Galois group of quartic with only complex roots Suppose we have a separable, irreducible quartic that has 4 complex roots, and so it has 2 complex conjugate pairs. Does this then mean that the Galois group of this polynomial over the rationals cannot contain a 3-cycle, since this would not preserve the property that each of these roots have a complex conjugate pair?
I am confused about what properties are preserved by the galois group.
 A: Let us view the Galois group $G$ as a group of permutations on the four roots.
No, the fact that the roots form two complex conjugate pairs does NOT let us conclude that the group has no elements of order three (I will produce an example if necessary). All it implies is that the group $G$ contains a product of two disjoint 2-cycles.
What is confusing you is probably the fact that the complex conjugation does not necessarily belong to the center of $G$. In other words, there may be automorphisms $\tau$ that don't commute with complex conjugation. For such an automorphism $\overline{\tau(z)}$ may be different from $\tau(\overline{z}).$
A: The polynomial $x^4+8x+12$ has Galois Group $A_4$, and so has elements of order 3. A root of the polynomial is $\sqrt(-2*cos(2*\pi/9))+\sqrt(-2*cos(8*\pi/9))+\sqrt(-2*cos(14*\pi/9))$ . The cosine terms (without the minus signs) are roots of $x^3-3x+1$ , which has Galois group $Z_3$, so a cubic extension is required and in addition two quadratic extensions (not three, since one of the square roots depends on the other two) are needed to bring the root (and its conjugates) into the field. That's an extension of order 12. Since $S_4$ has only one group of order 12, this group, $A_4$, has to be the Galois group of the equation. All four roots are complex.
