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...)


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: 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 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 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.

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  • 1
    $\begingroup$ (I deleted some comments after thinking things through.) I believe what you say in your second sentence is literally true if $n$ is a power of $p$ and generally not otherwise. (So it's true with $\mathbb{Z}$ replaced by $\mathbb{Z}_p$.) But the general idea looks very good. Thanks a lot, and I'll report back after I work it out. $\endgroup$ – Pete L. Clark Jul 27 '13 at 0:20

Here is my attempt at fleshing out Peter Mueller's answer. If anyone sees any inaccuracies (or worse) here, I would be grateful if you could let me know. If this answer seems correct to you, please be sure to upvote Peter's answer. $\newcommand{\GL}{\operatorname{GL}}$ $\newcommand{\ra}{\rightarrow}$ $\newcommand{\Z}{\mathbb{Z}}$

If $G$ is a group, then we say a group $S$ is a subquotient of $G$ if there are subgroups $N \subset H \subset G$ with $N$ normal in $H$ such that $H/N \cong S$.

Step 1: Let $G$ be a finite group. We claim that every simple subquotient of $G$ is a subquotient of some Jordan-Holder factor of $G$. By induction it suffices to show that if $N$ is a normal subgroup of $G$, then every simple subquotient of $G$ is a subquotient of either $N$ or $G/N$. Let $G'$ be a subgroup of $G$, let $N'$ be a normal subgroup of $G'$, and let $\pi: G \ra G/N$ be the quotient map. Then $G'$ is an extension of $\pi(G')$ by $G' \cap N$, so the Jordan-Holder factor $G'/N'$ of $G'$ must appear up to isomorphism as a Jordan-Holder factor of either $G' \cap N$ or of $\pi(G')$, and it follows that $G'/N'$ is a subquotient of either $N$ or $G/N$.

Step 2: Let $m \in \Z^+$. Let $n \in \Z^+$ have prime power factorization $n = p_1^{a_1} \cdots p_r^{a_r}$. Then $\GL_m(\Z/n\Z) \cong \prod_{i=1}^r \GL_m(\Z/p_i^{a_i} \Z)$. Further, for $1 \leq i \leq r$, the kernel of the natural map $\GL_m(\Z/p_i^{a_i} \Z) \ra \GL_m(\Z/p_i \Z)$ is a $p_i$-group. Thus every noncyclic simple subquotient of $\GL_m(\Z/n\Z)$ is a simple subquotient of $\GL_m(\Z/p_i \Z)$ for some $i$. Thus we assume henceforth that $n = p$ is prime.

Step 3: We apply the following theorem of Larsen-Pink: there is a positive integer $J'(m)$ such that every subgroup $\Gamma \subset \GL_m(\Z/p\Z)$ admits a subnormal series $\Gamma_3 \subset \Gamma_2 \subset \Gamma_1 \subset \Gamma$ such that:
$\bullet$ $[\Gamma:\Gamma_1] \leq J'(m)$;
$\bullet$ $\Gamma_1/\Gamma_2$ is a direct product of finite simple groups of Lie type;
$\bullet$ $\Gamma_2/\Gamma_3$ is commutative; and
$\bullet$ $\Gamma_3$ is a $p$-group.
In view of Step 1, the result follows immediately from this and the fact that for all $k \geq 9$, $A_k$ is not isomorphic to a finite simple group of Lie type.

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