluation Linear Transformation/Module Theory"
] |
proofwiki-1285 | Evaluation Linear Transformation is Bilinear | Let $R$ be a commutative ring.
Let $G$ be an $R$-module.
Let $G^*$ be the algebraic dual of $G$.
Let $\innerprod x t$ be the evaluation linear transformation from $G$ to $G^{**}$.
Then the mapping $\phi: G \times G^* \to R$ defined as:
:$\forall \tuple {x, t} \in G \times G^*: \map \phi {x, t} = \innerprod x t$
satisfi... | {{begin-eqn}}
{{eqn | n = 1
| q = \forall x, y \in G: \forall t \in G^*
| l = \innerprod {x + y} t
| r = \map t {x + y}
| c = {{Defof|Evaluation Linear Transformation/Module Theory}}
}}
{{eqn | r = \map t x + \map t y
| c = $t$ is a Linear Transformation
}}
{{eqn | r = \innerprod x t + \in... | Let $R$ be a [[Definition:Commutative Ring|commutative ring]].
Let $G$ be an [[Definition:Module over Ring|$R$-module]].
Let $G^*$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G$.
Let $\innerprod x t$ be the [[Definition:Evaluation Linear Transformation/Module Theory|evaluation linear transformation]] fr... | {{begin-eqn}}
{{eqn | n = 1
| q = \forall x, y \in G: \forall t \in G^*
| l = \innerprod {x + y} t
| r = \map t {x + y}
| c = {{Defof|Evaluation Linear Transformation/Module Theory}}
}}
{{eqn | r = \map t x + \map t y
| c = $t$ is a [[Definition:Linear Transformation|Linear Transformation]... | Evaluation Linear Transformation is Bilinear | https://proofwiki.org/wiki/Evaluation_Linear_Transformation_is_Bilinear | https://proofwiki.org/wiki/Evaluation_Linear_Transformation_is_Bilinear | [
"Linear Transformations",
"Evaluation Linear Transformations (Module Theory)"
] | [
"Definition:Commutative Ring",
"Definition:Module over Ring",
"Definition:Algebraic Dual",
"Definition:Evaluation Linear Transformation/Module Theory",
"Definition:Mapping",
"Definition:Bilinear Mapping"
] | [
"Definition:Linear Transformation",
"Definition:Linear Transformation",
"Definition:Linear Transformation"
] |
proofwiki-1286 | Zero Vector Space Product iff Factor is Zero | Let $\struct {F, +_F, \times_F}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {\mathbf V, +_V, \ast}_F$ be a vector space over $F$, as defined by the vector space axioms.
Let $\mathbf v \in \mathbf V, \lambda \in F$.
Then:
:$\lambda \ast \mathbf v = \bszero \iff \paren {\lambda = 0_F \lor x = \... | A vector space is a module, so all results about modules also apply to vector spaces.
So from Scalar Product with Identity it follows directly that:
:$\lambda = 0_F \lor \mathbf v = e \implies \lambda \ast \mathbf v = \bszero$
Next, suppose $\lambda \ast \mathbf v = \bszero$ but $\lambda \ne 0_F$.
Then:
{{begin-eqn}}
{... | Let $\struct {F, +_F, \times_F}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $\struct {\mathbf V, +_V, \ast}_F$ be a [[Definition:Vector Space|vector space]] over $F$, as defined by the [[Axiom:Vector Space... | A [[Definition:Vector Space|vector space]] is a [[Definition:Module over Ring|module]], so all results about [[Definition:Module over Ring|modules]] also apply to [[Definition:Vector Space|vector spaces]].
So from [[Scalar Product with Identity]] it follows directly that:
:$\lambda = 0_F \lor \mathbf v = e \implies \l... | Zero Vector Space Product iff Factor is Zero/Proof 1 | https://proofwiki.org/wiki/Zero_Vector_Space_Product_iff_Factor_is_Zero | https://proofwiki.org/wiki/Zero_Vector_Space_Product_iff_Factor_is_Zero/Proof_1 | [
"Zero Vector Space Product iff Factor is Zero",
"Zero Vectors",
"Vector Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Vector Space",
"Axiom:Vector Space Axioms"
] | [
"Definition:Vector Space",
"Definition:Module over Ring",
"Definition:Module over Ring",
"Definition:Vector Space",
"Scalar Product with Identity",
"Zero Vector Scaled is Zero Vector"
] |
proofwiki-1287 | Zero Vector Space Product iff Factor is Zero | Let $\struct {F, +_F, \times_F}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {\mathbf V, +_V, \ast}_F$ be a vector space over $F$, as defined by the vector space axioms.
