6
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ In other words, you are asking if $Sp(2n,2)$ is maximal in $SL(2n,2)$? $\endgroup$
    – verret
    Jun 11, 2018 at 1:05
  • $\begingroup$ @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 $n=2$)? $\endgroup$
    – BlueLemon
    Jun 11, 2018 at 6:01

1 Answer 1

Reset to default
8
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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.) $\endgroup$
    – Noam D. Elkies
    Jun 13, 2018 at 0:53
  • $\begingroup$ 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 $n > 1$). $\endgroup$
    – Derek Holt
    Jun 13, 2018 at 7:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.