Some basic properties of holomorphs

Let $H$ be a group of order $n$, let $K = \mathsf{Aut}(H)$, and form $G = \mathsf{Hol}(H) = H \rtimes_1 K$. Let $G$ act by left multiplication on the left cosets of $K$ in $G$ and let $\pi : G \rightarrow S_{G/K}$ be the induced permutation representation.

1. Prove that the elements of $H$ are coset representatives for the left cosets of $K$ in $G$ and that with this choice of representatives, $\pi|_H$ is the left regular representation of $H$.
2. Prove that $\mathsf{im}\ \pi$ is the normalizer in $S_{G/K}$ of $\pi[H]$. Deduce that under the left regular representation of a finite group $H$ of order $n$, the noramlizer in $S_n$ of the image of $H$ is isomorphic to $\mathsf{Hol}(H)$. [Hint: Show that $|G| = |N_{S_n}(\pi[H])|]$ using §5.5 #1 and §5.5 #2.
3. Deduce that the normalizer of the group generated by an $n$-cycle in $S_n$ is isomorphic to $\mathsf{Hol}(Z_n)$ and has order $n \varphi(n)$, where $\varphi$ denotes the Euler totient function.

1. Recall that $(h,k) = (h,1)(1,k)$. It suffices to show that if $h_1,h_2 \in H$ such that $h_1K = h_2K$, then $h_1 = h_2$. To that end, if $(h_1,1) = (h_2,k)$, then $h_1 = h_2$. There are $n$ left cosets of $K$, and $H$ maps to these injectively. Thus $H$ is a set of coset representatives of $G/K$. Define $\theta : S_{G/K} \rightarrow S_H$ by $\theta(hK) = h$, and let $\rho : H \rightarrow S_H$ be the left regular representation of $H$. Then $\rho(h)(a) = ha = (ha)K$ $h(aK)$ $= \pi(h)(aK)$ $= (\pi(h) \circ \theta)(a)$. Hence $\rho(h) = \pi(h) \circ \theta$; thus $\pi[H]$ is the left regular representation of $H$ up to an isomorphism.
2. First, let $\pi(g) \in \pi[G]$. Now $\pi(g)\pi[H]\pi(g)^{-1} = \pi[gHg^{-1}]$ $= \pi[H]$. Thus $\pi(g) \in N_{S_{G/K}}(\pi[H])$.

Now we prove a lemma.

Lemma [Adapted from Wikipedia]: If $\sigma \in N_{S_{G/K}}(\pi[H])$ and $\sigma(1) = 1$, then $\sigma \in \mathsf{Aut}(H)$. Proof: Since $\sigma$ normalizes $\pi[H]$, for every $h \in H$ there exists $k \in H$ such that $\sigma \circ \pi(h) = \pi(k) \circ \sigma$. Applying this mapping to 1, we see that $\sigma(h) = k$. Thus, if $\sigma(1) = 1$, then for all $h \in H$ we have $\sigma \circ \pi(h) = \pi(\sigma(h)) \circ \sigma$. Now for all $h,k \in H$, we have $\sigma \circ \pi(h) \circ \pi(k) = \pi(\sigma(h)) \circ \sigma \circ \pi(k)$ $= \pi(\sigma(h)) \circ \pi(\sigma(k)) \circ \sigma$ $= \pi(\sigma(h) \sigma(k)) \circ \sigma$, which is equal (since $\pi$ is a homomorphism) to $\sigma \circ \pi(hk) = \pi(\sigma(h,k)) \circ \sigma$. Since $\sigma$ is invertible and $\pi$ is injective, we have $\sigma(hk) = \sigma(h) \sigma(k)$; hence $\sigma$ is a homomorphism $H \rightarrow H$. Since $\sigma$ is also bijective, $\sigma \in \mathsf{Aut}(H)$. $\square$

$(\supseteq)$ Now let $\sigma \in N_{S_{G/K}}(\pi[H])$. By the argument above, $\pi(\sigma(1)^{-1})$ also normalizes $\pi[H]$. Since the normalizer is a subgroup, we have $\pi(\sigma^{-1}(1)) \circ \sigma \in N_{S_{G/K}}(\pi[H])$. Moreover, $(\pi(\sigma^{-1}(1)) \circ \sigma)(1) = \pi(\sigma(1)^{-1})(\sigma(1)) = 1$; by the lemma, $\beta = \pi(\sigma^{-1}(1)) \circ \sigma$ is an automorphism of $H$. Finally, note that $\pi(\beta)(h) = \beta(h)$, so that $\beta = \pi(\beta)$. Thus we have $\sigma = \pi(\sigma(1)^{-1})^{-1} \circ \beta$ $= \pi(\sigma(1)) \circ \pi(\beta)$ $= \pi(\sigma(1),\beta)$. Hence $\sigma \in \pi[G]$.

Since $\pi$ is injective, we conclude that $\mathsf{Hol}(H) \cong N_{S_H}(\mathsf{im}\ \rho)$.

3. Let $\tau$ be an $n$-cycle in $S_n$. Up to a renaming, $\langle \tau \rangle$ is the image of $Z_n$ in $S_{Z_n}$ via the permutation representation $\pi : \mathsf{Hol}(Z_n) \rightarrow S_n$ used above. Now $N_{S_n}(\langle \tau \rangle) \cong \mathsf{Hol}(Z_n)$, and $\mathsf{Hol}(Z_n) = Z_n \rtimes \mathsf{Aut}(Z_n)$. We know that $|\mathsf{Aut}(Z_n)| = \varphi(n)$, where $\varphi$ denotes the Euler totient. Thus $|N_{S_n}(\langle \tau \rangle)| = n\varphi(n)$.
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