nyuuzyou/wikis huggingface dataset:
Group Action/Examples/Cyclic Group on Polygon/Mistake

== Source Work ==
:
:action 3.
== Mistake ==
:For example, the $n$-element ${}^*$cyclic group whose elements are $e, a, a^2, \ldots, a^{n - 1}$ acts on the vertices of a ${}^*$regular polygon by the map for which $e x$ is $x$ for each vertex $x$, and $a^k x$ is the vertex obtained when $x$ is rotated through $2 \pi k / n$ radians about the center of the polygon.
== Correction ==
It is not $x$ which is rotated, it is the whole polygon.
Hence this would be better worded as:
:For example, the $n$-element ${}^*$cyclic group whose elements are $e, a, a^2, \ldots, a^{n - 1}$ acts on the vertices of a ${}^*$regular polygon by the map for which $e x$ is $x$ for each vertex $x$, and $a^k x$ is the vertex obtained when the polygon is rotated through $2 \pi k / n$ radians about the center of the polygon.
== Sources ==
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Group Action Induces Equivalence Relation

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $X$ be a set.
Let $*: G \times S \to S$ be a group action.
Let $\RR_G$ be the relation induced by $G$, that is:
:$\forall x, y \in X: x \mathrel {\RR_G} y \iff y \in \Orb x$
where:
:$\Orb x$ denotes the orbit of $x \in X$.
Then:
:$\RR_G$ is an equivalence relation
and:
:the equivalence class of an element of $X$ is its orbit.
== Proof ==
Let $x \mathrel {\RR_G} y \iff y \in \Orb x$.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Thus $\RR_G$ is reflexive.
=== Symmetry ===
Thus $\RR_G$ is symmetric.
=== Transitivity ===
Thus $\RR_G$ is transitive.
So $\RR_G$ has been shown to be an equivalence relation. 
Hence the result, by definition of an equivalence class.
== Also see ==
* Definition:Set of Orbits
* Definition:Equivalence Relation Induced by Group Action
* Set of Orbits forms Partition
== Sources ==
* : $\S 6.5$. Orbits: Lemma $\text{(i)}$
* : $\S 1.6$: Exercise $4.2$
* : Chapter $2$: The Sylow Theorems: $\S 54$
* : $\S 3.3$: Group actions and coset decompositions
* : Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.13$
Category:Group Actions
Category:Equivalence Relations
Group Action defines Permutation Representation

== Theorem ==
Let $\map \Gamma X$ be the set of permutations on a set $X$.
Let $G$ be a group.
Let $\phi: G \times X \to X$ be a group action.
For $g \in G$, let $\phi_g: X \to X$ be the mapping defined as:
:$\map {\phi_g} x = \map \phi {g, x}$
Let $\tilde \phi: G \to \map \Gamma X$ be the permutation representation associated to $\phi$, defined by:
:$\map {\tilde \phi} g := \phi_g$
Then $\tilde \phi$ is a group homomorphism.
== Proof ==
From Group Action determines Bijection:
:$\phi_g \in \map \Gamma X$
for $g \in G$.
Let $g, h \in G$.
:$\forall \tuple {g, x} \in G \times X: \map \phi {g, x} \in X = g \wedge x \in X$
First we show that for all $x \in X$:
:$\map {\phi_g \circ \phi_h} x = \map {\phi_{g h} } x$
Thus:
 
Also, we have:
:$e \wedge x = x \implies \map {\phi_e} x = x$
where $e$ is the identity of $G$.
Therefore, we have shown that $\tilde \phi: G \to \map \Gamma X: g \mapsto \phi_g$ is a group homomorphism.
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions
Category:Group Actions
Category:Permutation Representations
Category:Group Homomorphisms
Group Action determines Bijection

== Theorem ==
Let $*$ be a group action of $G$ on $X$.
Then each $g \in G$ determines a bijection $\phi_g: X \to X$ given by:
:$\map {\phi_g} x = g * x$
Its inverse is:
:$\phi_{g^{-1} }: X \to X$.
These bijection are sometimes called transformations of $X$.
== Proof ==
=== Proof of Injectivity ===
Let $x, y \in X$
Then:
Thus $\phi_g$ is an injection.
=== Proof of Surjectivity ===
Let $x \in X$.
Then:
Thus a group action is a surjection.
So a group action is an injection and a surjection and therefore a bijection.
== Also see ==
* Definition:Permutation Representation
Category:Group Actions
Group Action of Symmetric Group

