$\mathbb Z^n$ as a quotient of $\mathbb Z^m$ It is quite obvious that if $n\le m$, the group $\mathbb Z^n$ can be obtained as a quotient of $\mathbb Z^m$. But is the converse statement also true? That is, if $\mathbb Z^n$ is a quotient of $\mathbb Z^m$, is $n\le m$? In that case, is there an easy proof?
 A: A basis of $\mathbb{Z}^m$ has $m$ elements. Therefore every quotient of $\mathbb{Z}^m$ is generated by $m$ (or fewer) elements. If $\mathbb{Z}^n$ is a quotient of $\mathbb{Z}^m$, it is therefore generated by at most $m$ elements. And if $\mathbb{Z}^n$ is generated by $k$ elements $\{a_1,\dotsc,a_k\}$, the images $\{b_1,\dotsc,b_k\}$ of these elements in $Q = \mathbb{Z}^n/(2\mathbb{Z}^n)$ generate $Q$. We can view $Q$ in a natural way as a vector space over $\mathbb{F}_2$, since every element of $Q\setminus \{0\}$ has order $2$. On the one hand, we know that $Q$ contains $2^n$ elements, so the dimension is $n$. On the other hand, it is generated as a group, hence as an $\mathbb{F}_2$-vector space, by the $k$ elements $\{b_1,\dotsc,b_k\}$, hence $n = \dim_{\mathbb{F}_2} Q \leqslant k$.
A: If we have a surjective homomorphism $f\colon\mathbb{Z}^m\to\mathbb{Z}^n$, then this homomorphism splits, so
$$
\mathbb{Z}^m\cong\ker f\oplus\mathbb{Z}^n.
$$
Tensoring with $\mathbb{Q}$ gives
$$
\mathbb{Q}^m\cong(\ker f\otimes\mathbb{Q})\oplus\mathbb{Q}^n
$$
and therefore $n\le m$.
