Let G be a finite p-group and H<G. Prove that H<NG(H).
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...
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 $\displaystyle N_G(H)=H.$
we know that in a p-group $\displaystyle |Z(G)| > 1.$ clearly $\displaystyle Z(G) \subseteq N_G(H)=H.$ let $\displaystyle G_1=\frac{G}{Z(G)}.$ then $\displaystyle |G_1| < |G|.$ now draw a contradiction!
Remark: the above result holds in a much larger family of groups, e.g. any finitely generated nilpotent group.