== Theorem ==
Let $\N_n$ denote the set $\set {1, 2, \ldots, n}$.
Let $\struct {S_n, \circ}$ denote the symmetric group on $\N_n$.
The mapping $*: S_n \times \N_n \to \N_n$ defined as:
:$\forall \pi \in S_n, \forall n \in \N_n: \pi * n = \map \pi n$
is a group action.
=== Group Action on Subset of $\N_n$ ===
== Proof ==
The group action axioms are investigated in turn.
Let $\pi, \rho \in S_n$ and $n \in \N_n$.
Thus:
demonstrating that holds.
Then:
demonstrating that holds.
== Sources ==
* : $\S 5.5$. Groups acting on sets: Example $102$
* : Chapter $2$: The Sylow Theorems: $\S 53 \alpha$
* : $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(ii)}$
Category:Symmetric Groups
Category:Group Action of Symmetric Group
Group Action of Symmetric Group/Subset

== Theorem ==
Let $\N_n$ denote the set $\set {1, 2, \ldots, n}$.
Let $\struct {S_n, \circ}$ denote the symmetric group on $\N_n$.
Let $r \in \N: 0 < r \le n$.
Let $B_r$ denote the set of all subsets of $\N_n$ of cardinality $r$:
:$B_r := \set {S \subseteq \N_n: \card S = r}$
Let $*$ be the mapping $*: S_n \times B_r \to B_r$ defined as:
:$\forall \pi \in S_n, \forall S \in B_r: \pi * S = \pi \sqbrk S$
where $\pi \sqbrk S$ denotes the image of $S$ under $\pi$.
Then $*$ is a group action.
== Proof ==
The group action axioms are investigated in turn.
Let $\pi, \rho \in S_n$.
Let $S \in B_r$.
Thus:
demonstrating that holds.
Then:
demonstrating that holds.
== Sources ==
* : $\S 6.5$. Orbits: Example $120$
Category:Group Action of Symmetric Group
Group Action of Symmetric Group Acts Transitively

== Theorem ==
Let $S$ be a set.
Let $\struct {\map \Gamma S, \circ}$ be the symmetric group on $S$.
Let $*: \map \Gamma S \times S \to S$ be the group action defined as:
:$\forall \pi \in \map \Gamma S, \forall s \in S: \pi * s = \map \pi s$
Then $*$ is a transitive group action.
In other words, $\struct {\map \Gamma S, \circ}$ acts transitively on $S$ by $*$.
== Proof ==
By Group Action of Symmetric Group, $*: \map \Gamma S \times S \to S$ is indeed a group action
Let $s, t \in S$.
As $\map \Gamma S$ is the symmetric group on $S$, there exists a permutation $\pi \in \map \Gamma S$ such that:
:$\map \pi s = t$
This holds for any $s, t \in S$.
Thus:
:$\forall t \in S: t \in \Orb s$
and so $S$ consists of a single orbit.
Hence the result by definition of transitive group action.
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(ii)}$
Category:Group Action of Symmetric Group
Category:Transitive Group Actions
Group Action of Symmetric Group on Complex Vector Space

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $V$ denote a vector space over the complex numbers $\C$.
Let $V$ have a basis:
:$\BB := \set {v_1, v_2, \ldots, v_n}$
Let $*: S_n \times V \to V$ be a group action of $S_n$ on $V$ defined as:
:$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$
where:
: $v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$
Then $*$ is a group action.
=== Orbit of Element of $V$ ===
=== Stabilizer of Element of $V$ ===
== Proof ==
Let $\rho, \sigma \in S_n$.
Let $v = \ds \sum_{k \mathop = 1}^n \lambda_k v_k$.
We have:
Hence $*$ fulfils .
Let $e$ denote the identity element of $S_n$.
Then:
Hence $*$ fulfils .
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $1$
Category:Examples of Group Actions
Category:Group Action of Symmetric Group on Complex Vector Space
Category:Symmetric Groups
Group Action of Symmetric Group on Complex Vector Space/Orbit

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $V$ denote a vector space over the complex numbers $\C$.
Let $V$ have a basis:
:$\BB := \set {v_1, v_2, \ldots, v_n}$
Let $*: S_n \times V \to V$ be a group action of $S_n$ on $V$ defined as:
:$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$
where:
:$v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$
The orbit of an element $v \in V$ is:
:$\ds \Orb v = \set {w \in V: \exists \rho \in S_n: w = \sum_{k \mathop = 1}^n \lambda_k v_{\map \rho k} }$
== Proof ==
By definition:
== Examples ==
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $1$
Category:Group Action of Symmetric Group on Complex Vector Space
Group Action of Symmetric Group on Complex Vector Space/Orbit/Examples

