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I am currently taking Abstract Algebra II and am stuck on the last question of my assignment. I asked the Professor about it and he said that he didn't remember how to do it and that there might be a typo. Would anyone be able to help answer the question (or provide a counterexample if there is a typo)?

Let $P$ be a Sylow p-subgroup of a group $G$. If $H$ is a subgroup of $G$ and $N(P) \subseteq H$, show that $N(H) = H$.

This is what I have so far - I'm not entirely sure if it's useful:

Since $H\subseteq N(H)$, all we need to do is prove that $N(H) \subseteq H$. Let $a\in N(H)$, then $aHa^{-1}=H$. But $N(P)\subseteq H$ so $N(P)\subseteq aHa^{-1}$, or $aN(P)a^{-1} \subseteq H$. Also, $P\subseteq N(P) \subseteq H$ so $aPa^{-1}\subseteq H$. But everything of the form $aPa^{-1}$ is also a Sylow p-subgroup by Sylow's second theorem.

Thanks!

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I almost can't believe someone teaching material this basic can forget such a basic result... –  DonAntonio Oct 31 '12 at 3:05
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2 Answers 2

up vote 4 down vote accepted

This is just the Frattini argument.

Let $N=N_G(H)$, in which $H$ is obviously normal and $P$ is a Sylow $p$-subgroup. Thus, an application of the Frattini argument gives $N_N(P)H=N$, but $N_N(P)\subseteq N_G(P)=H$. Therefore, $N_G(H)\subseteq H$ as required.

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Thank you so much!! –  Samuel Reid Oct 31 '12 at 4:19
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Another nice property of any subgroup $H$ with $N_G(P) \subseteq H \subseteq G$ with $P \in Syl_p(G)$ is that the index $[G:H] \equiv 1$ mod $p$.

This follows from the fact that in $G$ we have $|Syl_p(G)|=[G:N_G(P)] \equiv 1$ mod $p$. Similarly, since obviously $P \in Syl_p(H)$ also the index $[H:N_H(P)] \equiv 1$ mod $p$. But $N_H(P) = H \cap N_G(P)= N_G(P)$, so $[G:N_G(P)]=[G:H]\cdot[H:N_G(P)]=[G:H]\cdot[H:N_H(P)]$ and taking all this mod $p$ the congruence follows.

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