If $H$ is a subgroup of $G$ of order $p^n$, prove that $H$ is a characteristic subgroup of $G$.

Question

Suppose that $G$ is an abelian group of order $p^nm$ where $p\nmid m$ is a prime. If $H$ is a subgroup of $G$ of order $p^n$, prove that $H$ is a characteristic subgroup of $G$.

Under the assumption that $H$ is not characteristic, I have proven that $p^n|m$, does this lead to $p|m$?

Answer 1

Hint: first note that $H$ is maximal w.r.t. having $p$-power order. Secondly what can be said about the order of the subgroup $H\alpha[H]$, if $\alpha \in Aut(G)$?

Answer 2

By Sylow theorems, the subgroups of $G$ of order $p^{n}$ are all coniugate, but $G$ is abelian, and so there is an unique sylow subgroup of order $p^{n}$, namely $H$.

If $\phi \in $Aut($G$), then $\phi(H)$ is a subgroups of $G$ of order $p^{n}$, and so $$\phi(H) = H$$ Then $H$ is characteristic.