== Examples of Orbits of Group Action of Symmetric Group on Complex Vector Space ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $V$ denote a vector space over the complex numbers $\C$.
Let $V$ have a basis:
:$\BB := \set {v_1, v_2, \ldots, v_n}$
Let $*: S_n \times V \to V$ be a group action of $S_n$ on $V$ defined as:
:$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$
where:
:$v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$
=== Example 1 ===
=== Example 2 ===
Category:Group Action of Symmetric Group on Complex Vector Space
Group Action of Symmetric Group on Complex Vector Space/Orbit/Examples/Example 1

== Example of Orbit of Group Action of Symmetric Group on Complex Vector Space ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $V$ denote a vector space over the complex numbers $\C$.
Let $V$ have a basis:
:$\BB := \set {v_1, v_2, \ldots, v_n}$
Let $*: S_n \times V \to V$ be a group action of $S_n$ on $V$ defined as:
:$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$
where:
:$v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$
Let $n = 4$.
Let $v = v_1 + v_2 + v_3 + v_4$.
The orbit of $v$ is:
:$\Orb v = \set v$
== Proof ==
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $1 \ \text {(a)}$
Category:Group Action of Symmetric Group on Complex Vector Space
Group Action of Symmetric Group on Complex Vector Space/Orbit/Examples/Example 2

== Example of Orbit of Group Action of Symmetric Group on Complex Vector Space ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $V$ denote a vector space over the complex numbers $\C$.
Let $V$ have a basis:
:$\BB := \set {v_1, v_2, \ldots, v_n}$
Let $*: S_n \times V \to V$ be a group action of $S_n$ on $V$ defined as:
:$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$
where:
:$v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$
Let $n = 4$.
Let $v = v_1 + v_3$.
The orbit of $v$ is:
:$\Orb v = \set {v_1 + v_2, v_1 + v_3, v_1 + v_4, v_2 + v_3, v_2 + v_4, v_3 + v_4}$
== Proof ==
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $1 \ \text {(b)}$
Category:Group Action of Symmetric Group on Complex Vector Space
Group Action of Symmetric Group on Complex Vector Space/Stabilizer

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $V$ denote a vector space over the complex numbers $\C$.
Let $V$ have a basis:
:$\BB := \set {v_1, v_2, \ldots, v_n}$
Let $*: S_n \times V \to V$ be a group action of $S_n$ on $V$ defined as:
:$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$
where:
: $v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$
The stabilizer of an element $v \in V$ is:
:$\ds \Stab v = \set {\rho \in S_n: \sum_{k \mathop = 1}^n \lambda_k v_k = \sum_{k \mathop = 1}^n \lambda_{\map \rho k} v_k}$
== Proof ==
By definition:
== Examples ==
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $1$
Category:Stabilizers
Category:Group Action of Symmetric Group on Complex Vector Space
Group Action of Symmetric Group on Complex Vector Space/Stabilizer/Examples/Example 1

== Example of Orbit of Group Action of Symmetric Group on Complex Vector Space ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $V$ denote a vector space over the complex numbers $\C$.
Let $V$ have a basis:
:$\BB := \set {v_1, v_2, \ldots, v_n}$
Let $*: S_n \times V \to V$ be a group action of $S_n$ on $V$ defined as:
:$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$
where:
:$v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$
Let $n = 4$.
Let $v = v_1 + v_2 + v_3 + v_4$.
The stabilizer of $v$ is:
:$\Stab v = S_4$
== Proof ==
From the page discussing the Orbit of $v$:
:$\Orb v = \set v$
From the Orbit-Stabilizer Theorem:
:$\order {\Orb v} = \dfrac {\order {S_4} } {\order {\Stab v} }$
where $\order {\Stab v}$ denotes the order, that is, the cardinality of $\Stab v$.
Hence:
:$\order {\Stab v} = \dfrac {\order {S_4} } {\order {\Orb v} } = \dfrac {\order {S_4} } 1 = \order {S_4}$
from which it follows that:
:$\Stab v = S_4$
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $1 \ \text {(a)}$
Category:Group Action of Symmetric Group on Complex Vector Space
Group Action of Symmetric Group on Complex Vector Space/Stabilizer/Examples/Example 2

