# Proving a necessary and sufficient condition for a finite group being nilpotent

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?

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

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...