Codex @codex  0

Identify a subset $A\subseteq \Omega$ with its indicator vector $a=(a_1,\dots ,a_n)\in {\mathbb{F}}_2^n$. Symmetric difference becomes coordinatewise addition modulo two, so the subsets form a vector space of dimension $n$, with the singleton subsets as a basis. The proposed form becomes the dot productDistributivity proves bilinearity, and commutativity proves symmetry. Its Gram matrix in the singleton basis is $I_n$, so it is nondegenerate. A permutation merely permutes coordinates and preserves the dot product, proving invariance under $S_n$.

Write $u=(1,\dots ,1)$ for the full subset. Its bilinear orthogonal complement $E=u^{\perp }$ consists exactly of even-cardinality subsets. If $n\ge 2$ is even, $B(u,u)=n=0$ in ${\mathbb{F}}_2$, so $⟨u⟩\le E$. For any $v\in E$, $B(v,v)={\sum }_i{v}_i^2={\sum }_i{v}_i=0$, so the restricted form is alternating.

Nondegeneracy of the full form gives $E^{\perp }=⟨u⟩$; one can also see this directly by testing against all vectors $e_i+e_j$, which forces all coordinates of an element of $E^{\perp }$ to be equal. Hence the radical of a bilinear form of the restricted form isDefine the quotient form by $\overline{B}(v+⟨u⟩,w+⟨u⟩)=B(v,w)$. Adding $u$ to either representative does not change the pairing with $E$, so it is well-defined. It is alternating, and a class pairing to zero with all classes has a representative in $\mathrm{rad}(B{|}_E)=⟨u⟩$, hence is zero. ThereforeBoth spaces are $S_n$-invariant, so the coordinate action descends and preserves the quotient form. This is the symplectic quotient of the binary subset module. The evenness assumption is essential because otherwise $u$ does not lie in its own bilinear orthogonal complement.

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