For a positive integer $m$, let $\mathcal{A}(m)$ be the set of all integers $k \geq 5$ such that: there is a positive integer $n$ and a subgroup $G \subset \operatorname{GL}_m(\mathbb{Z}/n\mathbb{Z})$ such that the alternating group $A_k$ is a composition factor of $G$ (i.e., writing $G$ as an iterated extension of finite simple groups, at least one of them is isomorphic to $A_k$).

Is $\mathcal{A}(m)$ is finite for all $m \in \mathbb{Z}^+$?

I have asked several questions here recently of the form "It would be great if $X$ were true. It seems unlikely, but I might as well ask." This time it would be *nice* if this were true, and I would be quite surprised if it were false.

(Some motivation: let $K$ be a field, and let $A_{/K}$ be an abelian variety. Suppose we want to build nontorsion points on $A(\overline{K})$. This cannot be done in general -- certainly it cannot be done when $K$ is algebraic over a finite field. But suppose that $K$ is a Hilbertian field, so for all $k \in \mathbb{Z}^+$ there is an $A_k$-Galois extension $L_k/K$. Then an affirmative answer to the above shows that $K(A(\overline{K})[\operatorname{tors}])$ contains only finitely many of the fields $L_k/K$. There will be points in $A(\overline{K})$ whose field of definition contains $L_k$, and these points cannot be torsion points. In fact if this is true then one can deduce the precise structure of $A(\overline{K})$ for any field $K$.)

There was a previous question on this site that was less ambitious but still interesting and relevant: it asked for examples of groups which cannot be subgroups of $\operatorname{GL}_2(\mathbb{Z}/p\mathbb{Z})$ for a prime $p$. The accepted answer used the structure of Sylow $p$-subgroups, which was something I hadn't thought of. It is plausible to me that one might be able to answer the question by showing that for all $m$, if $k$ is large enough then for all $n$, the $2$-Sylow subgroups ($2$ here is playing the role of a prime which is much smaller than $k$) of $\operatorname{GL}_m(\mathbb{Z}/n\mathbb{Z})$ have nilpotency class smaller than the nilpotency class of the $2$-Sylow subgroups of $A_k$. This would be sufficient, I believe.

If the answer turns out to be false, then I would be interested in hearing if you can make it true by replacing $A_k$ by a different infinite set of finite simple groups, especially a set each element of which is known to occur as a Galois group over every Hilbertian field.

**Added**: Finally an affirmative answer: hooray. What I really didn't know was the magnificent Larsen-Pink Theorem. As Peter Mueller says, that really makes things easy. Still, it makes it look like the result requires 21st century technology, and I strongly suspect that that is not the case. Although the proof I wrote below is the one I'll use in my paper, I am still interested to see other, more elementary proofs. (And I wonder if the idea of using nilpotency classes actually works...)