# Galois group of quintic polynomial with 4 complex solution

Suppose we have an irreducible quintic polynomial $f(x)\in \mathbb{Q}[x]$ with 4 complex solutions, say for e.g. $x^5+x^4+x^3-2x^2-2x+5$. It is easy to see that the Galois group $Gal(E/\mathbb{Q})$ has a $5$- cycle.

Now if we label the complex roots as $a_1, a'_1, a_2, a'_2$ where $a'_i$ is the complex conjugate of $a_i$ and look at the automorphisms, $f:a_1 \to a'_1$ (fixing the rest)and $g:a_2\to a'_2$ (fixing the rest). But this does not give any information whether the $Gal(E/\mathbb{Q})\subset S_n$ or $A_n$.

So, what can we do further to get more information about how the Galois group of the polynomial looks like?

• Why does it contain the two automorphisms $f$ and $g$ you describe? – Prahlad Vaidyanathan Mar 23 '15 at 2:25
• One way to check if $G \subseteq A_n$ is to compute the discriminant. This paper discusses the topic, see p. 5. – André 3000 Mar 23 '15 at 2:25
• @PrahladVaidyanathan: I am trying to permute the roots and try to get some information on the cycle structure – Rutherford Mark Mar 23 '15 at 4:02
• @RutherfordMark: What I mean is, how do you know you can conjugate one pair of roots and fix the rest? As I see it, a priori, you can only assume that conjugation induces an element of the Galois group. This would give you the element $(12)(34) \in S_5$ – Prahlad Vaidyanathan Mar 23 '15 at 10:11
• @PrahladVaidyanathan: That would mean that it has a $4$- cycle but it is not enough to conclude that the Galois group is $S_5$ – Rutherford Mark Mar 24 '15 at 2:15

Check all the transitive subgroups of $\mathfrak{S}_5$ you will have five groups (up to conjugation).
You have first the subgroup generated by one 5-cycle. Then its normalizer N in $\mathfrak{A}_5$, in $\mathfrak{S}_5$, $\mathfrak{A}_5$ and $\mathfrak{S}_5$.