Let $\mathbf v \in \mathbf V, \lambda \in F$.
Then:
:$\lambda \ast \mathbf v = \bszero \iff \paren {\lambda = 0_F \lor x = \... | The sufficient condition is proved in Vector Scaled by Zero is Zero Vector, and in Zero Vector Scaled is Zero Vector.
The necessary condition is proved in Vector Product is Zero only if Factor is Zero.
{{qed}} | Let $\struct {F, +_F, \times_F}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $\struct {\mathbf V, +_V, \ast}_F$ be a [[Definition:Vector Space|vector space]] over $F$, as defined by the [[Axiom:Vector Space... | The [[Definition:Sufficient Condition|sufficient condition]] is proved in [[Vector Scaled by Zero is Zero Vector]], and in [[Zero Vector Scaled is Zero Vector]].
The [[Definition:Necessary Condition|necessary condition]] is proved in [[Vector Product is Zero only if Factor is Zero]].
{{qed}} | Zero Vector Space Product iff Factor is Zero/Proof 2 | https://proofwiki.org/wiki/Zero_Vector_Space_Product_iff_Factor_is_Zero | https://proofwiki.org/wiki/Zero_Vector_Space_Product_iff_Factor_is_Zero/Proof_2 | [
"Zero Vector Space Product iff Factor is Zero",
"Zero Vectors",
"Vector Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Vector Space",
"Axiom:Vector Space Axioms"
] | [
"Definition:Conditional/Sufficient Condition",
"Vector Scaled by Zero is Zero Vector",
"Zero Vector Scaled is Zero Vector",
"Definition:Conditional/Necessary Condition",
"Vector Product is Zero only if Factor is Zero"
] |
proofwiki-1288 | Homomorphic Image of Vector Space | Let $\struct {K, +_K, \times_K}$ be a division ring.
Let $\struct {V, +_V, \circ_V}_K$ be a $K$-vector space.
Let $\struct {W, +_W, \circ_W}_K$ be a $K$-algebraic structure.
Let $\phi: V \to W$ be a homomorphism, that is, a linear transformation.
Then the homomorphic image of $\phi$ is a $K$-vector space. | Let us write $\phi \sqbrk V$ for the homomorphic image of $\phi$.
From Homomorphic Image of R-Module is R-Module, $\phi \sqbrk V$ is a $K$-module.
It thus suffices to show that $\phi \sqbrk V$ is unitary, since then it will be a $K$-vector space.
To this end, let $1_K$ be the unity of $K$.
Then for any $\map \phi {\mat... | Let $\struct {K, +_K, \times_K}$ be a [[Definition:Division Ring|division ring]].
Let $\struct {V, +_V, \circ_V}_K$ be a [[Definition:Vector Space|$K$-vector space]].
Let $\struct {W, +_W, \circ_W}_K$ be a [[Definition:R-Algebraic Structure|$K$-algebraic structure]].
Let $\phi: V \to W$ be a [[Definition:R-Algebraic... | Let us write $\phi \sqbrk V$ for the [[Definition:Homomorphic Image|homomorphic image]] of $\phi$.
From [[Homomorphic Image of R-Module is R-Module]], $\phi \sqbrk V$ is a [[Definition:Module over Ring|$K$-module]].
It thus suffices to show that $\phi \sqbrk V$ is [[Definition:Unitary Module over Ring|unitary]], sinc... | Homomorphic Image of Vector Space | https://proofwiki.org/wiki/Homomorphic_Image_of_Vector_Space | https://proofwiki.org/wiki/Homomorphic_Image_of_Vector_Space | [
"Homomorphisms (Abstract Algebra)",
"Linear Algebra"
] | [
"Definition:Division Ring",
"Definition:Vector Space",
"Definition:R-Algebraic Structure",
"Definition:R-Algebraic Structure Homomorphism",
"Definition:Linear Transformation/Vector Space",
"Definition:Homomorphism (Abstract Algebra)/Image",
"Definition:Vector Space"
] | [
"Definition:Homomorphism (Abstract Algebra)/Image",
"Homomorphic Image of R-Module is R-Module",
"Definition:Module over Ring",
"Definition:Unitary Module over Ring",
"Definition:Vector Space",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Linear Transformation/Vector Space",
"Definition:Vec... |
proofwiki-1289 | Direct Product of Vector Spaces is Vector Space | Let $K$ be a field.
