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?

Suppose $m\neq n$,
$$\Bbb Z^m\cong\langle x_1,\dots, x_m\mid \{x_ix_j=x_jx_i: i,j\in\{1,\dots m\}\}\rangle,$$
and
$$\Bbb Z^n\cong\langle y_1,\dots, y_n\mid \{y_ry_s=y_sy_r: r,s\in\{1,\dots n\}\}\rangle.$$
In each presentation, kill $z^2$ for each element $z$. Then we get, by abelianness,
$$G:=\Bbb Z^m/\langle \{x_i^2:i\in\{1,\dots, m\} \rangle \cong\langle x_1,\dots, x_m\mid \{x_ix_j=x_jx_i: i,j\in\{1,\dots m\}\}\cup\{x_i^2: i\in\{1,\dots, m\}\}\rangle$$
and
$$H:=\Bbb Z^n/\langle \{y_r^2:r\in\{1,\dots, n\} \rangle \cong\langle y_1,\dots, y_n\mid \{y_ry_s=y_sy_r: i,j\in\{1,\dots n\}\}\cup\{y_r^2: r\in\{1,\dots, n\}\}\rangle.$$
But then
$$G\cong\Bbb Z_2^m\not\cong \Bbb Z_2^n\cong H$$
by this standard result, since $m\neq n$; but we did the same thing to each presentation and yet we got different results. Thus
$$\Bbb Z^m\not\cong\Bbb Z^n.$$
This is the contrapositive of the result in question, so we are done.
