Galois group of a degree 5 irreducible polynomial with two complex roots.

Question

Let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree five with exactly three real roots and let $K$ be the splitting field of $f$. Prove that Gal$(K/Q) \cong S_5$.

My attempt: The splitting field of an irreducible polynomial of degree $5$ has Galois group which is a transitive subgroup of $S_5$. Hence if we can show that the Galois group has elements of cycle type $[2][1][1][1]$ and $[3][1][1][1]$ then $S_5$ is the only such subgroup possible. (We are eliminating the possible groups and subgroups of $F_{20}$ and $A_5$.)

Conjugation is a field automorphism that fixes $\mathbb{Q}$. It is in thus in the Galois Group and has cycle type $[2][1][1][1]$.

How do I show that there is an element of order three in the Galois group?

Answer 1

Let the roots be $a,b,c,d,e$ with $a,b $ non-real. Your subgroup is transitive and has at least one transposition $(a b)$. Since it is transitive you can conjugate your transposition by some element to get a transposition which moves $c$ (conjugation preserves the cycle structure), and likewise one which moves $d$ or $e$.

Case 1. If this is $(bc)$ or $(ac)$ you can find your 3-cycle as a product of transpositions.

Case 2. If it is $(cd)$ then you can use conjugation to find a transposition which moves $e$ and whichever one you get, you can find your 3-cycle by composing it with either $(ab)$ or $(cd)$.