# Why is a nontrivial finite group is nilpotent if every maximal subgroup is normal?

In the following proof I understand the above proof up to the part where it says "by Sylow theory $N(M)=M$", could someone explain to me why is this true. We have just started learning about group theory and few days ago we started learning about the first and second Sylow theorms but I am not sure which one of them implies that $N(M)=M$

Thanks.

• The Sylow theory they're referencing is that any subgroup containing the normalizer of a Sylow subgroup is self-normalizing. The proof follows from Frattini's argument. Read the third application at the bottom. – Ben West May 21 '16 at 4:56

## 1 Answer

This is due to the following.

Lemma Let $G$ be a finite group, $P \in Syl_p(G)$ and $H \leq G$ with $N_G(P) \subseteq H$. Then $N_G(H)=H$.

Proof Let $x \in N_G(H)$. Since $P \subseteq N_G(P) \subseteq H$, $P \in Syl_p(H)$. Since $x$ normalizes $H$, $P^x \in Syl_p(H)$, so by Sylow theory, $P^x=P^h$ for some $h \in H$. This means $xh^{-1} \in N_G(P)$, which implies $x \in hN_G(P) \subseteq H$. Conversely, $H \subseteq N_G(H)$ follows from the definition. Hence we are done.