Let $\mathbb{Z}^n,\mathbb{Z}^m$ be a free abelian group, $m<n.$
Prove $\mathbb{Z}^n \not\cong \mathbb{Z}^m$
Is it possible that the problem is trivial since $\mathbb{Z}^n$ has $n$ generators and $\mathbb{Z}^m$ has $m$ generators ,$n\neq m$ ,hence $\mathbb{Z}^n\not\cong\mathbb{Z}^m$?