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Abstract algebra A Kursobjekt for Math 331

Aaron Greicius

Section 1.17 Group actions: stabilizer

Definition 1.17.1. G-sets.

Let \(G\) be a group. A G-set is a nonempty set \(A\) together with a group action \(G\times A\rightarrow A\text{.}\)
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Given \(G\)-sets \(A\) and \(B\text{,}\) a \(G\)-set morphism is a function \(f\colon A\rightarrow B\) satisfying
\begin{equation} f(g\cdot a)=g\cdot f(a)\tag{1.17.1} \end{equation}
for all \(g\in G\) and \(a\in A\text{.}\) A \(G\)-set isomorphism is a bijective \(G\)-set morphism.
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Definition 1.17.2. Stabilizers and kernels of actions.

Let \(G\) be a group, and let \(A\) be a \(G\)-set. Given a \(G\)-set \(A\) and nonempty subset \(X\subseteq A\text{,}\) the stabilizer of \(X\) is the set \(G_X\) of all elements of \(G\) that fix every element of \(X\text{:}\) i.e.,
\begin{equation} G_X=\{g\in G\mid g\cdot x=x \text{ for all } x\in X\}\text{.}\tag{1.17.2} \end{equation}
In the case of a singleton \(\{a\}\text{,}\) we write \(G_a\) instead of \(G_{\{a\}}\text{.}\)
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The kernel of a a group action \(G\times A\rightarrow A\) is
\begin{equation*} G_A=\{g\in G\mid g\cdot a=a \text{ for all } a\in A\}\text{.} \end{equation*}
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Proof.

  1. Left as an exercise.
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  2. Let \(a'=g\cdot a\text{.}\) We have
    \begin{align*} g'\in G_{a'} \amp \iff g'\cdot (g\cdot a)=g\cdot a \\ \amp \iff (g^{-1}g'g)\cdot a=a\\ \amp \iff g^{-1}g'g\in G_a\\ \amp \iff g'\in gG_ag^{-1}\text{.} \end{align*}
    This proves \(G_{a'}=gG_ag^{-1}\text{.}\)
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  3. The fact that \(G_A=\ker\phi\) follows essentially from the definition of \(G_A\) and \(\phi\text{:}\) \(g\in G_A\) if and only if \(g\cdot a=a\) for all \(a\in A\text{,}\) if and only if its associated permutation \(\phi_g\in S_A\text{,}\) defined as \(\phi_g(a)=g\cdot a\) for all \(a\in A\text{,}\) is the identity map.
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Definition 1.17.4. Permutation representation and faithfulness.

Given a group action \(G\times A\rightarrow A\text{,}\) the corresponding homomorphism
\begin{align*} \phi\colon G \amp \rightarrow S_A\\ g \amp \mapsto \phi_g\text{,} \end{align*}
where \(\phi_g(a)=g\cdot a\) for all \(a\in A\text{,}\) is called the permutation representation associated to the group action. The homomorphism
\begin{equation*} \phi\colon G\rightarrow S_G \end{equation*}
associated to the the action of \(G\) on itself by left multiplication is called the regular permutation representation.
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A group action \(G\times A\rightarrow A\) is faithful if \(G_A=\{e\}\text{:}\) equivalently, if its associated permutation representation is injective.
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Proof.

  1. Given any elements \(gH, g'H\in G/H\text{,}\) we have
    \begin{equation*} g'H=(g'g^{-1})\cdot gH\text{.} \end{equation*}
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  2. We have
    \begin{align*} G_A \amp =\bigcap_{g\in G}G_{gH}\\ \amp = \bigcap_{g\in G}gG_Hg^{-1}\\ \amp = \bigcap_{g\in G}gHg^{-1}\text{,} \end{align*}
    where the last line follows from the fact that \(g'\cdot H=H\) if and only if \(g'\in H\text{.}\)
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    Lastly, if \(N\) is a normal subgroup of \(H\text{,}\) then by definition we must have \(N=gNg^{-1}\leq gHg^{-1}\) for all \(g\in G\text{.}\) It follows that
    \begin{equation*} N\leq \bigcap_{g\in G}gHg^{-1}=G_A\text{.} \end{equation*}
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Proof.

Consider the action of \(G\) on itself by left multiplication, and corresponding regular permutation representation
\begin{equation*} \phi\colon G\rightarrow S_G\text{.} \end{equation*}
Since this action is faithful, \(\ker\phi=\{e\}\) and \(\phi\) is injective. We conclude that \(\phi\) is an isomorphism between \(G\) and the subgroup \(\phi(G)\leq S_G\text{.}\)
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In the case where \(\abs{G}=n< \infty\text{,}\) composing \(\phi\) with an isomorphism \(\alpha\colon S_G\rightarrow S_n\) yields an isomorphism between \(G\) and a subgroup of \(S_n\text{.}\)
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Proof.

Consider the left multiplication action of \(G\) on the coset space \(A=G/H\) and the corresponding permutation representation
\begin{equation*} \phi\colon G\rightarrow S_A\text{.} \end{equation*}
By the first isomorphism theorem, we have
\begin{equation*} G/\ker\phi\cong \phi(G)\text{,} \end{equation*}
where \(\phi(G)\) is a subgroup of \(S_A\text{.}\) Note that since \(A=G/H\text{,}\) we have \(\abs{A}=[G\colon H]=p\text{,}\) and thus \(\abs{S_A}=p!\text{.}\)
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We claim \(\ker\phi=H\text{,}\) whence the result follows. From Theorem 1.17.3 we know that \(\ker\phi\leq H\text{.}\) It follows that
\begin{equation*} \abs{\phi(G)}=[G\colon \ker\phi]=[G\colon H][H\colon \ker \phi=pq\text{.} \end{equation*}
Next, since \(pq=\abs{\phi(G)}\mid p!\text{,}\) \(q\mid (p-1)!\text{.}\) Since furthermore \(q\mid \abs{G}\text{,}\) we see that \(q\) is a common divisor of \((p-1)!\) and \(\abs{G}\text{.}\) But since all prime divisors of \((p-1)!\) are less than \(p\text{,}\) we see that \((p-1)!\) and \(\abs{G}\) are relatively prime! Thus \(q=1\) and \(K=\im\phi\text{,}\) as claimed.
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