$p\mid [G:H]$ then $p\mid [N_G(H):H]$

Question

I encountered the following problem for the first time. I sketched a proof for it. I will be thankful if I know it is correct or not. Thanks.

$p$ is a prime and $H$ is a $p$-subgroup of a finite group $G$ such that $p\mid [G:H]$ . Prove that $p\mid [N_G(H):H]$.

I assume $|G|=p^\alpha m, (p,m)=1$ and $|H|=p^\beta, \beta\lneqq\alpha$. According to Sylow's theorem, there is a $p$-sylow subgroup of $G$ including $H$ as a subgroup, say $K$. I see that $H<K$ $\mathrm{so^{(1)}}$ one theorem tells me $H<N_G(H)$ or $p\mid [N_G(H):H]$.

(1): Once $H<K$, then in agreement with a theorem, $H<N_K(H)$. But obviously, $N_K(H)\leq N_G(H)$ so $H<N_G(H)$ which means that $p\mid [N_G(H):H]$.

Answer 1

It's correct. That's exactly what I thought before I read your solution.

The important thing in your solution is the lemma that says "Non-Sylow $p$-groups grow in their normalizers".

Answer 2

Even more is true: in general, if $H$ is a $p$-subgroup of $G$ then $[G:H] \equiv [N_G(H):H]$ mod $p$. See for instance http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/sylowpf.pdf for a proof. The proof is not difficult and depends on the action of $H$ on the left coset space $G/H$ by left multplication.

Answer 3

Define an action of H on the set of cosets of H in G by multiplication. Then, from the class equation we deduce that $|G:H| \equiv |K| +\Sigma |O_i| \pmod p$, where $K$ is the set of fixed points, and $O_i$ are orbits of orders greater than 1. But $|O_i|=|H|/|S_i|$, where $S_i$ is the stabilizer of an element in $O_i$, thus $p$ divides $|O_i|$ and then $|G:H| \equiv |K| \pmod p$. However, $|K|$ is just $|N_GH:H|$, so the result follows.

Answer 4

$H$ is $p$-subgroup, and $p|[G\colon H]$ means $H$ is proper subgroup of Sylow-$p$ subgroup of $G$, say $K$. Now, in the Sylow-$p$ subgroup $K$, $H$ is normalized by some element outside $H$; hence $p|[N_K(H)\colon H]$, and this implies your conclusion.