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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 $p\mid [N_G(H):H]$.

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@BabakSorouh Sure you can, but it would be helpful if you just duplicated the corrected text and then correcting it, instead of deleting the old text entirely. – rschwieb Aug 27 '12 at 15:09
I changed the notation to something I personally find clearer, but feel free to reverse it if you don't agree. – M Turgeon Aug 27 '12 at 15:13
When I first read the title, I thought the question had something to do with a prime exactly dividing the order of a group. – M Turgeon Aug 27 '12 at 15:21
@Babak: while you proof is correct, it is more natral to use teh result to prove Sylow's theorem, rather than using Sylow's theorem to prove it. For if we are trying to prove existence of Sylow $p$-subgroups, $H \neq 1$ is a maximal $p$-subgroup of $G$, then we may suppose that $[G:H]$ is divisible by $p.$ Then $[N_{G}(H):H]$ is divisible by $p.$ But then $H$ is not even a maximal $p$-subgroup of $N_{G}(H),$ a contradiction. – Geoff Robinson Aug 27 '12 at 17:17
@GeoffRobinson: You mean, we put the Sylow's theorem aside, and then try to prove the problem without using it? If so, then the awllower's answer below would be our approach. – S. Snape Aug 27 '12 at 17:30
up vote 2 down vote accepted

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".

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Thanks for your time. :) – S. Snape Aug 27 '12 at 15:08

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 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.

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Honestly, what you noted here is what I need to prove some exercise. Thanks. :) – S. Snape Aug 27 '12 at 15:28
Yes, it also appears as Exercise 1.A.10 in Marty Isaacs' Finite Group Theory (a wonderful book by the way!). The exercise should be made without appeal to Sylow theory, which he develops in the subsequent chapters. – Nicky Hekster Aug 27 '12 at 20:02
+1 for the great link. – awllower Aug 28 '12 at 9:54

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.

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I posted before I saw the answer of Nicky. Sorry for the duplication. – awllower Aug 27 '12 at 15:19
This is another aproach. Thanks for noting me it. – S. Snape Aug 27 '12 at 15:29

$H$ 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.

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