 Solution

Let $q$ be a prime power and let $V={\mathbb{F}}_q^{2m}$ carry a nondegenerate alternating bilinear form $B$. The symplectic group over a finite field isIn a symplectic basis its matrix description is $g^{\mathsf{T}}Jg=J$, where $J=\left(\begin{array}{cc}0& I_m\\ -I_m& 0\end{array}\right)$. The definition is valid in characteristic two: alternating means $B(v,v)=0$, and then the form is also symmetric.

Count ordered symplectic bases $(e_1,f_1,\dots ,e_m,f_m)$ with $B(e_i,f_j)={\delta }_{ij}$ and all $e$-$e$ and $f$-$f$ pairings zero. There are $q^{2m}-1$ choices for $e_1\ne 0$. Nondegeneracy makes $v\mapsto B(e_1,v)$ a nonzero linear functional, so there are $q^{2m-1}$ choices for $f_1$ satisfying $B(e_1,f_1)=1$.

The span $P=⟨e_1,f_1⟩$ is nondegenerate. Thus $V=P\oplus P^{\perp }$, and its bilinear orthogonal complement is nondegenerate of dimension $2m-2$. For completeness, a vector in the radical of a bilinear form of $P^{\perp }$ is orthogonal both to $P$ and to $P^{\perp }$, hence to all $V$ and therefore zero. Recursing constructs and counts every symplectic basis. If $b_m$ is their number, $b_m=(q^{2m}-1)q^{2m-1}{b}_{m-1}$ with $b_0=1$.

The symplectic group over a finite field acts freely and transitively on symplectic bases: there is a unique linear map sending one ordered basis to another, and its equality of basis pairings ensures it preserves $B$. Its order is consequently $b_m$, givingThe exponent is $1+3+\cdots +(2m-1)=m^2$. The counting is valid for every prime power $q$, including $q=2$.