Show that there exist a subgroup $H\cong \mathbb{Z}_p\oplus\mathbb{Z}_p$ $p$ prime

Question

Let $G$ be a finite, abelian, non-cyclic group. Show that there exist a subgroup $H\leq G$ such that $H\cong \mathbb{Z}_p\oplus\mathbb{Z}_p$ $p$ prime.

My progress before getting stuck:

Since $G$ is non-cyclic $|G|=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ where there is at least one $p_j^{a_j}$ with $a_j\neq1$.

I consider the $p_j$-Sylow subgroup of $G$, call it $S$.
Then take the subgroup $H\leq S$ that has order $p^2_j$, by another result having order $p^2$ where p is prime implies $H\cong \mathbb{Z}_{p^2}$ or $H\cong \mathbb{Z}_p\times\mathbb{Z}_p$

I get stuck there, don't know how to show that $H$ can't be cyclic. Any advice is welcomed.

Answer 1

Hint: $G$ is isomorphic to the product $\Pi_i\mathbb{Z}/p_i^{n_i}$

if all the $p_i$ are distinct, $G$ is cyclic (Chinese remainder).

If $p_i=p_j=p$, $\mathbb{Z}/p_i^{n_i}$ and $\mathbb{Z}/p_i^{n_i}$ have subgroup isomorphic to $\mathbb{Z}/p$.