Prove that the group isomorphism $\mathbb{Z}^m \cong \mathbb{Z}^n$ implies that $m = n$ 
Let $m$ and $n$ be two nonnegative integers. Assume that there is a group isomorphism $\mathbb{Z}^m \cong \mathbb{Z}^n$. Prove that $m = n$.

I tried using a contrapositive, ($m \neq n$ implies $\mathbb{Z}^m \ncong \mathbb{Z}^n$), and I think the problem is that there won't be a homomorphism, but did not get anywhere. Is there a better approach to this problem?

Let $G=\mathbb{Z}^n$ and $H=\mathbb{Z}^m$. Then $G/2G = (\mathbb{Z}/2\mathbb{Z})^n$. In particular, $|G/2G| = 2^n$. By a similar argument $|H/2H| = 2^m$. If they are isomorphic then $2^n = 2^m$. 
