Prove that G has a subgroup of order p^n

If $\displaystyle G$ is a finite abelian group and $\displaystyle p$ is a prime such that $\displaystyle p^n$ divides $\displaystyle |G|$, then prove that $\displaystyle G$ has a subgroup of order $\displaystyle p^n$.

Note: This proof is very similar to the First Sylow Theorem, but in this one, G is abelian.

Attempt at the proof:

If $\displaystyle p^n||G|$, then $\displaystyle |G| = p^nm, m \in \mathbb{Z}, (p,m) = 1$.

Consider a group $\displaystyle H$.

Claim: $\displaystyle H \subset G$ is a group such that $\displaystyle |H| = p^n$.

I know that I have to use Lagrange's Theorem to show that $\displaystyle [G:H] = m$. I just don't know how to do this. Can anyone help?