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

Question

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.

Answer 1

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.