TITLE: Holomorph is isomorphic to normalizer of subgroup of symmetric group?
QUESTION [5 upvotes]: Let $G$ be the holomorph of $H$, and thus the semidirect product of $H$ and $K=Aut(H)$. If $H$ has order $n$, then I have already shown that, letting $G$ act on the $n$ left cosets of $K$ by left multiplication gives a homomorphism $\pi$ from $G$ into $S_n$. In addition, since the elements of $H$ can act as coset representatives (which I have shown), the image of $H$ is just the left regular representation of $H$. Now I need to show that the normalizer in $S_n$ of $\pi(H)$ is isomorphic to $G$. 
Now, since $G$ normalizes $H$, presumably $\pi(G)$ normalizes $\pi(H)$ (assuming this representation is faithful), so I at least have that $\pi(G)$ is contained in the normalizer. But how do I show the reverse containment?
Thanks!

REPLY [5 votes]: Denote by $[h]$ the left multiplication by $h\in H$, i.e., $[h]:H\to H$, $[h]:x\mapsto hx$. Let $g\in S_n$ be a bijection $g:H\to H$ that normalizes $[H]$. Then we have an automorphism $\overline g\in\text{Aut}[H]$ given by the rule $\overline g:[h]\mapsto g[h]g^{-1}$. As $h\mapsto[h]$ is an isomorphism $H\to[H]$, we obtain the corresponding automorphism $k\in K=\text{Aut}H$ given by $\big[k(h)\big]=\overline g\big([h]\big)$. In other words, for any $x\in H$ and $h\in H$, we have $k(h)x=g[h]g^{-1}(x)$, that is, $k(h)x=g\big(hg^{-1}(x)\big)$. Taking $x:=e:=g(1)\in H$, we obtain $k(h)e=g(h)$ for all $h\in H$. Let
$k'\in K=\text{Aut}H$ be given by $h\mapsto e^{-1}he$. Then $ek'k\in G=HK$ acts on the left cosets of $K$ exactly as $g$. Indeed, in terms of $G=HK$, $g(h)=khk^{-1}e=ek'khk^{-1}{k'}^{-1}$, hence, $g(hK)=ek'khK$ and $g=\pi(ek'k)$.