Let \(f\in\QQ[x]\) be an irreducible quartic polynomial with exactly two real roots. Show that the Galois group of its splitting field over \(\QQ\), viewed as a permutation group on the four roots, is isomorphic to \(S_4\) or \(D_4\).
Solution
Let \(G\) be the Galois group of the splitting field of \(f\) over \(\QQ\).
1. The action of \(G\) on the four roots is transitive.
Because \(f\) is irreducible over \(\QQ\), the Galois group acts transitively on its roots: any two roots are \(\QQ\)-conjugate and some \(\QQ\)-automorphism of the splitting field sends one to the other.
2. Up to conjugacy, the transitive subgroups of \(S_4\) are\[ C_4,\qquad V_4,\qquad D_4,\qquad A_4,\qquad S_4. \]
If \(H\le S_4\) is transitive, orbit-stabilizer implies \(4\mid |H|\). Hence \[ |H|\in\{4,8,12,24\}. \] If \(|H|=24\), then \(H=S_4\). If \(|H|=12\), then \(H\) has index \(2\), so \(H=A_4\). If \(|H|=8\), then \(H\) is a Sylow \(2\)-subgroup of \(S_4\), hence isomorphic to the dihedral group \(D_4\) of order \(8\). Finally, a transitive subgroup of order \(4\) is either cyclic \(C_4\) or the transitive Klein four group \(V_4\).
3. Complex conjugation acts on the roots as a single transposition.
The polynomial has exactly two real roots, which complex conjugation fixes, and one nonreal conjugate pair, which it swaps. Thus, in the permutation action on the four roots, complex conjugation has cycle type \[ (ab), \] a transposition.
4. The group \(G\) is not \(A_4\), \(C_4\), or \(V_4\).
The alternating group \(A_4\) contains no transpositions because they are odd permutations.
A transitive cyclic subgroup \(C_4\) is generated by a \(4\)-cycle; its unique element of order \(2\) is the square of that generator, which is a product of two disjoint transpositions, not a single transposition.
The transitive Klein four subgroup is \[ \{e,(12)(34),(13)(24),(14)(23)\}, \] which likewise contains no transposition.
By 3, \(G\) contains a transposition, so none of these three groups is possible.
5. Therefore\[ G\cong S_4\quad\text{or}\quad G\cong D_4. \]
By 2, the five listed groups exhaust the transitive possibilities. By 4, only \(S_4\) and \(D_4\) remain.