Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 2 / 6

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 isIt 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 $e_1\ne 0$. 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 $|S{p}_0(q)|=1$, yieldsThe construction also proves the existence of the required symplectic bases, in characteristic two as well as odd characteristic.

Represent a subset by the coordinate vector of its indicator function in ${\mathbb{F}}_2^n$. Symmetric difference then becomes coordinatewise addition, and the proposed pairing is the dot productIt is a symmetric bilinear form, and it is nondegenerate because its Gram matrix in the singleton basis is the identity. Coordinate permutations preserve the sum, so the natural $S_n$ action preserves the bilinear form.

Let $j=(1,\dots ,1)$, the vector representing the full set. Thenis the even-weight subspace. For even $n$, $B(j,j)=n=0$ in ${\mathbb{F}}_2$, so $j\in W$. Moreover $B(x,x)={\sum }_i{x}_i^2={\sum }_i{x}_i=0$ for $x\in W$, making the restricted bilinear form alternating. Nondegeneracy on the ambient space implies $W^{\perp }=⟨j⟩$, so the radical of $B{|}_W$ is exactlyThus it descends to an alternating bilinear form on $W/⟨j⟩$, independent of the representatives, and that quotient form has zero radical. The induced form is nondegenerate, and its dimension is $n-2$. This is the symplectic quotient of the binary subset module; at $n=2$ it is the zero-dimensional nondegenerate space.

Take $n=6$ in 6(b). The symplectic quotient of the binary subset module has dimension four, so the action gives a group homomorphismIt is injective. If $\sigma$ acts trivially on the quotient, every two-element subset $A$ has $\sigma A+A\in \{\varnothing ,\Omega \}$. Hence $\sigma A$ equals $A$ or its complement. The latter has size four, whereas $\sigma A$ has size two; therefore $\sigma$ fixes every two-element subset. Given $i$, choose distinct $j,k\ne i$. It then fixes the intersection $\{i,j\}\cap \{i,k\}=\{i\}$, so it fixes each point and is the identity.

Finally 6(a) givesAn injective group homomorphism between these finite groups of the same order is surjective. Thus