Is there a surjective group homomorphism $\operatorname{GL}_{n}(k) \to \operatorname{GL}_{m}(k)$ where $n > m$?

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.

This is far from a complete answer, but it's a start and it's too long for a comment.

Let $k:=\Bbb{F}_q$ be a finite field of $q$ elements and let $$\rho:\ \operatorname{GL}_n(k)\ \longrightarrow\ \operatorname{GL}_m(k),$$ be a surjective group homomorphism. Then $\ker\rho\unlhd\operatorname{GL}_n(k)$, so we have either $$\ker\rho\subset Z(\operatorname{GL}_n(k))\qquad\text{ or }\qquad\operatorname{SL}_n(k)\subset\ker\rho.$$ In the latter case $|\operatorname{im}\rho|$ divides $q-1$, contradicting the surjectivity of $\rho$. So the elements of $\ker\rho$ are diagonal matrices and therefore $|\ker\rho|$ divides $(q-1)^n$. Note that $$|\operatorname{GL}_n(k)|=\prod_{i=0}^{n-1}(q^n-q^i)=q^{\tfrac{n(n-1)}{2}}\prod_{i=1}^n(q^i-1),$$ which shows that $|\ker\rho|$ divides the product and so $|\operatorname{im}\rho|$ is divisible by $q^{\tfrac{n(n-1)}{2}}$. But $\rho$ is surjective so $\operatorname{im}\rho=\operatorname{GL}_m(k)$, where $$|\operatorname{GL}_m(k)|=\prod_{i=0}^{m-1}(q^m-q^i)=q^{\tfrac{m(m-1)}{2}}\prod_{i=1}^m(q^i-1),$$ is clearly not divisible by $q^{\tfrac{n(n-1)}{2}}$, a contradiction. So no such group homomorphism exists for finite fields.

I claim that no such homomorphism can exist with the conditions stated. Due to Servaes's answer, we are free to assume in the following that $k$ is infinite.

Suppose that $\phi\colon\mathrm{GL}_n(k)\to\mathrm{GL}_m(k')$ is a surjective homomorphism with $n>m>1$ and $k$ and $k'$ infinite fields.

Observe that a surjection $\mathrm{GL}_n(k)\to\mathrm{GL}_m(k')$ gives a surjection $\mathrm{PGL}_n(k)\to\mathrm{PGL}_m(k')$. Since the target is non-trivial (since $m>1$) and $\mathrm{PGL}_n(k)$ is simple, this implies that $\mathrm{PGL}_n(k)\cong\mathrm{PGL}_m(k')$. Then, by Theorem 1.3 of the Borel--Tits article, we deduce that $k\cong k'$ and $m=n$ since $\mathrm{PGL}_n$ being isogenous to $\mathrm{PGL}_m$ implies that $m=n$.

EDIT: Below is a more complicated attempt, but I'll leave it up in case someone else finds it interesting

Let us consider the induced homomorphism $\varphi$ given by the composition of the following maps

$$\mathrm{SL}_n(k)\hookrightarrow \text{GL}_n(k)\to\mathrm{GL}_m(k)\to\mathrm{PGL}_m(k).$$

Lemma: The map $\varphi$ has Zariski dense image and is non-trivial.

Proof: Let us first show that $\varphi$ has a Zariski dense image. Let $U\subseteq \mathrm{PGL}_m$ be a non-empty Zariski open subset. Note that since the multiplication-by-$m$ map $[m]\colon \mathrm{PGL}_m\to\mathrm{PGL}_m$ is continuous, we have that $[m]^{-1}(U)$ is Zariski open. We claim this set is also non-empty. To do this, it suffices to show that there is an $\overline{k}$-point of $U$ which is an $m^\text{th}$ power. Note though that the regular semisimple elements$^{(\ast)}$ of $\mathrm{PGL}_m$ form a dense open subset (e.g. see [Steinberg, 2.14]), and so $U(\overline{k})$ must contain such an element. But, strongly regular elements belong to a maximal torus $T\cong\mathbb{G}_m^n$ of $\mathrm{PGL}_{n,\overline{k}}$ which are easily seen to have $m^\text{th}$ powers.

