I've been stuck on this problem for a while now.
Let $G$ be finite and abelian. Suppose $\exists x \in G$ such that $x$ has non-square-free order, i.e., $|x| = p^km$ with $p$ prime and gcd$(p,m)=1$ and $k > 1$. Then, show that there exists an element of order $p^2$.
My attempt (using the erroneous assumption that $|G|$ was square-free, instead of $|x|$):
Let $G$ be as above. By Sylow, there exists a subgroup $H$ with order $p^2$ as $p^2$ divides the order of $G$
Furthermore, as $|H| = p^2$, $H$ is abelian. Thus, $H \cong C_p \times C_p$ or $H \cong C_{p^2}$.
I'm not sure how to proceed here; I had the thought that, if I was able to show that H was cyclic, I'd be done. We have the fundamental theorem of abelian groups and Sylow, but not too much else.