Let $P$ be a Sylow $p$-subgroup of a finite group $G$, for some prime $p$. Prove that if $H$ is a subgroup of $G$ that contains all Sylow $p$-subgroups of $G$, then $G = HN_G(P)$.

Here's what I have so far:

I know that the normalizer of $P$ in $G$ is $$N_G(P) = \left\{g \in G \,\mid\, gAg^{-1} = A\right\}$$

where $$gAg^{-1} = \left\{ gag^{-1}\,\mid\, a \in A \right\} $$

and that $HN_G(P)$ refers to the product of subgroups.

If the order of $G$ is $p^n$ for some $n \in \mathbb{N}$, then the multiplicity of $p$ is $n$. This means $G$ is a Sylow $p$-subgroup.

If we choose $P = G$, then $H$ must contain $P$, so $H = G$. $N_G(G) = G$, so $$HN_G(P) = GG = G$$

This isn't remotely a proof, only a case in which what I'm trying to prove is true. A hint would be great. Thanks in advance.

  • 1
    $\begingroup$ Why do you think $\;HN_G(P)\;$ is a direct product? It is a product of groups, not necessarily direct. $\endgroup$ – DonAntonio Dec 14 '13 at 16:12
  • $\begingroup$ Thanks - should be a group product. I'll edit the question. $\endgroup$ – Zubin Mukerjee Dec 14 '13 at 16:14

You could "copy" the proof of Frattini's Argument: let $\;g\in G\;$ and let $\;P\;$ be some Sylow subgroup of $\;G\;$ , so $\;P\le H\;$ . But also $\;P^g:=g^{-1}Pg\;$ is a Sylow subgroup and thus also $\;P^g\le H $ .

Note that all the Sylow subgroups of $\;G\;$ are also Sylow subgps. of $\;H\;$ , so by Sylow's Theorems

$$\exists\;h\in H\;\;s.t.\;\;P^g=P^h\implies P^{gh^{-1}}=P\implies gh^{-1}\in N_G(P)$$

and now end the proof.


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