---
tag: 4CMPPTSJNP
kind: statement
role: theorem
name: Galois group of a prime degree polynomial with two nonreal roots
msc: [12F10, 12D10]
uses: [00X2A9G9W3, 0JZY3R908H, 0RJMNEN3BJ, 0JM7PA288B, 02DTRYKGE8, 0BJV24C4SY, 0SN0MAEKYX]
about: [0JZY3R908H]
sources:
  - url: https://en.wikipedia.org/wiki/Galois_group
---

## Statement

Let $p$ be a [[prime number|0SN0MAEKYX]] and let $f \in \Q[x]$ be an [[irreducible|00X2A9G9W3]] polynomial of degree $p$.
Assume that $f$ has exactly two nonreal roots in $\C$.
Then the [[Galois group|0JZY3R908H]] of $f$ over $\Q$, acting on the $p$ roots of $f$, is the full [[symmetric group|0RJMNEN3BJ]] $S_p$.

## Remarks

Let $G$ be the Galois group, seen as a group of permutations of the $p$ roots.
Since $f$ is irreducible, adjoining one root gives a field of degree $p$, so $p$ divides $\abs{G}$.
By [[Cauchy's theorem|0JM7PA288B]], $G$ contains an element of order $p$, which is a $p$-cycle.
[[Complex conjugation|0BJV24C4SY]] restricts to an element of $G$ that swaps the two nonreal roots and fixes the others, so $G$ contains a [[transposition|02DTRYKGE8]].
For $p$ prime, a $p$-cycle and any transposition generate $S_p$.
Example: $x^5 - 4x + 2$ has Galois group $S_5$; see the [[quintic equation|S1WS18QEGJ]].
The prime degree matters: in degree $4$, the group generated by $(1\,2\,3\,4)$ and $(1\,3)$ is a dihedral group of order $8$, not $S_4$.

## Explanations

### L1 human

The theorem gives an easy test for when the symmetries of the roots of a polynomial include every permutation.
Count the real roots: if the degree is prime and exactly two roots are not real, the group is as big as possible.
This is the usual way to produce explicit equations that cannot be solved by radicals.

### L2 human

Combined with [[Galois' theorem on solvability|VHCWCFA3MX]] and the fact that [[$S_p$ is not solvable for $p \ge 5$|0J53NE674D]], the theorem proves the second part of the [[Abel–Ruffini theorem|163A2WQQJX]].
The number of real roots can be found with calculus or with [[Sturm's theorem|R44CV0KD9T]].
The proof uses only that $\Q \subseteq \R$; it works for any subfield of $\R$ in place of $\Q$.