An extension corresponding to a subgroup of Galois group

Question

Let $G$ be the Galois group of $f(x)=x^6-2x^4+2x^2-2$ over $\mathbb{Q}$. Describe an extension corresponding to any of it's proper subgroups of maximal order (i.e. find generators of this extension).

I know that $G\simeq C_2\wr S_3\simeq S_4\times S_2$. Hence, $S_4$ and $S_3\times S_2$ are it's maximal subgroups.

I also know that Galois group of $g(x)=x^3-2x^2+2x-2$ is isomorphic to $S_3$ (notice that $f(x)=g(x^2)$) and I know the generators of it's splitting field (one of the roots and the discriminant).

How can I find generators of such an extension?

Let $x_1, x_2, x_3$ be the roots of $g(x)$. Then the splitting field of $f(x)$ is $\mathbb{Q}(\sqrt x_1,\sqrt x_2,\sqrt x_3)$. Is it enough to replace one of $\sqrt x_i$ with $x_i$?

Answer 1

I used a "visualization" of this Galois group to find these fixed field. That is, the minimal intermediate fields between $\Bbb{Q}$ and the splitting field $L=\Bbb{Q}(\sqrt{x_1},\sqrt{x_2},\sqrt{x_3})$. Namely, $G=C_2\wr S_3$ is the group of all signed permutations of three coordinates. Or the group of monomial $3\times3$ matrices with a single $\pm1$ on each row/column together with six zeros. Or, the group of symmetries of a cube. To make the linear transfromations related to the monomial matrices symmetries of the cube we can place the vertices of the cube at the points $(\pm1,\pm1,\pm1)$, all the eight sign combinations occur.

I find the last one particularly useful because it allows us to describe the maximal subgroups as permutation of the six roots.

After all, the six roots come in pairs, $\pm\sqrt{x_i},i=1,2,3,$ just like the three pairs of opposite faces of a cube. So, if we label the six faces of a cube with the roots in such a way that opposite faces get labelled by roots with oppposite signs, then $G$ becomes exactly the group of symmetries of the cube.

As the OP pointed out we can also view $G$ as a direct product $S_4\times C_2$. As symmetries of the cube the factor $C_2=Z(G)$ is generated by the symmetry $\tau:(x,y,z)\mapsto (-x,-y,-z)$ that obviously commutes with the other linear transformations because its matrix is $-I_3$. Because $\det (-I_3)=-1$ we see that $G$ is the direct product of $C_2=\langle \tau\rangle$ and the subgroup $H$ consisting of the monomial matrices with determinant $+1$ (= the subgroup of orientation preserving symmetries of the cube). The group $G$ acts on the set of four 3-dimensional diagonals of the cube, and it is easy to see that $\tau$ is the only non-trivial symmetry that maps each and every one of those diagonals to itself. Therefore $H\simeq S_4$.

Next we proceed to find the index two subgroups of $G$. They are all normal, so they are the kernels of surjective homomorphisms $f:G\to \{\pm1\}$. To classify such homomorphisms we observe that $G$ is also a Coxeter group generated by the three reflections: $$ \begin{aligned} s_1&:(x,y,z)\mapsto (y,x,z),\\ s_2&:(x,y,z)\mapsto (x,z,y),\\ s_3&:(x,y,z)\mapsto (x,y,-z), \end{aligned} $$ We see that $s_1s_2$ is of order three, so we must choose $f(s_1)=f(s_2)$, but $s_1$ and $s_3$ commute, and $s_2s_3$ has even order, so we can choose $f(s_3)$ independently from $f(s_1)$. The Coxeter relations imply that there are no other constraints to constructing $f$. Therefore we get three different homomorphisms, $f_1,f_2,f_3$, and consequently three different maximal subgroups $H_i=\operatorname{ker}(f_i)$ of index two.

