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 $q=2$?



Special linear groups contained in symplectic groups
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](http://symomega.wordpress.com/2010/07/17/irreducible-cyclic-groups-of-classical-groups/). 


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.


