# Subgroups of Galois group?

Set $$f(X) = X^4 − 6X^2 − 2$$ and denote by $$K$$ the splitting field of $$f$$ over $$\mathbb{Q}$$. I am interested in finding the subgroups of $$\operatorname{Gal}(K : Q)$$ and hence finding all subfields of $$K$$.

I have already found the roots of $$f$$ to be $$\pm\alpha$$ and $$\pm\beta$$, where $$\alpha=\sqrt{3+\sqrt{11}}$$ and $$\beta=\sqrt{3-\sqrt{11}}$$, and $$K=\mathbb{Q}(\alpha,\alpha\beta)$$. The extension $$K:\mathbb{Q}$$ is then Galois, non abelian with order 8. I just don't know how to do the last step to find these subgroups and subfields.

• Computing the discriminant should help narrow down the Galois group. – take008 Nov 22 '19 at 14:11

The Galois group contains the following three maps: \begin{align} \alpha\mapsto - \alpha&, \beta\mapsto \beta\\ \alpha\mapsto \alpha&, \beta\mapsto - \beta\\ \alpha\mapsto\beta&, \beta\mapsto\alpha \end{align} Call these three maps $$n_\alpha, n_\beta$$ and $$s$$ ("Negate $$\alpha$$", "Negate $$\beta$$", and "Swap") respectively. They all have oder $$2$$, but $$sn_\alpha$$ and $$sn_\beta$$ have order $$4$$. Also, $$n_\alpha$$ and $$n_\beta$$ commute, but neither of them commute with $$s$$ (we have $$n_\alpha s = sn_\beta$$), although $$n_\alpha n_\beta$$ does commute with $$s$$.

This gives us the 8 elements $$e, n_\alpha, n_\beta, n_\alpha n_\beta,\\ s, sn_\alpha, sn_\beta, sn_\alpha n_\beta$$ It doesn't take much work from here to prove that this is the dihedral group $$D_4$$.

And with that, we can start to look at what subgroups we have and what fixed subfields they give rise to. First the order $$2$$ subgroups, giving rise to degree $$4$$ extensions:

• $$\langle n_\alpha\rangle$$ clearly fixes $$\beta$$, so $$\Bbb Q(\beta)$$ is the corresponding fixed field
• $$\langle n_\beta\rangle$$ clearly fixes $$\alpha$$, so $$\Bbb Q(\alpha)$$ is the corresponding fixed field
• $$\langle n_\alpha n_\beta\rangle$$ fixes $$\alpha\beta = i\sqrt2$$ as well as $$\alpha^2 = 3 + \sqrt{11}$$, so the corresponding fixed field is $$\Bbb Q(\alpha\beta, \alpha^2) = \Bbb Q(i\sqrt2, \sqrt{11})$$
• $$\langle s\rangle$$ fixes $$\alpha + \beta$$, so the fixed field is $$\Bbb Q(\alpha + \beta)$$
• $$\langle sn_\alpha n_\beta)$$ fixes $$\alpha - \beta$$, so the fixed field is $$\Bbb Q(\alpha - \beta)$$

Then the order $$4$$ subgroups:

• $$\langle sn_\alpha\rangle$$ fixes $$\alpha\beta(\alpha^2-\beta^2) = 2i\sqrt{22}$$, so the fixed field is $$\Bbb Q(i\sqrt{22})$$
• $$\langle n_\alpha, n_\beta\rangle$$ fixes $$\alpha^2-\beta^2 = 2\sqrt{11}$$, so the fixed field is $$\Bbb Q(\sqrt{11})$$
• $$\langle s, sn_\alpha n_\beta\rangle$$ fixes $$\alpha\beta=i\sqrt2$$, so the fixed field is $$\Bbb Q(i\sqrt2)$$

And finally, of course, the trivial subgroup fixes all of $$\Bbb Q(\alpha, \beta)$$, and the whole group fixes only $$\Bbb Q$$.