Large(ish) finite non-abelian subgroups of $\operatorname{GL}_n \mathbb C$ for $n>70$

Question

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\SU{SU}\newcommand{\C}{\mathbb{C}}$My question is about large order finite non-abelian subgroups of $\GL_n\C$ without an abelian normal subgroup (or "small" abelian normal subgroup). I could look at $\SU_n$ as well since all finite subgroups of $\GL_n\C$ can be conjugated to an $\SU_n$ subgroup. For low $n$ it seems there isn't a regular pattern, for $\SU_2$ we have the binary icosahedral group, for $\SU_3$ there is the Valentiner group, for $\SU_4$ we have the result from Hanany and He (arXiv link) which gives a full classification. But Jordan's and Collins' well-known results say there might be something happening at $n=70$. I also understand that a full classification of all finite subgroups is hopeless for general $n$.

So let's limit ourselves to $n>70$, maybe we can find a pattern for these. Is it possible to give a finite non-abelian subgroup, without an (or "small") abelian normal subgroup, parametrized by $n$, such that for any $n$ it's a “large” subgroup? What I mean is that it doesn't have to have the largest order (if finding the largest order finite non-abelian subgroup is impossible), but reasonably large number of elements, for example exponential in $n$. Or if having no abelian normal subgroup is difficult, let's have as small an abelian normal subgroup as possible.

I know this is rather vague. But the reason I'm hoping something like this is possible is that groups $\PSL_kq$ do show up for various $n$ and maybe, perhaps for only even or only odd $n$, one can have explicitly $k$ and $q$ as a function of $n$ such that the order of $\PSL_kq$ is “large” and is a subgroup of $\GL_n\C$.

In other words, are there specific embeddings of $\PSL_kq$ into $\GL_n\C$ which work for all sufficiently large $n$?

For example the group of signed permutation matrices on $n$ elements has order $2^n n!$, so I'd think it should be possible to find other, possibly larger subgroups, with exponential order in $n$.

Is there a subgroup, for sufficiently large $n$, the order of which is exponential in $n$ and larger than $2^n n!$ ?

I went through the following, which partly inspired the question:

[1] The finite subgroups of SU(n)

[2] Finite simple groups and $ \operatorname{SU}_n $

[3] Closed Lie subgroups of $\mathrm{SU}(3)$

[4] Finite maximal closed subgroups of Lie groups

Answer 1

Your question is indeed somewhat vaguely formulated, but here is an attempt to state some known and fairly up to date facts. If $G$ is a finite subgroup of ${\rm GL}(n,\mathbb{C})$ and $n > 151$, then $G$ has an Abelian normal subgroup $A$ with $[G:A] \leq 60^{n-1}$ UNLESS $G$ has a composition factor $A_{m}$ (alternating group) with $m > 151.$

On the other hand, for large $n$, it is possible to find finite nilpotent subgroups $G$ of ${\rm GL}(n,\mathbb{C})$ with $|G| > c^{n-1}$ for some constant $c > 1$. Also, for any finite nilpotent subgroup $H$ of ${\rm GL}(n,\mathbb{C}),$ there is an Abelian normal subgroup $A$ of $H$ with $[H:A] \leq 2^{n-1}.$

For a finite simple group $G$ of Lie type, it is a (slight extension of a ) Theorem of E. Artin that $G$ has a nilpotent subgroup $U$ with $|G| < |U|^{3}.$

From these facts, it is possible to deduce (among other things) that there is a single fixed constant $h$ such that if $G$ is a finite simple group of Lie type which is a subgroup of ${\rm GL}(n,\mathbb{C})$ for some $n$, then $|G| \leq h^{n-1}$. Since there are only finitely many sporadic simple groups, we can enlarge the constant $h$ to cover sporadic simple groups.

This is basically the reason why we need large Alternating composition factors for $G$ if we are to be unable to bound $[G:A]$ (for $G$ a finite subgroup of ${\rm GL}(n,\mathbb{C})$ and $A$ an Abelian normal subgroup of $G$ of maximal order) by a bound of the form $[G:A] < d^{n}$ for a chosen constant $d$.