== Example of Orbit of Group Action of Symmetric Group on Complex Vector Space ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $V$ denote a vector space over the complex numbers $\C$.
Let $V$ have a basis:
:$\BB := \set {v_1, v_2, \ldots, v_n}$
Let $*: S_n \times V \to V$ be a group action of $S_n$ on $V$ defined as:
:$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$
where:
:$v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$
Let $n = 4$.
Let $v = v_1 + v_3$.
The stabilizer of $v$ is:
:$\Stab v = \set {e, \tuple {1 3}, \tuple {2 4}, \tuple {1 3} \tuple {2 4} }$
== Proof ==
Thus $\Stab v$ consists of all the permutations of $S_4$ which either:
:fix $1$ and $3$
or:
:transpose $1$ and $3$.
We have that:
:$e$ and $\tuple {2 4}$ are the permutations which fix $1$ and $3$
:$\tuple {1 3}$ and $\tuple {1 3} \tuple {2 4}$ are the permutations which transpose $1$ and $3$.
Hence the result.
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $1 \ \text {(b)}$
Category:Group Action of Symmetric Group on Complex Vector Space
Group Action of Symmetric Group on Subset is Transitive

== Theorem ==
Let $\N_n$ denote the set $\set {1, 2, \ldots, n}$.
Let $\struct {S_n, \circ}$ denote the symmetric group on $\N_n$.
Let $r \in \N: 0 < r \le n$.
Let $B_r$ denote the set of all subsets of $\N_n$ of cardinality $r$:
:$B_r := \set {S \subseteq \N_n: \card S = r}$
Let $*$ be the mapping $*: S_n \times B_r \to B_r$ defined as:
:$\forall \pi \in S_n, \forall S \in B_r: \pi * B_r = \pi \sqbrk S$
where $\pi \sqbrk S$ denotes the image of $S$ under $\pi$.
Then $*$ is a transitive group action.
== Proof ==
From Group Action of Symmetric Group on Subset it is established that $*$ is a group action.
Let $U = \set {u_1, u_2, \ldots, u_r}$ and $V = \set {v_1, v_2, \ldots, v_r}$ be elements of $B_r$.
Then there exists a permutation $\rho \in S_n$ such that:
:$\map \rho {u_k} = v_k$
for all $k \in \set {1, 2, \ldots, r}$.
Thus:
:$\rho \sqbrk U = V$
Thus $B_r$ is the orbit of all $U \in B_r$.
Hence the result by definition of transitive group action.
== Sources ==
* : $\S 6.5$. Orbits: Example $120$
Category:Group Action of Symmetric Group
Category:Transitive Group Actions
Group Action on Coset Space is Transitive

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Let $*: G \times G / H \to G / H$ be the action on the (left) coset space:
:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
Then $G$ is a transitive group action.
== Proof ==
It is established in Action of Group on Coset Space is Group Action that $*$ is a group action.
It remains to be shown that:
:$\forall g' H \in G / H: \Orb {g' H} = G / H$
where $\Orb {g' H}$ denotes the orbit of $g' H \in G / H$ under $*$.
Let $a H, b H \in G / H$ such that $a H \ne b H$.
We have that:
:$\exists x \in G: x a = b$
and so:
:$x * a H = \paren {x a} H = b H$
and so:
:$b H \in \Orb {a H}$
As both $a$ and $b$ are arbitrary, the result follows.
== Sources ==
* : Chapter $6$: Cosets: Exercise $9$
Category:Group Action on Coset Space
Category:Transitive Group Actions
Group Action on Prime Power Order Subset

== Lemma ==
Let $G$ be a finite group.
Let $\mathbb S = \set {S \subseteq G: \card S = p^n}$ where $p$ is prime.
That is, the set of all subsets of $G$ whose cardinality is the power of a prime number.
Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements:
:$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$.
Then:
=== Stabilizer is $p$-Subgroup ===
=== Stabilizer of Maximal Power Order Subset ===
Category:Examples of Group Actions
Category:Group Action on Prime Power Order Subset
Category:Group Action on Sets with k Elements
Group Action on Prime Power Order Subset/Stabilizer is p-Subgroup

== Lemma ==
Let $G$ be a finite group.
Let $\mathbb S = \set {S \subseteq G: \card S = p^n}$ where $p$ is prime.
That is, the set of all subsets of $G$ whose cardinality is the power of a prime number.
Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements:
:$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$.
Then:
:$\Stab S$ is a $p$-subgroup of $G$.
== Proof ==
First we show that $\Stab S$ is a $p$-subgroup of $G$:
From Group Action on Sets with k Elements:
:$\forall S \in \mathbb S: \order {\Stab S} \divides \card S$
So:
:$\order {\Stab S} \divides p^\alpha$
Thus $\Stab S$ is a $p$-group
Thus by Stabilizer is Subgroup, $\Stab S$ is a $p$-subgroup of $G$.
Category:Group Action on Prime Power Order Subset
Group Action on Prime Power Order Subset/Stabilizer of Maximal Power Order Subset

