The question:
Determine the Galois group of $K = \mathbb{Q}(\sqrt{2}, \sqrt{11}) / \mathbb{Q}$ and how it works on $\alpha = \sqrt{6 + \sqrt{11}} \to a\sqrt{2} + b\sqrt{11}$.
I'm not sure if I'm approaching this correctly and how to check the way it works on $\alpha$.
My attempt:
$K$ is the splitting field of the polynomial $(X^2 - 2)(X^2 -11)$.Then it is the splitting field of a separable polynomial, thus it is a Galois extension. It is trivial that $[K : \mathbb{Q}] = 4$, so the Galois group has order 4.
Also, since $K$ is a splitting field, the action is transitive so we have a group of order 4 that is a subgroup of $S_4$. Now, I've tried looking at the subgroups of order 4 of $S_4$ but I haven't really been able to see what group the Galois group now is. How do I find this group?
The Galois group is $\cong\Bbb Z/2\times \Bbb Z/2$, its generators are $\sigma, \tau$, commuting field isomorphisms of order two with: $$ \begin{aligned} \sigma\sqrt 2 &= -\sqrt 2\ ,\\ \sigma\sqrt {11} &= +\sqrt {11}\ ,\\[2mm] \tau\sqrt 2 &= +\sqrt 2\ ,\\ \tau\sqrt {11} &= -\sqrt {11}\ . \end{aligned} $$ The elements of the Galois group are thus $1, \sigma,\tau,\sigma\tau$.
The complicated given element can be rewritten $$ \sqrt {6+\sqrt{11}} = \frac 1{\sqrt 2}\cdot \sqrt {12+2\sqrt{11}} =\pm\frac 1{\sqrt 2}(1+\sqrt {11})\ , $$ and it should be clear how $\sigma,\tau$ act on it.