Suppose we have a quartic irreducible $x^4+ax^2+b\in\mathbb{Q}[x]$.
We wee that its roots are symmetric in origin: $\alpha\mapsto -\alpha$ is an automorphism of the roots, so it is a member of the Galois group of the polynomial.
Since this antipodal map commutes with every permutation of roots, i.e. it is an automorphism of order $2$ commuting with every automorphism of the roots in the Galois group, can we conclude that Galois group of the polynomial can not be $A_4$ or $S_4$?
Best Answer
We can construct the splitting field of $f=X^4+aX^2+b$ as $\mathbb{Q}(\xi,\sqrt{\xi},\sqrt{-a-\xi})$, where $\xi$ is a zero of $g=X^2+aX+b$. So it is of degree $4$ or $8$ (it can't be any less since $f$ is irreducible), which means that the Galois group is $C_4$, $V_4$, or $D_4$ (all these cases can occur), and in particular is not $A_4$ or $S_4$.