Let $V_1, V_2, \ldots, V_n$ be $K$-vector spaces.
Let $\struct {V, + , \circ}_K$ be their direct product.
Then $\struct {V, + , \circ}_K$ is a $K$-vector space. | This follows directly from Finite Direct Product of Modules is Module and the definition of vector space. | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $V_1, V_2, \ldots, V_n$ be [[Definition:Vector Space|$K$-vector spaces]].
Let $\struct {V, + , \circ}_K$ be their [[Definition:Direct Product of Vector Spaces|direct product]].
Then $\struct {V, + , \circ}_K$ is a [[Definition:Vector Space|$K$-vector s... | This follows directly from [[Finite Direct Product of Modules is Module]] and the definition of [[Definition:Vector Space|vector space]]. | Direct Product of Vector Spaces is Vector Space | https://proofwiki.org/wiki/Direct_Product_of_Vector_Spaces_is_Vector_Space | https://proofwiki.org/wiki/Direct_Product_of_Vector_Spaces_is_Vector_Space | [
"Direct Product of Vector Spaces",
"Linear Algebra",
"Direct Product of Vector Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Direct Product of Vector Spaces",
"Definition:Vector Space"
] | [
"Finite Direct Product of Modules is Module",
"Definition:Vector Space"
] |
proofwiki-1290 | Complex Numbers form Vector Space over Reals | Let $\R$ be the set of real numbers.
Let $\C$ be the set of complex numbers.
Then the $\R$-module $\C$ is a vector space. | Recall that Real Numbers form Field.
Thus by definition, $\R$ is also a division ring.
Thus we only need to show that $\R$-module $\C$ is a unitary module, by demonstrating the module properties:
$\forall x, y, \in \C, \forall \lambda, \mu \in \R$:
:$(1): \quad \lambda \paren {x + y} = \paren {\lambda x} + \paren {\lam... | Let $\R$ be the set of [[Definition:Real Number|real numbers]].
Let $\C$ be the set of [[Definition:Complex Number|complex numbers]].
Then the [[Definition:Module over Ring|$\R$-module]] $\C$ is a [[Definition:Vector Space|vector space]]. | Recall that [[Real Numbers form Field]].
Thus by definition, $\R$ is also a [[Definition:Division Ring|division ring]].
Thus we only need to show that [[Definition:Module over Ring|$\R$-module]] $\C$ is a [[Definition:Unitary Module over Ring|unitary module]], by demonstrating the module properties:
$\forall x, y, \... | Complex Numbers form Vector Space over Reals | https://proofwiki.org/wiki/Complex_Numbers_form_Vector_Space_over_Reals | https://proofwiki.org/wiki/Complex_Numbers_form_Vector_Space_over_Reals | [
"Linear Algebra",
"Complex Analysis",
"Examples of Vector Spaces"
] | [
"Definition:Real Number",
"Definition:Complex Number",
"Definition:Module over Ring",
"Definition:Vector Space"
] | [
"Real Numbers form Field",
"Definition:Division Ring",
"Definition:Module over Ring",
"Definition:Unitary Module over Ring",
"Complex Multiplication Distributes over Addition",
"Complex Multiplication is Associative",
"Complex Multiplication Identity is One",
"Definition:Multiplicative Identity"
] |
proofwiki-1291 | Division Ring is Vector Space over Prime Subfield | Let $\struct {K, +, \times}$ be a division ring.
Let $\struct {S, +, \times}$ be the prime subfield of $K$
Then $\struct {K, +, \times_S}_S$ is an $S$-vector space, where $\times_S$ is the restriction of $\times$ to $S \times K$. | Because $K$ is a division ring, it satisfies the vector space axioms for addition, in particular it is an abelian group. The distribuitivty and associativity of multiplication follow from the rules for multiplciation in any ring. Also, any prime field contains the multiplicative identity.
{{tidy}}
{{qed}} | Let $\struct {K, +, \times}$ be a [[Definition:Division Ring|division ring]].