So, since $\mathrm{PGL}_m(k)$ is Zariski dense in $\mathrm{PGL}_m$ (e.g. see [Milne, Theorem 17.93]) and so there is a point $g$ in the $k$-points of $[m]^{-1}(U)$. By definition this is of the form $g=h^m$ for $g\in U(k)$ and $h\in\mathrm{PGL}_m(k)$. Note though that $\mathrm{PGL}_m=\mathrm{PSL}_m$ and so one has an exact sequence $$1\to \mu_m(k)\to \mathrm{SL}_m(k)\to \mathrm{PGL}_m(k)\to H^1_\mathrm{fppf}(\text{Spec}(k),\mu_m).$$ Since $H^1_\text{fppf}(\mathrm{Spec}(k),\mu_m)$ is $m$-torsion we see that $g=h^m$ must have trivial image in $H^1_\mathrm{fppf}(\mathrm{Spec}(k),\mu_m)$ and therefore must be in the image of $\mathrm{SL}_m(k)\to\mathrm{PGL}_m(k)$. Note though that since $\phi$ is surjective, it induces a surjection $$\mathrm{SL}_n(k)=\mathrm{GL}_n(k)^\mathrm{der}\to\mathrm{GL}_m(k)^\mathrm{der}=\mathrm{SL}_m(k)$$ (where these equalities hold since $k$ is infinite). This gives the desired density.

The fact that $\varphi$ is non-trivial since $\mathrm{SL}_m(k)\to\mathrm{PGL}_m(k)$ is non-trivial since $m>1$. $\blacksquare$

Note that $\mathrm{SL}_n$ is connected, split, absolutely simple, and simply connected. From this we see that $\mathrm{SL}_n=\mathrm{SL}_n^+$ in the sense of [Borel--Tits] (see Remark 1.4.(i) of op. cit.). Moreover, $\mathrm{PGL}_m$ is connected and absolutely simple. Finally, $\varphi$ is not trivial since $m>1$.

So, by [Borel--Tits, Theorem 1.5] (note that they are missing the assumption that $\varphi$ is not trivial, cf. Theorem 1.2 of op. cit.) one may write $\varphi=\beta\circ\phi$ where $\phi$ is an automorphism of $k$ (as a ring) and $\beta$ is an isogeny $\mathrm{SL}_n\to\mathrm{PGL}_m$. But, this is a contradiction since there are no non-zero isogenies $\mathrm{SL}_n\to\mathrm{PGL}_m$ (since such morphisms are finite, and $\mathrm{dim}(\mathrm{SL}_n)>\mathrm{dim}(\mathrm{PGL}_m)$).

$(\ast)$ Regular semisimple elements are the analogue for general algebraic groups of 'having distinct eigenvalues' for the general linear group.

General comment on difficulty: Most of the 'content' above, beyond the Borel--Tits reference, is due to the annoying fact that $\mathrm{PSL}_n(k)\ne \mathrm{SL}_n(k)/\mu_n(k)$ for a general field $k$. In fact it's equal to the group $\mathrm{PGL}_n(k)=\mathrm{GL}_n(k)/\mathbb{G}_m(k)$. If one is willing to assume that $m$ is invertible in $k$, and that $k^\times=(k^\times)^m$ then basically you can ignore most of the above, and if you are happy to assume that $k$ is algebraically closed (and $m$ is invertible in $k$) then you can basically ignore everything.

General comment on scope: There are almost certainly more 'down-to-earth' ways to attack this literal problem, but it should be pointed out that Borel--Tits reference (and the same logic as above) should in fact show that there aren't surjections $\mathrm{GL}_n(k)\to\mathrm{GL}_m(k')$ where $k$ and $k'$ are any two (infinite?) fields.

References:

[Borel--Tits] Borel, A. and Tits, J., 2018. On “abstract” homomorphisms of simple algebraic groups. ALGEBRAIC GEOMETRY, p.78.

[Milne] Milne, J.S., 2017. Algebraic groups: the theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.

[Steinberg] Steinberg, R., 1965. Regular elements of semi-simple algebraic groups. Publications Mathématiques de l'IHÉS, 25, pp.49-80.

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.