 Solution

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