# Thread: Help with finite p-group

1. ## Help with finite p-group

Let G be a finite p-group and H<G. Prove that H<NG(H).

2. I think the proof goes as follows (though I'm used to asking questions here, not helping ).

G is finite and with a prime order. Therefore it's cyclic (have you studied that? It's pretty basic).
A cyclic group is of course commutative (abelian).
Therefore, for every sub-group H in R, the normalizer in G N(H) is actually G itself...
So obviously H is a subgroup of N(H)=G...

3. Originally Posted by bogazichili
Let G be a finite p-group and H<G. Prove that H<NG(H).
suppose the claim is false for some p-groups. choose G to be one of those groups but with smallest possible order. so there exists a subgroup H of G such that $N_G(H)=H.$

we know that in a p-group $|Z(G)| > 1.$ clearly $Z(G) \subseteq N_G(H)=H.$ let $G_1=\frac{G}{Z(G)}.$ then $|G_1| < |G|.$ now draw a contradiction!

Remark: the above result holds in a much larger family of groups, e.g. any finitely generated nilpotent group.