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This is a similar question to this question, but I am stuck on a part of a proof for it that does not rely on the Frattini argument.

The statement is

Let $G$ be a finite group, $P \in \mathrm{Syl}_{p}(G)$, and $M$ a subgroup of $G$ with $N_{G}(P) \leq M$. Prove that $N_{G}(M) = M$.

The first step in this proof is to let $g \in N_{G}(M)$, so that $P^{g} \in \mathrm{Syl}_{p}(M)$. Why does it follow that $P^{g} \in \mathrm{Syl}_{p}(M)$?

Since $G$ is finite, we write $|G| = p^{\alpha}t$ with $p \nmid t$ so that $|P| = p^{\alpha}$. Similarly, since $M \leq G$, we have that $|M| \,\big|\,|G|$ so that $|M| \,\big|\,p^{\alpha}$ or $|M| \,\big|\, t$. I tried to use the fact that $N_{G}(P) \leq M$, but all I could get from this was that any $g \in G$ such that $P^{g} = P$ also lies in $M$. If $g \in N_{G}(M)$, then $M^{g} = M$, but I'm not sure how to use this either to get the result.

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    $\begingroup$ Because $\;|P|=|P^g|\;$ ...In fact, both are isomorphic. $\endgroup$ – DonAntonio May 4 '18 at 19:42
  • $\begingroup$ But I don't see why $|M| = p^{\alpha}s$ with $s \leq t$ yet. $\endgroup$ – Bill Wallis May 4 '18 at 19:45
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First a quick correction, It is not necessary that $|M|\mid p^\alpha$ or $|M|\mid t$. If say $\alpha=1$, $p=2$, $t=15$, you might have $|M|=6$.

Any subgroup of $G$ is contained in it's own normaliser, so $P\le N_G(P)\le M$. If $g\in N_G(M)$ then $M^g=M$, so $P^g\le M^g=M$. Since $|M|$ divides $|G|$, $p^\alpha$ must be the largest power of $p$ dividing $|M|$, so $P$ and $P^g$ are Sylow $p$-subgroups of $M$.

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  • $\begingroup$ Perfect, that cleared everything up. Thank you! $\endgroup$ – Bill Wallis May 4 '18 at 19:51

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