Let $G$ be an abelian group of order $(p^n)m$, where $p$ is a prime and $m$ is not divide by $p$. Let $P = \{a\in G\mid a^{p^k} = e\text{ for some $k$ depending on $a$}\}$.
Prove that
(a) $P$ is a subgroup of $G$. 
(b) $G/P$ has no elements of order $p$. 
(c) order of $P=p^n$.

I like the rest above, so I just wanted to suggest an argument for b).
Suppose there is an $\overline{x} \in G/P$ with order $p$, then $(x+P)^p=P$.  But this means $x^p \in P$, so $(x^p)^{p^k}=e$ for some $k$.  But this means $x\in P$ and thus $\overline{x}$ is actually the identity in $G/P$.  But the identity cannot have order $p$, so this is not possible.
