Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 18H (Galois Theory)

The Klein four-group $V_4$ is normal in $S_4$, and $S_4/V_4\cong S_3$ through the action on the three partitions into pairs. A subgroup containing $V_4$ is the full preimage of its image in this quotient. Up to conjugacy the subgroups of $S_3$ have orders $1,2,3,6$. Their preimages give the complete listwith respective orders $4,8,12,24$, where $D_8$ denotes the dihedral group of order eight. The order-eight preimage is a Sylow two-subgroup of $S_4$, generated for example by a four-cycle and a reflection reversing it.

The three resolvent cubic roots correspond to the partitions $12|34$, $13|24$, $14|23$. Permuting the four roots of the quartic permutes these three quantities and defines a homomorphism $G\to S_3$. They are distinct: because the quartic has zero cubic coefficient, their differences includesince characteristic zero and irreducibility imply separability. Thus the kernel of the action is exactly $V_4$. The splitting field of the resolvent is the subfield generated by its roots; restriction from the quartic splitting field is surjective onto its Galois group, since automorphisms of a normal separable intermediate extension extend to the splitting field. HenceIn the four cases these groups are respectively the trivial group, $C_2$, $C_3$ and $S_3$. Computing the symmetric functions also gives $Q(X)=X^3+2b{X}^2+(b^2-4d)X-c^2$ with the source's root convention.

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.

Over $K=\mathbb{Q}$, takeIts roots are $\pm \sqrt{2}\pm \sqrt{3}$. For $\alpha =\sqrt{2}+\sqrt{3}$, its inverse is $\sqrt{3}-\sqrt{2}$, so both square roots belong to $\mathbb{Q}(\alpha )$. Since $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree four, $P$ is irreducible and this field is its splitting field. Independently changing the signs of the two square roots gives its Galois group $C_2\times C_2=V_4$, which is abelian and satisfies the hypothesis.