Let $\struct {S, +, \times}$ be the [[Definition:Prime Subfield|prime subfield]] of $K$
Then $\struct {K, +, \times_S}_S$ is an [[Definition:Vector Space|$S$-vector space]], where $\times_S$ is the [[Definition:Restriction of Operation|res... | Because $K$ is a [[Definition:Division Ring|division ring]], it satisfies the [[Definition:Vector Space|vector space]] axioms for addition, in particular it is an abelian group. The distribuitivty and associativity of multiplication follow from the rules for multiplciation in any [[Definition:Ring|ring]]. Also, any pri... | Division Ring is Vector Space over Prime Subfield | https://proofwiki.org/wiki/Division_Ring_is_Vector_Space_over_Prime_Subfield | https://proofwiki.org/wiki/Division_Ring_is_Vector_Space_over_Prime_Subfield | [
"Division Rings",
"Subfields",
"Linear Algebra"
] | [
"Definition:Division Ring",
"Definition:Prime Subfield",
"Definition:Vector Space",
"Definition:Restriction/Operation"
] | [
"Definition:Division Ring",
"Definition:Vector Space",
"Definition:Ring"
] |
proofwiki-1292 | Subspaces of Dimension 2 Real Vector Space | Take the $\R$-vector space $\left({\R^2, +, \times}\right)_\R$.
Let $S$ be a subspace of $\left({\R^2, +, \times}\right)_\R$.
Then $S$ is one of:
: $(1): \quad \left({\R^2, +, \times}\right)_\R$
: $(2): \quad \left\{{0}\right\}$
: $(3): \quad$ A line through the origin. | Let $S$ be a non-zero subspace of $\struct {\R^2, +, \times}_\R$.
Then $S$ contains a non-zero vector $\tuple {\alpha_1, \alpha_2}$.
Hence $S$ also contains $\set {\lambda \times \tuple {\alpha_1, \alpha_2}, \lambda \in \R}$.
From Equation of Straight Line in Plane, this set may be described as a line through the origi... | Take the [[Definition:Real Vector Space|$\R$-vector space]] $\left({\R^2, +, \times}\right)_\R$.
Let $S$ be a [[Definition:Vector Subspace|subspace]] of $\left({\R^2, +, \times}\right)_\R$.
Then $S$ is one of:
: $(1): \quad \left({\R^2, +, \times}\right)_\R$
: $(2): \quad \left\{{0}\right\}$
: $(3): \quad$ A line thr... | Let $S$ be a non-zero [[Definition:Vector Subspace|subspace]] of $\struct {\R^2, +, \times}_\R$.
Then $S$ contains a [[Definition:Zero Vector|non-zero vector]] $\tuple {\alpha_1, \alpha_2}$.
Hence $S$ also contains $\set {\lambda \times \tuple {\alpha_1, \alpha_2}, \lambda \in \R}$.
From [[Equation of Straight Line ... | Subspaces of Dimension 2 Real Vector Space/Proof 1 | https://proofwiki.org/wiki/Subspaces_of_Dimension_2_Real_Vector_Space | https://proofwiki.org/wiki/Subspaces_of_Dimension_2_Real_Vector_Space/Proof_1 | [
"Linear Algebra",
"Subspaces of Dimension 2 Real Vector Space"
] | [
"Definition:Real Vector Space",
"Definition:Vector Subspace"
] | [
"Definition:Vector Subspace",
"Definition:Zero Vector",
"Equation of Straight Line in Plane",
"Definition:Zero Vector",
"Definition:Vector",
"Definition:Simultaneous Equations/Linear Equations"
] |
proofwiki-1293 | Subspaces of Dimension 2 Real Vector Space | Take the $\R$-vector space $\left({\R^2, +, \times}\right)_\R$.
Let $S$ be a subspace of $\left({\R^2, +, \times}\right)_\R$.
Then $S$ is one of:
: $(1): \quad \left({\R^2, +, \times}\right)_\R$
: $(2): \quad \left\{{0}\right\}$
: $(3): \quad$ A line through the origin. | Follows directly from Dimension of Proper Subspace is Less Than its Superspace.
{{qed}} | Take the [[Definition:Real Vector Space|$\R$-vector space]] $\left({\R^2, +, \times}\right)_\R$.
Let $S$ be a [[Definition:Vector Subspace|subspace]] of $\left({\R^2, +, \times}\right)_\R$.
