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Let $H$ be a group of order $n$, $K$ = Aut($H$), and $G$ = $H \rtimes K$ where $\rtimes$ is the semidirect product with respect to the identity homomorphism $\varphi$. Let $G$ act on the set $X$ of left cosets of $K$ in $G$ by left multiplication, inducing a permutation representation $\pi$ from $G$ into $S_n$. Why do we have that \begin{equation*} \pi(G) = N_{S_n} (\pi(H))? \end{equation*} I've shown that the left side is a subset of the right, but I can't get the reverse inclusion. The textbook hints that \begin{equation*} C_K (H) = \text{ker} \varphi = \{ 1 \} \end{equation*} \begin{equation*} C_H (K) = N_H (K) \end{equation*} could be helpful for proving \begin{equation*} |G| = |N_{S_n} (\pi(H))|, \end{equation*} but I can't see why. I did figure out that $\pi$ is injective. Furthermore, if you let $X$ have coset representatives which are exactly the elements of $H$, then $\pi$ restricted to $H$ is the left regular representation of $H$. Thanks in advance for any help.

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Since $H\cong\pi(H)\trianglelefteq N_{S_n}(\pi(H))$, conjugation by $N_{S_n}(\pi(H))$ preserves and therefore acts as an automorphism on $\pi(H)$, the embedded copy of $H$ in $S_n$. Using $\mathrm{Aut}(\pi(H))\cong\mathrm{Aut}(H)=K$, we can say that for any $\sigma\in N_{S_n}(\pi(H))$, conjugation by $\sigma$ acts precisely as $\pi(k)$ for some $k\in K$. Now,


in $H\rtimes K$, so conjugation by $\sigma$ and conjugation by $\rho=\pi(0,k)$ agree with each other, and so $\sigma \rho^{-1}$ acts trivially by conjugation on $\pi(H)$. The only element that does that is the identity, so $\sigma=\rho$ and we deduce the inclusion $N_{S_n}(\pi(H))\le \pi(G)$.

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@JoshuaCiappara Do you have any questions about this answer? – anon Dec 14 '12 at 0:24
Sorry for the late response; no, I don't. Thanks for your help. – Joshua Ciappara Jan 19 '13 at 5:31

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