Exercise 4.1.2 (Kernel of a permutation group)

Question

Let $G$ be a {\bf permutation group} on the set $A$ (i.e., $ G \leq S_A$), let $\sigma \in G$ and let $a \in A$. Prove that $\sigma G_a \sigma^{-1} = G_{\sigma(a)}$. Deduce that if $G$ acts transitively on $A$ then
$$\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = 1.$$

Richard Ganaye · Accepted Answer

Let $\mathrm{id}_A$ denote the map
$$
\mathrm{id}_A\ 
\left\{
\begin{array}{ccl}
A & 	o & A\
a & \mapsto &a.
\end{array}
ight.
$$
(Then $\mathrm{id}_A = 1_G = 1$ is the neutral element of $G$.)
\begin{proof} 
We know that $G \leq S_A$ acts on $A$ by $\sigma\cdot i = \sigma(i)$. Let $\sigma \in G$ and let $a \in A$. By Exercise 1, if $b = \sigma \cdot a$, then $G_b = \sigma G_a \sigma^{-1}$, so
$$\sigma G_a \sigma^{-1} = G_{\sigma(a)}.$$
Let $\varphi : G 	o G$ be the permutation representation induced by this action. Then $\varphi = 1_G = \mathrm{id}_A$, since for all $\sigma \in G$, and for all $a \in A$, $\varphi(\sigma)(a) = \sigma \cdot a = \sigma(a)$, so $\varphi(\sigma) =\sigma$ for all $\sigma \in G$, so $\varphi = 1_G$. The kernel of the action is the kernel of $\mathrm{id}_A$, which is
$$\ker(\varphi) = \{1\}.$$

By Exercise 1 anew
$$\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1}  = \ker(\varphi) = \{1\}.$$

\end{proof}

Exercise 4.1.2 (Kernel of a permutation group)

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Exercise 4.1.2 (Kernel of a permutation group)

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