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Let $V$ have dimension $2m$ over ${\mathbb{F}}_q$, with a nondegenerate alternating bilinear form $B$. The symplectic group isIt acts freely and transitively on ordered symplectic bases. Choose the first vector $e_1\ne 0$ in $q^{2m}-1$ ways, then choose $f_1$ with $B(e_1,f_1)=1$ in $q^{2m-1}$ ways. Their span is nondegenerate, so choose the remaining symplectic basis in its orthogonal complement. This gives the recurrenceand therefore

For the permutation module ${\mathbb{F}}_2^r$ with $r$ even, put $\mathbf{t}=(1,\dots ,1)$ andIt is invariant under coordinate permutations and has codimension one. The dot product restricted to $W$ is alternating because $B(x,x)={\sum }_i{x}_i^2={\sum }_i{x}_i=0$. Since the ambient dot product is nondegenerate, $W^{\perp }=⟨\mathbf{t}⟩$. Evenness of $r$ ensures $\mathbf{t}\in W$, soHere the word symplectic must be understood as an alternating form before quotienting: a form with a nonzero radical is degenerate. The nondegenerate symplectic form is induced on $\overline{W}=W/⟨\mathbf{t}⟩$, of dimension $r-2$.

Now put $r=2m+2$. Coordinate permutations preserve the induced form and give a homomorphism $S_r\to S{p}_{r-2}(2)$. To see it is faithful for $r\ge 6$, suppose a permutation $\sigma$ acts trivially on $\overline{W}$. For every pair of distinct indices,The two weight-two vectors could differ by $\mathbf{t}$ only if the complement of a two-element set also had size two. For $r\ge 6$ that is impossible. Thus $\sigma$ preserves every two-element set, and hence fixes every index: intersect two such sets with one common element. ThereforeFor $m=2$, both groups have order $720$, and the injection is onto. This provesThe lower bound matters: for $r=4$ the action on the quotient has the Klein four-group as group homomorphism kernel. The construction is the deleted permutation module in characteristic two.