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 $A\_k$). 



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> Is $\mathcal{A}(m)$ is finite for all $m \in \mathbb{Z}^+$?
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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](https://mathoverflow.net/questions/91649/subgroups-of-gl2-q) 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...)



Restricting the composition factors of subgroups of GL_m(Z/nZ)
**Latest Edit:** Below I put a sketch on how to replace the deep Larsen-Pink Theorem by easier and more familiar results. 


**Edit:** There was a silly mistake in showing that we may assume that $n$ is a prime. Still, it is true, see Pete L. Clark's fix in his answer, which also enhances my rather sketchy answer. Here is the rest of my original answer (with $p=n$):


Next, by passing to a smaller group in the composition series of $G$, you may assume that $A\_k$ is a homomorphic image of $G$. However, that bounds $k$ in terms of $m$. There should be an easy argument. A sledgehammer argument uses a positive characteristic analogue of Jordan's upper bounds on linear groups by [Larsen and Pink](http://www.ams.org/journals/jams/2011-24-04/S0894-0347-2011-00695-4/home.html): There is a constant $j$ depending only on $m$, not on $p$, such that $G$ has normal subgroups $M\le N\le G$ where $[G:N]\le j$, $N/M$ is a direct product of simple Lie type groups of characteristic $p$, and $N$ is solvable.


**Avoiding Larsen-Pink:** Let $G/N$ be isomorphic to $A\_k$, and suppose that $G\le GL\_m(p)$. We need to show that $k$ is bounded in terms of $m$. Given $k,m$ and $p$, suppose that $\lvert G\rvert$ is minimal. Then each maximal subgroup $H$ of $G$ is contained in $N$, for otherwise $G=HN$, hence $H/(H\cap N)=A\_k$, and we could replace $G$ by $H$. Thus $N$ is contained in the intersection $\Phi(G)$ of all maximal subgroups of $G$. Note that $\Phi(G)$ is called the [Frattini subgroup](http://en.wikipedia.org/wiki/Frattini_subgroup) of $G$, and it is well known (and easy to prove) that $\Phi(G)$ is nilpotent. In particular $N$ is nilpotent. (Actually $N=\Phi(G)$.) Let $s$ be a prime divisor of $\lvert N\rvert$, and $S$ be a Sylow $s$-subgroup of $N$. Note that all Sylow subgroups of $N$ are normal in $N$, and so they are normal in $G$ too. If $s$ is bigger than $k$, then $S$ is a normal Sylow subgroup of $G$. But then $G=S\rtimes H$ for a subgroup $H$ of $G$ by Schur-Zassenhaus, contrary to $HN\lt G$. So each prime divisor of $\lvert G\rvert$ is $\le k$. On the other hand, as $G\le GL\_m(p)$, we have $k!\le p^{m^2}$. For fixed $m$, and $k$ big enough, this forces $p>k$, so $p$ does not divide the order of $G$. A standard result shows that $G$ can be lifted to characteristic $0$, that means there is a field $K$ of characteristic $0$ with $G\le GL\_m(K)$. See e.g. [Corollary 3.8](http://books.google.de/books?hl=de&id=Nv3uAAAAMAAJ&q=corollary%203.8#search_anchor) in Dixon's *The structure of linear groups*. Finally we can apply the old theorem of Jordan: There is a function $J$ on the positive integers such that if $G\le GL\_m(K)$, then $G$ has an abelian normal subgroup $A$ with $\lvert G/A\rvert\le J(m)$, and the result follows.


