# How much do we need to add to the generating set of the symplectic group to get $SL(2n,2)$?

Here $Sp(2n,\mathbb{F}_2)$ means the group of matrices preserving the form $\Omega = \left( \begin{array}{cc} 0&I \\ -I&0& \end{array} \right)$, i.e. the symplectic group over an even characteristic ($-1=1$). It is well known that, $Sp(2n,\mathbb{F}_2) \leq SL(2n,\mathbb{F}_2)$. Additionally both groups are generated by transvections, but for $Sp(2n,\mathbb{F}_2)$ the transvections need to preserve the form $\Omega$.

Say $S$ is a generating set such that $Sp(2n,\mathbb{F}_2) = \langle S \rangle$, and $X \in SL(2n,\mathbb{F}_2)$ but$X \notin Sp(2n,\mathbb{F}_2)$, then does $\langle S, X \rangle = SL(2n, \mathbb{F}_2)$ hold?

It's clear that $SL(2,\mathbb{F}_2) = Sp(2,\mathbb{F}_2)$, yet I'm not sure about the general case. It's tedious, but possible to show it for $n=2$ with elementary methods, but the induction seems unclear. I am thankful for suggestions.

• In other words, you are asking if $Sp(2n,2)$ is maximal in $SL(2n,2)$? – verret Jun 11 '18 at 1:05
• @verret Almost, if $Sp(2n,2)$ is maximal in $SL(2n,2)$, that is a sufficient answer. But if not, then I'm still interested if we can say something about $\langle S,X\rangle$ and $X$, especially if we can restrict our choice of $X$ such that $⟨S,X⟩=SL(2n,2)$ still holds, for example is it enough to restrict $X$ to operations on $2\times2$ bits (case $n=2$)? – BlueLemon Jun 11 '18 at 6:01

For general $q$, it's not quite maximal. For $n>1$, the maximal $M$ subgroup of $G={\rm SL}(2n,q)$ containing $H={\rm Sp}(2n,q)$ contains $H$ as a subgroup of index $\gcd(q-1,n)$. So, as Noam D. Elkies points out in a comment, in the case $q=2$ it is indeed maximal for all $n>1$.
We have $M = \langle H, X \rangle$, where $X$ contains firstly scalar matrices in ${\rm SL}(2n,q)$, and secondly, when $n$ is even and $q$ is odd, there is an element in the conformal group (i.e. maps the fixed form to a scalar multiple of itself) that induces an outer automorphism of $H$ of order $2$. (When $n$ is odd there is no such conformal element with determinant $1$.)
You can find the details in the book "The subgroup structure of the finite classical groups" by P. Kleidman and M. Liebeck. The structure of $M$, which is the normalizer of $H$ in $G$ is described in Proposition 4.8.3, page 166 (although to understand that you need to know a lot of notation), and the maximality of $M$ in $G$ is done in Section 7.8, page 245.
• In the OP's case, $q=2$, so $\gcd(q-1,n) = \gcd(1,n) = 1$ and $H$ is maximal for all $n$. (Curiously for $n=2$, when $H \cong S_6$, the index-$2$ subgroup $A_6$ is contained in another proper subgroup of $G$, because in this case $G \cong A_8 > A_7 > A_6$. But $S_6 < A_8$ is still maximal.) – Noam D. Elkies Jun 13 '18 at 0:53
• Yes that's true! For some reason I had thought that the question was about ${\rm Sp}(2n,q) \le {\rm SL}(2n,q)$ for all $q$. The case ${\rm Sp}(4,2)$ is the only example which is not perfect (for $n > 1$). – Derek Holt Jun 13 '18 at 7:49