Do we have SL$(n,2) \leqslant$ Sp$(2n, 2)$? And better yet, $\operatorname{SL}(n,q) \leqslant$ Sp$(2n, q)$ or $\operatorname{GL}(n,q) \leqslant$ Sp$(2n, q)$?
I checked using Magma for small degrees and they hold. I was thinking since the symplectic basis consists of pairs, that if we pick only one from each pair, then we should have a linear group sitting inside? Has this been studied before? Thank you.
Yes. Let $V$ be a finite-dimensional vector space over a field $\mathbb{F}$, equipped with an non-degenerate alternating bilinear form $b: V \times V \rightarrow \mathbb{F}$.
You have a decomposition $V = W \oplus W'$, where $W$ and $W'$ are maximal totally isotropic subspaces of $V$. The elements of $Sp(V)$ that leave $W$ and $W'$ invariant form a subgroup isomorphic to $GL(W)$.
In terms of matrices, you can take a basis $e_1$, $\ldots$, $e_n$ of $W$ and $f_1$, $\ldots$, $f_n$ of $W'$ such that $b(e_i,f_j) = 1$ if $i = j$ and $0$ otherwise.
Then $$\left\{ \begin{pmatrix} A & 0 \\ 0 & A^{-T} \end{pmatrix} : A \in GL_n(\mathbb{F}) \right\}$$ is a subgroup of $Sp_{2n}(\mathbb{F})$ and isomorphic to $GL_n(\mathbb{F})$.
EDIT: So basically as noted in a comment: it comes down to $W \oplus W^*$ admitting a natural $GL(W)$-invariant alternating bilinear form.