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Let $P$ be a $p$-subgroup of a finite group $G$. Show that $P$ is a Sylow $p$-subgroup of $G$ iff $P$ is a Sylow $p$-subgroup of $N_G(P)$.

I proved one direction by using the fact $[G:P]=[G:N_G(P)][N_G(P):P]$. To prove other direction, since $p$ doesn't divide $[N_G(P):P]$, it is sufficient to show that $p$ doesn't divide $[G:N_G(P)]$. But I can not go on.

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    $\begingroup$ What have you tried? One direction is obvious (to me), so can you explain where you are stuck? $\endgroup$ – Prahlad Vaidyanathan Nov 14 '13 at 17:31
  • $\begingroup$ @PrahladVaidyanathan: It seems we are sitting here to solve others questions which contains no attempt. $\endgroup$ – mrs Nov 14 '13 at 17:36
  • $\begingroup$ @B.S. Fine with me, so long as the questions are interesting. But I dare say I may solve some without posting my solution if there's no evidence the OP has even thought about the question! $\endgroup$ – Brett Frankel Nov 14 '13 at 17:39
  • $\begingroup$ Note that $[G:N_G(P)]$ is precisely $n_p$, the number of $p-$Sylow subgroups, and $n_p \equiv 1\pmod{p}$. $\endgroup$ – Prahlad Vaidyanathan Nov 14 '13 at 17:44
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One direction should be easy, for the other direction you can use this:

If $G$ is a $p$-group and $H$ is a proper subgroup of $G$, then $H$ is a proper subgroup of $N_G(H)$.

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At Spin's answer (the "normalizers grow" principle) an additional hint: choose $S \in Syl_p(G)$ with $P \subseteq S$ (this is possible). Assume $P \subsetneq S$ and apply the "normalizers grow" in $S$ to arrive at a contradiction.

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  • $\begingroup$ Suppose $P \notin Sy{l_p}\left( G \right)$. Then, $\exists Q \in Sy{l_p}\left(G\right)$ such that $P\ne<Q$ By Spin's answer, it follows that $P \ne < {N_Q}\left( P \right) \le Q$. Note that ${N_Q}\left(P\right) \le {N_G}\left(P\right)$. Now, since $p$ doesn't divide $\left[ {{N_G}\left( P \right):P} \right] = \left[ {{N_G}\left( P \right):{N_Q}\left( P \right)} \right]\left[ {{N_Q}\left( P \right):P} \right]$, p doesn't divide $\left[ {{N_Q}\left( P \right):P} \right]$. So, ${{N_Q}\left( P \right)}$ is not a p-subgroup. But this contradicts ${N_Q}\left( P \right) \le Q$ $\endgroup$ – egrtomath Nov 16 '13 at 13:58
  • $\begingroup$ Mustafa, you got it! Well done!! $\endgroup$ – Nicky Hekster Nov 16 '13 at 14:46

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