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.

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


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