Let G be a group of order p^n suppose that H is a normal subgroup of G

Question

Prove that there exists a normal subgroup H' a subset of G such that [H':H] = p

Answer 1

The number of subgroups of $H$ with index $p$ is congruent to $1\pmod p$. We look at the orbits of these subgroups under conjugation by elements in $G$; of course, they form orbits of size $p^k$ for some non-negative integer $k$. Since the sum of the sizes of these orbits is congruent to $1\pmod p$, at least one of them must be $1$; in other words, it is normal in $G$.