# Prove that the Galois group of $f\left(x\right)$ over $\mathbb{Q}$ is not a simple group.

Let $$f\left(x\right)\in\mathbb{Q}\left[x\right]$$ be an irreducible polynomial of degree $$n>2$$ which has $$n-2$$ real roots and exactly one pair of complex roots. Prove that the Galois group of $$f\left(x\right)$$ over $$\mathbb{Q}$$ is not a simple group.

• @peter The Galois group need not be $S_n$: consider $x^4-2$. – Lord Shark the Unknown Mar 16 at 20:07
• Now, I remember , I forgot a crucial detail. – Peter Mar 17 at 15:00

Consider the action of complex conjugation. This induces an element of the Galois group which is a transposition on the roots of the polynomial, so an odd permutation. The Galois group $$G$$ is a transitive subgroup of $$S_n$$, but has $$A_n\cap G$$ as a proper normal subgroup.