Then $S$ is one of:
: $(1): \quad \left({\R^2, +, \times}\right)_\R$
: $(2): \quad \left\{{0}\right\}$
: $(3): \quad$ A line thr... | Follows directly from [[Dimension of Proper Subspace is Less Than its Superspace]].
{{qed}} | Subspaces of Dimension 2 Real Vector Space/Proof 2 | https://proofwiki.org/wiki/Subspaces_of_Dimension_2_Real_Vector_Space | https://proofwiki.org/wiki/Subspaces_of_Dimension_2_Real_Vector_Space/Proof_2 | [
"Linear Algebra",
"Subspaces of Dimension 2 Real Vector Space"
] | [
"Definition:Real Vector Space",
"Definition:Vector Subspace"
] | [
"Dimension of Proper Subspace is Less Than its Superspace"
] |
proofwiki-1294 | Subspace of Real Continuous Functions | Let $\mathbb J = \set {x \in \R: a \le x \le b}$ be a closed interval of the real number line $\R$.
Let $\map C {\mathbb J}$ be the set of all continuous real functions on $\mathbb J$.
Then $\struct {\map C {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$. | By definition, $\map C {\mathbb J} \subseteq \R^{\mathbb J}$.
Let $f, g \in \map C {\mathbb J}$.
By Two-Step Vector Subspace Test, it needs to be shown that:
:$(1): \quad f + g \in \map C {\mathbb J}$
:$(2): \quad \lambda f \in \map C {\mathbb J}$ for any $\lambda \in \R$
$(1)$ follows by Sum Rule for Continuous Real F... | Let $\mathbb J = \set {x \in \R: a \le x \le b}$ be a [[Definition:Closed Real Interval|closed interval]] of the [[Definition:Real Number Line|real number line]] $\R$.
Let $\map C {\mathbb J}$ be the set of all [[Definition:Continuous Real Function|continuous real functions]] on $\mathbb J$.
Then $\struct {\map C {\... | By definition, $\map C {\mathbb J} \subseteq \R^{\mathbb J}$.
Let $f, g \in \map C {\mathbb J}$.
By [[Two-Step Vector Subspace Test]], it needs to be shown that:
:$(1): \quad f + g \in \map C {\mathbb J}$
:$(2): \quad \lambda f \in \map C {\mathbb J}$ for any $\lambda \in \R$
$(1)$ follows by [[Sum Rule for Contin... | Subspace of Real Continuous Functions | https://proofwiki.org/wiki/Subspace_of_Real_Continuous_Functions | https://proofwiki.org/wiki/Subspace_of_Real_Continuous_Functions | [
"Vector Subspaces",
"Analysis"
] | [
"Definition:Real Interval/Closed",
"Definition:Real Number/Real Number Line",
"Definition:Continuous Real Function",
"Definition:Vector Subspace",
"Definition:Vector Space"
] | [
"Two-Step Vector Subspace Test",
"Combination Theorem for Continuous Functions/Real/Sum Rule",
"Combination Theorem for Continuous Functions/Real/Multiple Rule",
"Definition:Vector Subspace",
"Definition:Vector Space"
] |
proofwiki-1295 | Singleton is Linearly Independent | Let $K$ be a division ring.
Let $\struct {G, +_G}$ be a group whose identity is $e$.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space whose zero is $0_K$.
Let $x \in G: x \ne e$.
Then $\set x$ is a linearly independent subset of $G$. | The only sequence of distinct terms in $\set x$ is the one that goes: $x$.
Suppose $\exists \lambda \in K: \lambda \circ x = e$.
From Zero Vector Space Product iff Factor is Zero it follows that $\lambda = 0$.
Hence the result from definition of linearly independent set.
{{qed}} | Let $K$ be a [[Definition:Division Ring|division ring]].
Let $\struct {G, +_G}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\struct {G, +_G, \circ}_K$ be a [[Definition:Vector Space|$K$-vector space]] whose [[Definition:Zero Scalar|zero]] is $0_K$.
Let $x \in G: x \ne ... | The only [[Definition:Sequence of Distinct Terms|sequence of distinct terms]] in $\set x$ is the one that goes: $x$.
Suppose $\exists \lambda \in K: \lambda \circ x = e$.
From [[Zero Vector Space Product iff Factor is Zero]] it follows that $\lambda = 0$.
