= Solution
Let $V$ have dimension $2m$ over the <finite field> $\mathbb F_q$, with a <nondegenerate> <alternating bilinear form> $B$. The <symplectic group over a finite field> is
$$
Sp(V,B)=\{g\in GL(V):B(gx,gy)=B(x,y)\text{ for all }x,y\}.
$$
It acts simply transitively on ordered <symplectic bases>: a <linear map> between two such bases is uniquely specified and preserves all pairings, hence the whole <bilinear form>.
Count a first <hyperbolic pair> $(e_1,f_1)$. There are $q^{2m}-1$ choices for $e_1\ne0$. Nondegeneracy makes $B(e_1,-)$ a nonzero <linear functional>, so there are $q^{2m-1}$ choices of $f_1$ with $B(e_1,f_1)=1$. Their span is <nondegenerate>, and its <symplectic orthogonal complement> is a <symplectic vector space> of dimension $2m-2$. Recursing, with $|Sp_0(q)|=1$, yields
$$
\boxed{|Sp_{2m}(q)|=\prod_{i=1}^m(q^{2i}-1)q^{2i-1}
=q^{m^2}\prod_{i=1}^m(q^{2i}-1).}
$$
The construction also proves the existence of the required <symplectic bases>, in <characteristic> two as well as odd <characteristic>.
Back to article page
