H is a proper subgroup of p-group G. Show that normalizer of H, N(H) is strictly larger than H, and that H is contained in a normal subgroup index p.
Here's what I've got so far:
- If H is normal, N(H) is all of G and we are done.
- If H is not normal, then suppose for the sake of contradiction that N(H)=H. Then there is no element outside of H that fixes H by conjugation. But the center Z(G) of G does fix H, so Z must be in H.
Don't know if I'm going on the right path or not, but either way can't really think my way out of this one... Any help appreaciated.