Prove that $\mathbb Z^n$ is not isomorphic to $\mathbb Z^m$ for $m\neq n$ 
Prove that $\mathbb Z^n$ is not isomorphic to $\mathbb Z^m$ for $m\neq n$.

My try:
Let $\mathbb Z^n\cong \mathbb Z^m $. To show that $m=n$.
Case 1: Let $m>n$. Now that $\mathbb Z^m$ has $m$ generators whereas $\mathbb Z^n$ has $n$  generators and an isomorphism takes a generator to generator ; that is the contradiction.
Please correct me if I am wrong.
The case  $m<n$ also follows similarly.

$\mathbb{Z}^n \cong \mathbb{Z}^m$ implies $(\mathbb{Z}/2)^n \cong \mathbb{Z}^n / 2 \mathbb{Z}^n \cong \mathbb{Z}^m / 2 \mathbb{Z}^m \cong (\mathbb{Z}/2)^m$. By comparing the number of elements, we get $2^n=2^m$, i.e. $n=m$. (No linear algebra is necessary here!)
