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Let $G$ be a finite group of size $n$. Let $H$ be a normal subgroup of size $m$. Let $C(H)$ be the centralizer of the $H$, with size $k$. Prove $n/k$ divides $(m-1)!$

I'm not really sure how to start, my guess is that it uses the Orbit Stabilizer Theorem.

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Let $G$ act on $H$ by conjugation. Let $S_{m-1}$ act on the nonidentity elements of $H$ by permutation. The action of $G$ on $H$ induces a homomorphism $G\to S_{m-1}$ and the kernel is exactly $C(H)$ (because any element not in $C(H)$ induces a nontrivial permutation), so the image has size $n/k$. Thus there is a subgroup of $S_{m-1}$ of order $n/k$, so $n/k\mid (m-1)!$.

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