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

(1): Once $H, then in agreement with a theorem, $H$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.2012-08-27
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    I changed the notation to something I personally find clearer, but feel free to reverse it if you don't agree.2012-08-27
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    When I first read the title, I thought the question had something to do with a prime exactly dividing the order of a group.2012-08-27
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    @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.2012-08-27
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    @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.2012-08-27
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    Yes, that is what I meant. I think that the approach to proving Sylow's theorem this way goes all the way back to Sylow himself in fact.2012-08-27

4 Answers 4

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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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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.

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    Honestly, what you noted here is what I need to prove some exercise. Thanks. :)2012-08-27
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    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.2012-08-27
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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.2012-08-27
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    This is another aproach. Thanks for noting me it.2012-08-27
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$H$ 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.