Hence the result from definition of [[Definition:Linearly Ind... | Singleton is Linearly Independent | https://proofwiki.org/wiki/Singleton_is_Linearly_Independent | https://proofwiki.org/wiki/Singleton_is_Linearly_Independent | [
"Linear Independence"
] | [
"Definition:Division Ring",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Vector Space",
"Definition:Scalar Ring/Zero Scalar",
"Definition:Linearly Independent/Set"
] | [
"Definition:Sequence of Distinct Terms",
"Zero Vector Space Product iff Factor is Zero",
"Definition:Linearly Independent/Set"
] |
proofwiki-1296 | Linearly Dependent Sequence of Vector Space | Let $\struct {G, +}$ be a group whose identity is $\mathbf 0$.
Let $\struct {G, +, \circ}_K$ be a $K$-vector space.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of distinct non-zero vectors of $G$.
Then $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is linearly dependent {{iff}}:
:$\exists p \... | === Necessary Condition ===
Suppose $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is linearly dependent.
We have {{hypothesis}} that the set of all integers $r \in \closedint 1 n$ such that $\sequence {a_k}_{1 \mathop \le k \mathop \le r}$ is linearly dependent is not empty.
Let $p$ be its smallest element.
Then fr... | Let $\struct {G, +}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $\mathbf 0$.
Let $\struct {G, +, \circ}_K$ be a [[Definition:Vector Space|$K$-vector space]].
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Sequence|sequence]] of distinct [[Definition:Zero... | === Necessary Condition ===
Suppose $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is [[Definition:Linearly Dependent Sequence|linearly dependent]].
We have {{hypothesis}} that the [[Definition:Set|set]] of all [[Definition:Integer|integers]] $r \in \closedint 1 n$ such that $\sequence {a_k}_{1 \mathop \le k \math... | Linearly Dependent Sequence of Vector Space | https://proofwiki.org/wiki/Linearly_Dependent_Sequence_of_Vector_Space | https://proofwiki.org/wiki/Linearly_Dependent_Sequence_of_Vector_Space | [
"Linear Dependence",
"Vector Spaces"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Vector Space",
"Definition:Sequence",
"Definition:Zero Vector",
"Definition:Linearly Dependent/Sequence",
"Definition:Linear Combination"
] | [
"Definition:Linearly Dependent/Sequence",
"Definition:Set",
"Definition:Integer",
"Definition:Linearly Independent/Sequence",
"Definition:Empty Set",
"Definition:Smallest Element",
"Singleton is Linearly Independent",
"Definition:Linearly Independent/Sequence",
"Definition:Scalar/Vector Space",
"D... |
proofwiki-1297 | Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set | Let $K$ be a division ring.
Let $G$ be a finitely generated $K$-vector space.
Let $H$ be a linearly independent subset of $G$.
Let $F$ be a finite generator for $G$ such that $H \subseteq F$.
Then there is a basis $B$ for $G$ such that $H \subseteq B \subseteq F$. | Let $\mathbb S$ be the set of all $S \subseteq G$ such that $S$ is a generator for $G$ and that $H \subseteq S \subseteq F$.
Because $F \in \mathbb S$ {{hypothesis}}, it follows that:
:$\mathbb S \ne \O$
Because $F$ is finite, then so is every element of $\mathbb S$.
Let $R = \set {r \in \Z: r = \card S \land S \in \Bb... | Let $K$ be a [[Definition:Division Ring|division ring]].
Let $G$ be a [[Definition:Finitely Generated Module|finitely generated]] [[Definition:Vector Space|$K$-vector space]].
Let $H$ be a [[Definition:Linearly Independent Set|linearly independent subset]] of $G$.
Let $F$ be a [[Definition:Finite Set|finite]] [[Defi... | Let $\mathbb S$ be the [[Definition:Set|set]] of all $S \subseteq G$ such that $S$ is a [[Definition:Generator of Vector Space|generator]] for $G$ and that $H \subseteq S \subseteq F$.
