This might be a silly question, but do we have the following?
$Sp(2,q) < Sp(4,q) < Sp(6,q) < Sp(8,q) <....$
I checked the list of maximal subgroups of $Sp(n,q)$ for $n = 4,6,8,10,12$ and it seems to suggest the relation holds . But I haven't found a generalised statement so far. Thank you!
Let $V$ be a vector space of dimension $n > 2$, equipped with an alternating bilinear form. Then $n$ is even, and you can split $V$ into an orthogonal direct sum $$V = W \perp W'$$ where $\dim W = n-2$ and $\dim W' = 2$.
This gives you an obvious embedding $i: Sp(W) \rightarrow Sp(V)$. That is for $f \in Sp(W)$ define $i(f) = f + I_{W'}$. So $i(f)$ acts like $f$ does on $W$, and trivially on $W'$.
This explains why you have $Sp_{n-2}(F) < Sp_n(F)$ for any field $F$ and even integer $n > 2$.
Similarly you can see that $Sp_m(F) \times Sp_{m'}(F) < Sp_{m+m'}(F)$ for $m,m' > 0$ even. You also have similar inclusions for the orthogonal groups, unitary groups, etc.
