Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 18H / iii / Solution

Factorization over $K$ gives the orbits of the Galois group on the three resolvent roots. If $Q$ splits completely, the quotient is trivial and $G=V_4$. If $Q$ is a linear factor times an irreducible quadratic, the quotient is $C_2$ and $G=D_8$.

If $Q$ is irreducible, the quotient acts transitively on three roots, so it is $C_3$ or $S_3$. The product of their pairwise differences is a square root of the nonzero discriminant $\Delta$ and changes by the sign of the permutation. Thus $\Delta$ is a square in $K$ exactly when the quotient contains only even permutations. The criterion isThe assumed containment of $V_4$ is essential to reconstruct $G$ from the quotient by this table.