1. The choices $f_1(s_1)=f(s_2)=-1=f(s_3)$ mean that $f_1$ simply calculates the determinant. Hence its kernel $H_1=H$.
2. The choices $f_2(s_1)=1=f_2(s_2), f_2(s_3)=-1$ describe the group $H_2=\operatorname{ker}(f_2)$ consisting of permutations of the three coordinates with an even number of sign changes.
3. The choices $f_3(s_1)=f_3(s_2)=-1, f_3(s_3)=1$ yield $H_3=\operatorname{ker}(f_2)$, the group of such signed permutations that the permutation is even, in other words $C_2\wr A_3$. In addition to these index two subgroups we have:
4. Three conjugate subgroups $K_1,K_2,K_3$ that are stabilizers of a pair of opposite faces of the cube. They are all of index three in $G$ (by the orbit-stabilizer theorem). They obviously all contain $-I_3$, so we can also think of them as the groups $D_4\times C_2$, where $D_4$ is one of the Sylow $2$-subgroups of $S_4$, each isomorphic to the group of symmetries of a square.
5. Four conjugate subgroups $M_1,M_2,M_3,M_4$ of index four. These are the stabilizers of one of the diagonals. By the earlier discussion they are all isomorphic to $S_3\times C_2$, where $S_3$ is one of the point stabilizers of $S_4$ (and hence a maximal subgroup of $S_4$).

I believe these are all the maximal subgroups of $G$, but I am also prepared to have missed something.

Let's try and identify the corresponding fixed fields.

1. Describing $\operatorname{Inv}(H_1)$ is a bit awkward at this point. A generator can be selected as the product of the two square roots of integers from the next two items (that is $\sqrt{x_1x_2x_3\Delta}$. Anyway, the fixed field has to be $\Bbb{Q}(\sqrt{-22})$.
2. Any signed permutation of $\{\sqrt{x_1},\sqrt{x_2},\sqrt{x_3}\}$ with an even number of sign changes fixes $$\sqrt{x_1x_2x_3}=\sqrt2,$$ so $\operatorname{Inv}(H_2)=\Bbb{Q}(\sqrt2)$.
3. Any even permutation of $\{x_1,x_2,x_3\}$ fixes $\sqrt{\Delta}$, where $$\Delta=(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2=-44$$ is the discriminant of cubic polynomial $g(x)$. Therefore $\operatorname{Inv}(H_3)=\Bbb{Q}(\sqrt\Delta)=\Bbb{Q}(\sqrt{-11})$.
4. The subgroup $K_i$ fixes the pair $\pm\sqrt{x_i}$ of "opposite faces". Therefore obviously $x_i\in\operatorname{Inv}(K_i)$. As $g(x)$ is an irreducible cubic, we can conclude that $\operatorname{Inv}(K_i)=\Bbb{Q}(x_i)$ for all $i=1,2,3$.
5. The subgroups $M_j, j=1,2,3,4$, map a diagonal of the cube to itself. The diagonal is spanned by a vector $(1,\epsilon_2,\epsilon_3)$ with $\epsilon_{2,3}\in\{\pm1\}$. The endpoints of that diagonal can be thought of as the meeting points of three faces of the cube. Turning intersections into sums we associate $$\pm(\sqrt{x_1}+\epsilon_2\sqrt{x_2}+\epsilon_3\sqrt{x_3})$$ with those endpoints. Thinking about it this way immediately leads to the conclusion that the subgroup $M_j$ associated with the pair $(\epsilon_2,\epsilon_3)$ keeps the square $$ \begin{aligned} A_j&=(\sqrt{x_1}+\epsilon_2\sqrt{x_2}+\epsilon_3\sqrt{x_3})^2\\ &=x_1+x_2+x_3+2\epsilon_2\sqrt{x_1x_2}+2\epsilon_3\sqrt{x_1x_3}+2\epsilon_2\epsilon_3\sqrt{x_2x_3} \end{aligned} $$ fixed. Observe that here $x_1+x_2+x_3=2$ by Vieta, and that term can be ignored. Anyway, the $G$-conjugates of $$ a_1=\sqrt{x_1x_2}+\sqrt{x_2x_3}+\sqrt{x_3x_1}, $$ a fixed point of, say, $M_1$, are $a_2,a_3,a_4$, gotten from $a_1$ by changing an even number of signs. I cheated and fired up Mathematica to numerically calculate that their shared minimal polynomial is $$ h(T)=(T-a_1)(T-a_2)(T-a_3)(T-a_4)=T^4-4T^2-16T-12. $$ The polynomial $h(T)$ can be verified to be irreducible. For example, it is irreducible modulo $7$. Anyway, $\operatorname{Inv}(M_j)=\Bbb{Q}(a_j), j=1,2,3,4,$ for some suitable numbering of the zeros $a_j$ of $h(T)$.