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I was solving the following question :

A finite group $G$ is nilpotent if every proper maximal subgroup of $G$ is normal. [Hint: If $P$ is a Sylow $p$-subgroup of $G$, show that any subgroup containing $N_G(P)$ is its own normalizer.]

Here, $N_G(P)$ is the normalizer of $P$ in $G$.

I tried as follows :

It suffices to show that every Sylow subgroup of $G$ is normal in $G$ (by a theorem in the book). So let $P$ be a Sylow $p$-subgroup of $G$. By another theorem in the book, we have $N_G(P)=N_G(N_G(P))$. On the other hand, by a proposition in the book, every proper subgroup of a nilpotent group is a proper subgroup of its normalizer. Therefore $N_G(P)$ must be the whole group $G$. That is, $P$ is normal in $G$, so we are done.

So, why the hint is needed? I have no idea. Is my proof wrong?

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    $\begingroup$ Which book are you referring to? Dummit and Foote? (See Proposition 7, Chapter 6 of Dummit and Foote for a proof) $\endgroup$
    – cqfd
    Sep 12, 2019 at 17:52
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    $\begingroup$ @Thomas Shelby Actually, this is from a lecture note in my school, and probably the lecture note is mainly from Hungerford $\endgroup$
    – user302934
    Sep 12, 2019 at 17:57

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... by a proposition in the book, every proper subgroup of a nilpotent group is a proper subgroup of its normalizer. [...] Is my proof wrong?

What group do you suppose is nilpotent, so that you can apply this to it? It's not like you know $G$ is nilpotent yet: that's what you're trying to prove.

Suppose you believe the hint. Then in the case that $N_G(P)\neq G$, you argue that it's contained in a maximal subgroup of $G$, which is equal to its own normalizer. But that contradicts the hypotheses...

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  • $\begingroup$ Oh, I have done a terrible mistake.. Thanks $\endgroup$
    – user302934
    Sep 12, 2019 at 18:13

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