TITLE: Galois group of a degree 5 irreducible polynomial with two complex roots.
QUESTION [8 upvotes]: 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 subgroups of $F_{20}$ with the cycle of order three since $3 \nmid 20$ and $A_5$ since the transposition cannot be an even permutation).
Conjugation is a field automorphism that fixes $\mathbb{Q}$. It is in thus in the Galois Group and has cycle type $[2][1][1][1]$.
My question
How do I show that there is an element of order three in the Galois group?

REPLY [3 votes]: 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 $(cd)$.