When is the special linear group $SL_n(\Bbb F_q)$ over a finite field $\Bbb F_q$ solvable?

Question

Recall that the special linear group

$$SL_n(\Bbb F_q)=\{ A\in GL_n(\Bbb F_q)\mid \det(A)=1\},$$

where $\Bbb F_q$ is the field of $q$ elements for finite $q$.

A group $G$ is solvable if the derived series, defined by $G^{(0)}=G$ and $G^{(n)}=(G^{(n-1)})'$ (that is, the derived subgroup of $G^{(n-1)}$), terminates in the trivial subgroup of $G$.

The Question:

When is $SL_n(\Bbb F_q)$ solvable?

Context:

There's an idea I'm excited about that I cannot share (for scoop reasons) that is made easier to handle if I knew when $SL_n(\Bbb F_q)$ is solvable.

Thoughts:

I have the following data:

• $SL_2(\Bbb F_2)$ is solvable
• $SL_2(\Bbb F_3)$ is solvable
• $SL_2(\Bbb F_4)$ is not solvable
• $SL_2(\Bbb F_5)$ is not solvable
• $SL_3(\Bbb F_2)$ is not solvable
• $SL_2(\Bbb F_7)$ is not solvable

Further Context:

For two questions of mine on solvable groups, see:

1. There exist infinite solvable $p$-groups with trivial centre. (Use a hint.)
2. Two exercises by Robinson on supersolvable groups seem to contradict.

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Please help :)

Answer 1

As Jules Besson said in his answer, these groups are perfect, and hence non-solvable for all $n>1$ except for the two cases $(n,q)=(2,2),(2,3)$. The central quotients ${\rm PSL}(n,q)$ are simple. In fact this is all true for ${\rm SL}(n,K)$ and ${\rm PSL}(n,K)$ for all fields $K$, not just for finite fields.

The proof is perhaps a bit too long to write out in full here. The main steps in proving that ${\rm SL}(n,K)$ is perfect (except in the two exceptional cases) are to show that this group is generated by transvections, and that all transvections are conjugate. Then you just have to show that one transvection is contained in the commutator subgroup, which you can do by direct calculation. The proof of the simplicity of ${\rm PSL}(n,K)$, involves an application of Iwasawa's Lemma.

A convenient reference for the proofs is Chapter 8 of J. J. Rotman's textbook "An Introduction to the Theory of Groups".

Answer 2

So I only write an answer so it has a better layout than in a comment. In Daniel Perrin Cours d'algèbre, on the page 101, there is a theorem (3.1) with its proof:

Theorem:

• We have $$D(GL_n(k)) = SL_n(k)$$ except for one case: $n=2$ and $k = \mathbb{F}_2$.
• We have $$D(SL_n(k)) = SL_n(k)$$ except for two cases: $(n= 2, \, k = \mathbb{F}_2)$ and $(n= 2, \, k = \mathbb{F}_3)$.

Here $D(G)$ is the derived group (different notation).

The trick is to show that any transvection is a commutator, and for that you only need to show that one transvection is a commutator (check a result that states: the class of conjugation of a transvection contains all of them).