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Let $q$ be a power of prime $p$, and $n, m, k$ positive integers such that $mk=2n$ and $2\leq m<2n$. Let $\mathrm{Sp}(2n,q)$ be the symplectic group of dimension $2n$ over $\mathrm{GF}(q)$ and $\mathrm{SL}(m,q^k)$ the special linear group of dimension $m$ over $\mathrm{GF}(q^k)$. For what values of $m,k$ does $\mathrm{Sp}(2n,q)$ contain a subgroup isomorphic to $\mathrm{SL}(m,q^k)$?

It is not difficult to show that $\mathrm{Sp}(2n,q)$ contains $\mathrm{SL}(2,q^n)=\mathrm{Sp}(2,q^n)$. If $\mathrm{Sp}(2n,q)$ contains $\mathrm{SL}(m,q^k)$ then the the greatest common divisor of $m$ and $p-1$ divides 2.

When $p$ is odd, consideration of representations of the lowest degree shows that $\mathrm{Sp}(2n,q)$ cannot contain $\mathrm{SL}(m,q^k)$ when $m\geq3$ (assuming $mk=2n$). Is there a simple group theoretic argument? What is the conclusion when $q=2$?

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    $\begingroup$ Try the book by Kleidman, or papers of Kleidman and Liebeck $\endgroup$ Aug 15, 2014 at 9:14
  • $\begingroup$ @GeoffRobinson, I think you mean the book by Kleidman & Liebeck and the papers by Kleidman(!) I have e-copies of these, so the OP can email if s/he wants a copy. For low-rank groups the new book by Bray, Holt and Roney-Dougal would be very useful. $\endgroup$
    – Nick Gill
    Aug 15, 2014 at 15:05

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The question is this:

When can $SL(m,q^k)$ be a subgroup of $Sp(2n,q)$ with $mk=2n$?

As you point out, this is possible if $m=2$. There are many particular cases that can be ruled out by order considerations - using Zsigmondy primes. However to give a complete answer, one should observe that if such an embedding exists, then the subgroup $SL(m,q^k)$ contains a Singer cycle of $Sp(2n,q)$, i.e. a maximal irreducible cyclic subgroup.

The maximal subgroups of $Sp(2n,q)$ that contain a Singer cycle were found by Bereczky:

A. Bereczky, Maximal Overgroups of Singer Elements in Classical Groups, Journal of Algebra, Volume 234, Issue 1, 1 December 2000, Pages 187–206

I don't have access to Bereczky's paper but (a version of) the statement can be found at this great blog.

As you will see, there are a bunch of exceptional situations and there are the field-extension subgroups. One can step recursively through the field-extension subgroups and one will either end with a subgroup $Sp(2,q^k)=SL(2,q^k)$ as you describe, or else you will end up in the exceptional cases which can be checked by hand. Thus I believe that you are correct in asserting that, in general, $m=2$ - although there might be a finite number of exceptions.

(In fact one of the exceptional cases is an infinite family - but it is just the $O^-$ type orthogonal groups inside $Sp$ and, since Bereczky's result applies to orthogonal groups too, recursion can be applied here too.)

Added, thanks to comment below of the OP: In fact the result can be proved much more easily: $SL(m,q^k)$ contains a Singer cycle of order $((q^k)^m-1)/(q^k-1)$. This Singer cycle must act irreducibly on the vector space associated with $Sp(2n,q)$ and so must lie in a Singer cycle of $Sp(2n,q)$. But a Singer cycle of $Sp(2n,q)$ has order $q^n+1$. We conclude that $m=2$ as required.

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  • $\begingroup$ Thank you very much for the detailed reply. I think $\mathrm{Sp}(2n,q)$ should contain a Singer cycle of $\mathrm{SL}(m,q^k)$ instead of the other way around. According to Table 1 of the paper of Bereczky as you suggested (you may email me if you want a copy of this paper), this is possible only when $m=2$. So $\mathrm{SL}(m,q^k)$ (assuming $mk=2n$ and $2\leq m<2n$) can be a subgroup of $\mathrm{Sp}(2n,q)$ if and only if $m=2$. $\endgroup$ Aug 15, 2014 at 17:29
  • $\begingroup$ Yes, that is a more direct way of proving the result. I had forgotten (if I ever knew) that the symplectic Singer cycle would have smaller order in this case than the Singer cycle for $SL$. $\endgroup$
    – Nick Gill
    Aug 16, 2014 at 1:42

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