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?

Once again, I'll give you the outline, you fill in the details.

Suppose that $\mathbb{Z}^m\cong\mathbb{Z}^n$. Then, you see that $\mathbb{Z}^m\otimes_{\mathbb{Z}}\mathbb{Z}_2\cong\mathbb{Z}^n\otimes_{\mathbb{Z}}\mathbb{Z}_2$, which tells you that $\mathbb{Z}_2^m\cong\mathbb{Z}_2^n$, and so $2^m=2^n$--thus $m=n$.

EDIT: Since you don't know tensor products, perhaps this will be more understandable. If $\mathbb{Z}^m\cong\mathbb{Z}^n$, then $\text{Hom}_\mathbb{Z}(\mathbb{Z}^m,\mathbb{Q})\cong\text{Hom}_\mathbb{Z}(\mathbb{Z}^n,\mathbb{Q})$ as $\mathbb{Q}$-vector spaces. But, you can prove that as $\mathbb{Q}$-vector spaces one has that $\text{Hom}_\mathbb{Z}(\mathbb{Z}^k,\mathbb{Q})\cong\mathbb{Q}^k$, and so we have $\mathbb{Q}^m\cong\mathbb{Q}^n$, from where normal vector space theory tells us that $m=n$.

EDIT EDIT: You seemed to only take issue with the previous proof because of the vector spaces. In patticular,you seemed ok with the Hom manipulation. Here's a way to combine the first and the previous proof! Show first that $\text{Hom}(\mathbb{Z}^k,A)\cong A^k$ for any abelian group A. Then apply this fact to show that our problem implies $\mathbb{Z}_2^m\cong\mathbb{Z}_2^n$ again.

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 $2^n = 2^m$.

You can prove it in the same way that you prove that the dimension of a vector space is well-defined.

Suppose that $m\le n$ and $h:\Bbb Z^m\to\Bbb Z^n$ is an isomorphism. For $k=1,\dots,m$ let $e_k$ be the element $\langle a_1,\dots,a_m\rangle$ such that $a_k=1$ and $a_i=0$ for $i\ne k$. Note for any $\langle a_1,\dots,a_m\rangle\in\Bbb Z^m$,

$$\langle a_1,\dots,a_m\rangle=\sum_{k=1}^ma_ke_k\;,$$

and this representation is unique: if

$$\sum_{k=1}^ma_ke_k=\sum_{k=1}^mb_ke_k\;,$$ then $$\sum_{k=1}^m(a_k-b_k)e_k=\langle \underbrace{0,\dots,0}_m\rangle\;,$$ and therefore $a_1=b_1,\dots,a_m=b_m$. In other words, $\{e_1,\dots,e_m\}$ behaves very much like a basis for a vector space.

Now show that $\{h(e_1),\dots,h(e_n)\}$ behaves like a basis for $\Bbb Z^n$: each $z\in\Bbb Z^n$ can be written uniquely in the form $$z=\sum_{k=1}^ma_kh(e_k)$$ for some integers $a_k$, $k=1,\dots,m$.

Now get a contradiction when $n>m$ by considering the elements of $\Bbb Z^n$ that are analogous to the $e_k\in\Bbb Z^m$.

Embed $\Bbb Z$ in $\Bbb Q$ and thereby $\Bbb Z^n$ in $\Bbb Q^n$. Now clearly $\Bbb Z^n$ has an $n$-tuple of elements that are linearly independent over $\Bbb Z$ (the standard basis will do as an example), and it does not have any $(n+1)$-tuple that is linearly independent over $\Bbb Z$, since already $\Bbb Q^n$ doesn't admit any $(n+1)$-tuple that is linearly independent over $\Bbb Q$, let alone over $\Bbb Z$. Then you can recover the $n$ in $A\cong\Bbb Z^n$ as the maximum number of linearly independent elements over $\Bbb Z$ one can find in the free Abelian group $A$. Thus $\Bbb Z^m\cong\Bbb Z^n$ implies $m=n$.