Because $F \in \mathbb S$ {{hypothesis}}, it follows that:
:$\mathbb S \ne \O$
Because $F$ is [[Definition:Finite Set|finite]], then... | Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set | https://proofwiki.org/wiki/Vector_Space_has_Basis_Between_Linearly_Independent_Set_and_Finite_Spanning_Set | https://proofwiki.org/wiki/Vector_Space_has_Basis_Between_Linearly_Independent_Set_and_Finite_Spanning_Set | [
"Bases of Vector Spaces",
"Generators of Vector Spaces"
] | [
"Definition:Division Ring",
"Definition:Finitely Generated Module",
"Definition:Vector Space",
"Definition:Linearly Independent/Set",
"Definition:Finite Set",
"Definition:Generator of Vector Space",
"Definition:Basis of Vector Space"
] | [
"Definition:Set",
"Definition:Generator of Vector Space",
"Definition:Finite Set",
"Definition:Element",
"Definition:Set",
"Definition:Integer",
"Definition:Element",
"Definition:Generator of Vector Space",
"Definition:Subset",
"Definition:Element",
"Definition:Element",
"Subset of Module Cont... |
proofwiki-1298 | Finitely Generated Vector Space has Finite Basis | Let $K$ be a division ring.
Let $V$ be a finitely generated vector space over $K$.
Then $V$ has a finite basis. | This follows from Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set.
It suffices to find:
:A linearly independent subset $L\subset V$
:A finite generator $S\subset V$
with $L\subset S$.
By Empty Set is Linearly Independent, we make take $L = \O$ and $S$ any finite generator, which exists b... | Let $K$ be a [[Definition:Division Ring|division ring]].
Let $V$ be a [[Definition:Finitely Generated Module|finitely generated]] [[Definition:Vector Space|vector space]] over $K$.
Then $V$ has a [[Definition:Finite Set|finite]] [[Definition:Basis of Vector Space|basis]]. | This follows from [[Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set]].
It suffices to find:
:A [[Definition:Linearly Independent Set|linearly independent]] [[Definition:Subset|subset]] $L\subset V$
:A [[Definition:Finite Set|finite]] [[Definition:Generator of Vector Space|generator]] $S... | Finitely Generated Vector Space has Finite Basis | https://proofwiki.org/wiki/Finitely_Generated_Vector_Space_has_Finite_Basis | https://proofwiki.org/wiki/Finitely_Generated_Vector_Space_has_Finite_Basis | [
"Bases of Vector Spaces"
] | [
"Definition:Division Ring",
"Definition:Finitely Generated Module",
"Definition:Vector Space",
"Definition:Finite Set",
"Definition:Basis of Vector Space"
] | [
"Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set",
"Definition:Linearly Independent/Set",
"Definition:Subset",
"Definition:Finite Set",
"Definition:Generator of Vector Space",
"Empty Set is Linearly Independent",
"Definition:Finite Set",
"Definition:Generator of Vector ... |
proofwiki-1299 | Bases of Finitely Generated Vector Space have Equal Cardinality | Let $K$ be a division ring.
Let $G$ be a finitely generated $K$-vector space.
Then any two bases of $G$ are finite and equivalent. | Since a basis is, by definition, both linearly independent and a generator, the result follows directly from Size of Linearly Independent Subset is at Most Size of Finite Generator.
{{qed}} | Let $K$ be a [[Definition:Division Ring|division ring]].
Let $G$ be a [[Definition:Finitely Generated Module|finitely generated]] [[Definition:Vector Space|$K$-vector space]].
Then any two [[Definition:Basis of Vector Space|bases]] of $G$ are [[Definition:Finite Set|finite]] and [[Definition:Set Equivalence|equivale... | Since a [[Definition:Basis of Vector Space|basis]] is, by definition, both [[Definition:Linearly Independent Set|linearly independent]] and a [[Definition:Generator of Vector Space|generator]], the result follows directly from [[Size of Linearly Independent Subset is at Most Size of Finite Generator]].
{{qed}} | Bases of Finitely Generated Vector Space have Equal Cardinality | https://proofwiki.org/wiki/Bases_of_Finitely_Generated_Vector_Space_have_Equal_Cardinality | https://proofwiki.org/wiki/Bases_of_Finitely_Generated_Vector_Space_have_Equal_Cardinality | [
"Bases of Vector Spaces"
] | [
"Definition:Division Ring",
"Definition:Finitely Generated Module",
"Definition:Vector Space",
"Definition:Basis of Vector Space",
"Definition:Finite Set",
"Definition:Set Equivalence"
] | [
"Definition:Basis of Vector Space",
"Definition:Linearly Independent/Set",
"Definition:Generator of Vector Space",
"Size of Linearly Independent Subset is at Most Size of Finite Generator"
] |
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