Let \(K\) be a subfield of \(\mathbb{R}\) and let \(f \in K[x]\) be an irreducible quartic. If \(f\) has exactly 2 real roots, show that the Galois group of \(f\) is either \(S_{4}\) or \(D_{4}\).
Solution
Let \(F\subseteq\CC\) be the splitting field of \(f\) over \(K\), and let \[ G=\operatorname{Gal}(F/K). \] Label the four roots of \(f\) by \(\alpha_1,\alpha_2,\alpha_3,\alpha_4\).
1. The natural action of \(G\) on the four roots is faithful and transitive.
An automorphism of the splitting field is determined by its action on the roots, since those roots generate \(F\) over \(K\). Hence the action is faithful.
Because \(f\) is irreducible over \(K\), the Galois group of its splitting field acts transitively on its roots: for any two roots, the corresponding \(K\)-embeddings of their simple extensions extend to automorphisms of the splitting field. Thus \(G\) is a transitive subgroup of \(S_4\).
2. The group \(G\) contains a transposition.
Since \(K\subseteq\RR\), complex conjugation fixes \(K\). It preserves the splitting field \(F\), because it permutes the roots of the polynomial \(f\in K[x]\). Therefore complex conjugation restricts to an element of \(G\).
Exactly two roots of \(f\) are real, so conjugation fixes those two roots. The two nonreal roots form one conjugate pair and are interchanged. Thus, as a permutation of the four roots, complex conjugation is a single transposition.
3. A transitive subgroup of \(S_4\) is isomorphic to one of\[ C_4,\qquad V_4,\qquad D_4,\qquad A_4,\qquad S_4. \]
Let \(H\le S_4\) be transitive. Orbit-stabilizer shows that \(4\mid|H|\), so by Lagrange \[ |H|\in\{4,8,12,24\}. \]
If \(|H|=24\), then \(H=S_4\). If \(|H|=12\), it has index \(2\) in \(S_4\), hence is \(A_4\). If \(|H|=8\), it is a Sylow \(2\)-subgroup of \(S_4\), hence isomorphic to the dihedral group \(D_4\) of order \(8\).
If \(|H|=4\), then \(H\) is either cyclic or Klein four. Transitivity rules out the Klein subgroup containing ordinary transpositions and leaves, up to conjugacy, either the regular cyclic subgroup generated by a \(4\)-cycle or the regular Klein four subgroup \[ \{1,(12)(34),(13)(24),(14)(23)\}. \] Thus the abstract possibilities are \(C_4\) and \(V_4\).
4. Among the five groups in 3, only \(D_4\) and \(S_4\) contain atransposition in their transitive degree-\(4\) actions.
The cyclic group generated by a \(4\)-cycle has elements of cycle types \[ 1,\quad (4),\quad (2,2),\quad (4), \] so it contains no transposition. The regular Klein four group consists of the identity and three double transpositions, so it contains none either. The group \(A_4\) contains only even permutations and hence no transposition.
The dihedral subgroup \(D_4\) contains reflections across diagonals of the square; in the action on its four vertices these are transpositions. The group \(S_4\) contains every transposition.
5. Therefore\[ G\cong D_4\qquad\text{or}\qquad G\cong S_4. \]
By 1, \(G\) is transitive; by 2, it contains a transposition. Apply 3 and 4.