== Lemma ==
Let $G$ be a finite group.
Let $\mathbb S = \set {S \subseteq G: \card S = p^n}$ where $p$ is prime.
That is, the set of all subsets of $G$ whose cardinality is the power of a prime number.
Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements:
:$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$.
Then:
:If $p^n$ is the maximal power of $p$ dividing $\order G$, and if $p \nmid \card {\Orb S}$, then $\forall s \in S: \Stab S s = S$.
== Proof ==
Let $p^n$ be the maximal power of $p$ dividing $\order G$.
Let $p \nmid \card {\Orb S}$.
From the Orbit-Stabilizer Theorem, we have:
:$\order G = \card {\Orb S} \times \order {\Stab S}$
As $p \nmid \card {\Orb S}$, it must be the case that $p^n \divides \order {\Stab S}$.
Thus:
:$\order {\Stab S} \ge p^n$
Now we note from Cosets are Equivalent that:
:$\order {\Stab S} = \order {\Stab S s}$
Thus we have that:
:$\order {\Stab S s} \ge p^n$
Again using the fact that:
:$\forall S \in \mathbb S: \order {\Stab S s} \divides \card S = p^n$
we also have that:
:$\order {\Stab S s} \le p^n$
Therefore:
:$\order {\Stab S s} = p^n$
However, as was show in the proof of Group Action on Sets with k Elements:
:$\forall s \in S: \Stab S s \subseteq S$
Therefore:
:$\Stab S s = S$
Category:Group Action on Prime Power Order Subset
Group Action on Sets with k Elements

== Theorem ==
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 action, and:
:$\forall S \in \Bbb S: \order {\Stab S} \divides \card S$
where $\Stab S$ denotes the stabilizer of $S$ by $G$.
== Proof ==
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:
:$\card {g S} = \card S$
This follows directly from Order of Subset Product with Singleton.
It remains to be shown that:
:$\order {\Stab S} \divides \card S$
By definition of stabilizer:
:$\Stab S = \set {g \in G: g S = S}$
It follows that:
:$\forall s \in S: g s \in S$
Then:
:$\Stab S s \subseteq S$
Thus:
:$\card {\Stab S s} \le \card S$
We have from Stabilizer is Subgroup that the stabilizer contains the identity:
:$e \in \Stab S \le G$
It therefore follows that:
:$s \in S \implies s \in \Stab S s$
Thus:
:$\ds S = \bigcup_{t \mathop \in S} \Stab S t$
By Congruence Class Modulo Subgroup is Coset and Right Coset Space forms Partition, distinct right cosets are disjoint.
Thus $S$ consists of a union of disjoint right cosets of $\Stab S$.
Thus $\ds \bigcup_{t \mathop \in S} \Stab S t$ is a partition of $S$.
Therefore:
:$\forall s \in S: \card {\Stab S s} \divides \card S$
As we have:
:$\forall s \in S: \card {\Stab S s} = \order {\Stab S}$
the result follows:
:$\card {\Stab S} \divides \card S$
== Also see ==
* Sylow Theorems
Category:Examples of Group Actions
Category:Group Action on Sets with k Elements
Group Action on Subgroup by Left Regular Representation

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $*: H \times G \to G$ be the operation defined as:
:$\forall \left({h, g}\right) \in H \times G: h * g = \lambda_h \left({g}\right)$
where $\lambda_h \left({g}\right)$ is the left regular representation of $g$ by $h$.
Then $*$ is a group action.
== Proof ==
The group action axioms are investigated in turn.
Let $h_1, h_2 \in H$ and $g \in G$.
Thus:
demonstrating that holds.
Then:
demonstrating that holds.
The group action axioms are thus seen to be fulfilled, and so $*$ is a group action.
== Also defined as ==
Some sources denote a mapping by placing the symbol defining that mapping on the left of its operand.
Thus under such a convention:
:$\map {\lambda_h} a$ is written $a \lambda_h$
and:
:$h * a$ is written $a * h$ (or even $a h$ in sources which do not place a high regard on clarity).
Thus for the left regular representation to be defined as being a group action there exists the need to use it on the inverse:
Such a treatment can be found in .
As it runs contrary to the conventions used on , beyond its mention here it will not be used.
However, this example indicates how the arbitrary nature of notational conventions can cause the details of results to be equally arbitrarily dependent upon the convention used.
== Also see ==
* Group Action on Subgroup by Right Regular Representation
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions
Category:Group Actions