Proper subgroup of normalizer of a finite group impies nilpotent.

Question

So I came upon this exercise: Let $G$ be finite. If every proper subgroup $H$ of $G$ has the property $H < N_G(H)$, then $G$ is nilpotent.

I can prove the converse by induction on the nilpotency class of $G$, but I'm kind of stuck here.

D_S · Accepted Answer · 2017-06-11 02:57:37Z

3

It suffices to show that every Sylow $p$-subgroup $P$ of $G$ is normal. But $N_G(P) = N_G(N_G(P))$, so your hypothesis implies $N_G(P) = G$.

answered Jun 11 '17 at 2:57

D_S

24.6k6 gold badges26 silver badges78 bronze badges

$\begingroup$ How do you get to the conclusion that $N_G(P) = G$? If $P$ is properly contained in $G$, then by my hypothesis, $P$ is properly contained in $N_G(P)$, so the order of $N_G(P)$ is not divisible by p. How does that imply $N_G(P) = G$? $\endgroup$ – Sid Caroline Jun 11 '17 at 6:57
$\begingroup$ Let $H = N_G(P)$. $\endgroup$ – D_S Jun 11 '17 at 13:46

Add a comment |

1 Answer 1