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

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?

• Since the Galois group is transitive, it has order divisible by 5, and so contains an element of order 5. You can assume that the elements of order 2 and 5 are (1,2) and(1,2,3,4,5). Try and make an element of order 3 out of them! – Derek Holt Jun 3 '12 at 17:25
• @DerekHolt Since the Galois group is transitive, can we not say that the transpoistion $(13)\in G$ ? It switches between a complex root to a real root, so I'm not sure. However, if $(13)\in G$, then $(13(12345)(12)=(345)$. If we assume $(12345) \in G$ (by transitivy), then $(345)\in G$. However, why can't we go the other way around assuming that $(345)\in G$ (since it corresponds to a legitimate permutation of the real roots) and from that deduce $(12345) \in G$? Then we can use that $S_5$ is generated by $(12)$ and $(12345)$ and thus $(12),(12345) \in G \Rightarrow G \cong S_5$. – Zachi Evenor Jun 9 '12 at 14:32

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 $(cd)$.