There is no surjective homomorphism $\Bbb Z^m\to (\Bbb Z_n)^k$ for $n>1$ if $k>m$

Question

Let $n>1$ be an integer and let $G$ be the product of $k\geq 1$ cyclic groups of order $n$, i.e. $G=\Bbb Z_n\times \cdots \times \Bbb Z_n$. Of course there is a generating set of $G$ of order $k$, or equivalently there is a surjective homomorphism $\Bbb Z^k\to G$. However how can we show that there is no surjective homomorphism from $\Bbb Z^m$ to $G$ if $m<k$?

Answer 1

Pick $p|n$ where $p$ is a prime, and if such a map exists, combined with $(\mathbb Z/n\mathbb Z)^k\rightarrow (\mathbb Z/p\mathbb Z)^k$, we would have a surjective homomorphism $\mathbb Z^m\rightarrow (\mathbb Z/p\mathbb Z)^k$, that is $(\mathbb Z/p\mathbb Z)^k$ as an abelian group can be generated by $m$ elements. On the other hand, $(\mathbb Z/p\mathbb Z)^k$ if a vector space of dimension $k$ over the field $\mathbb Z/p\mathbb Z$, it cannot be generated by less than $k$ elements over $\mathbb Z/p\mathbb Z$ (hence over $\mathbb Z$). Therefore $m\ge k$.

Also check this out: Minimal number of generators for a finitely generated abelian $p$-group (where the above argument is given in the comment).

More generally, the same technique can be used to establish the minimal numbers of generators for a finitely generated abelian group is equal to the number of its invariant factors.