TITLE: Is there a surjective group homomorphism $\operatorname{GL}_{n}(k) \to \operatorname{GL}_{m}(k)$ where $n > m$?
QUESTION [16 upvotes]: Does there exist a field $k$, two positive integers $n > m > 1$, and a surjective group homomorphism $\operatorname{GL}_{n}(k) \to \operatorname{GL}_{m}(k)$?

Here $k$ can be any field, and $\operatorname{GL}_{n}(k)$ is viewed as an abstract group (as opposed to group scheme or Lie group), and this group homomorphism doesn't have to be "algebraic" or "smooth" in any sense. Note that if $m = 1$ then the determinant map gives a surjective map.

REPLY [2 votes]: Here is a solution:
Suppose that there exists a surjective map $GL_n(k) \rightarrow GL_m(k)$ for $n > m > 1$. This induces a surjective map from the commutator subgroup of $GL_n(k)$ to the commutator subgroup of $GL_m(k)$.
The commutator subgroup of $GL_n(k)$ is $SL_n(k)$ except when $m = 2$ and $k = \mathbb{F}_2$. In the case $k = \mathbb{F}_2$ we have $GL_n(k) = SL_n(k)$, so in any case we can assume that we have a surjective map $SL_n(k) \rightarrow SL_m(k)$.
Now $SL_n(k)$ has a composition series, and it has a unique nonabelian simple composition factor, namely $PSL_n(k) = SL_n(k) / Z(SL_n(k))$. Similarly there is a unique nonabelian simple composition factor of $SL_m(k)$, which is $PSL_m(k)$.
Therefore we must have an isomorphism $PSL_n(k) \cong PSL_m(k)$.
But this is a contradiction, by the following old result:

$PSL_n(k) \cong PSL_m(k')$ implies $n = m$ and $k \cong k'$, except in the following cases: $PSL_2(\mathbb{F}_7) \cong PSL_3(\mathbb{F}_2)$ and $PSL_2(\mathbb{F}_4) \cong PSL_2(\mathbb{F}_5)$.

See p. 106, §9, chapter IV, in "La géométrie des groupes classiques" by Dieudonné.