Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 3 / 1D / iii / Solution

Let $N$ be a normal subgroup and let $\pi :G\to G/N$ be the quotient group homomorphism. By the first part, $G=⟨a,b⟩$ for involutions $a,b$, so the quotient group is generated by $\pi (a),\pi (b)$, whose squares are the identity. If both images are nonidentity, they are involutions and the second part applies. If only one image is nonidentity, it generates $C_2$; choose it twice. Both images cannot be trivial because the quotient group is nontrivial. Hence every nontrivial quotient is dihedral in the stated extended convention. This is the nontrivial-quotient closure of two-involution groups.