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. 



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


