If $\mathbb Z^m\twoheadrightarrow\left(\mathbb Z\big /2\mathbb Z\right)^n$ is a surjective homomorphism, why is then $m=n$?

Question

If $\mathbb Z^m\twoheadrightarrow\left(\mathbb Z\big /2\mathbb Z\right)^n$ is a surjective homomorphism and $m\le n$, why does it force $m=n$ ?

Is there no appropriate choice for a map when $m< n$ ?

Answer 1

Note that the kernel of your group homomorphism $\varphi:\Bbb Z^m\twoheadrightarrow(\Bbb Z/2\Bbb Z)^n$ contains $(2\Bbb Z)^m$, hence $\varphi$ factors through the canonical mapping $\pi:\Bbb Z^m\twoheadrightarrow(\Bbb Z/2\Bbb Z)^m$, giving rise to a surjective group homomoprhism $\varphi':(\Bbb Z/2\Bbb Z)^m\twoheadrightarrow(\Bbb Z/2\Bbb Z)^n$. This is, in fact, an $\Bbb Z/2\Bbb Z$-linear mapping, hence by comparing degree as vector spaces over $\Bbb Z/2\Bbb Z$, we get $m\geq n$, hence $m=n$.

Answer 2

The target is a vector space of dimension $n$ over the field $\mathbb{F}_2$, so it can't be spanned by fewer than $n$ vectors. But the image of any map out of $\mathbb{Z}^m$ is going to consist of linear combinations of the images of the $m$ elements

$$(1, 0, 0, \ldots, 0), (0, 1, 0, \ldots, 0), \cdots, (0, 0, 0, \ldots, 1).$$

Answer 3

So in other words you ask: can $\mathbb{Z}_2^n$ have a generating set with less then $n$ elements? The answer is no. That's because if $\{g_i\}_{i=1}^k$ is such subset for $k<n$ then since $g_i=g_i^{-1}$ and $g_ig_j=g_jg_i$ then it follows that there are at most $2^k$ combinations of $g_i$'s (that produce different elements) and thus

$$2^n=|\mathbb{Z}_2^n|\leq 2^k$$

which is clearly a contradiction. This argument (unlike vector space argument) can be generalized to any finite product of cyclic groups of arbitrary (finite or not) orders.