Could you help me in finding the Galois group of $\mathbb{Q}\left(\sqrt{2}, \sqrt{3}, \sqrt{(2+\sqrt{2})(3+\sqrt{3}}\right)$ over $\mathbb{Q}$?
I can only say that $\mathbb{Q}(\sqrt{2}) /\mathbb{Q}$ and $\mathbb{Q}(\sqrt{3}) /\mathbb{Q}$ are two normal extentions and that $\mathbb{Q}\left(\sqrt{2}, \sqrt{3}, \sqrt{(2+\sqrt{2})(3+\sqrt{3})}\right)=\mathbb{Q}\left(\sqrt{2},\sqrt{(2+\sqrt{2})(3+\sqrt{3})}\right)$ is an eighth-degree extention over $\mathbb{Q}$.
thank you!
edit: Is the following a valid argument?
$\mathbb{E}=\mathbb{Q}\left(\sqrt{2}, \sqrt{3} \right) \subset \mathbb{Q}\left(\sqrt{2}, \sqrt{3}, \sqrt{(2+\sqrt{2})(3+\sqrt{3})}\right)=\mathbb{F}$ therefore $\mathbb{F}/\mathbb{E}$ is a second-degree galois extention where $\sigma: \sqrt{(2+\sqrt{2})(3+\sqrt{3})} \mapsto -\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ is the only automorphism in the galois group.
Therefore $Gal(\mathbb{F}/\mathbb{Q})=\langle \pi,\tau,\sigma\rangle$ where
$\pi:\sqrt{2}\mapsto -\sqrt{2}\\ \tau:\sqrt{3}\mapsto -\sqrt{3}\\ \sigma: \sqrt{(2+\sqrt{2})(3+\sqrt{3})} \mapsto -\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$
We can easily verify that $\pi (\tau ( \sigma)) \neq \sigma( \pi (\tau))$ but $\pi(\tau)=\tau( \pi)$ Therefore $\langle \pi,\tau \rangle$ is the center of the galois group which is not commutative and then $Gal (\mathbb{F}/\mathbb{Q})=\mathbb{Z}_2\times \mathbb{Z}_2 \rtimes \mathbb{Z}_2$
Let $\alpha = \sqrt{(2+\sqrt 2)(3+\sqrt 3)}$ and $K = \Bbb Q(\sqrt 2, \sqrt 3, \alpha)$. An automorphism $\sigma$ of $K$ over $L$ is determined by $\sigma(\sqrt 2),\sigma(\sqrt 3)$ and $\sigma(\alpha)$. Since $\sigma$ is an automorphism, those values have to satisfy the same algebraic relations of the original values, so $\sigma(\sqrt 2)^2 = 2, \sigma(\sqrt 3)^2 = 3$, and $\sigma(\alpha)^2 = (2+\sigma(\sqrt 2))(3+\sigma(\sqrt 3))$. In fact, $K = \Bbb Q(\alpha)$ so the image of $\alpha$ determines the images of $\sqrt 2$ and $\sqrt 3$.
So the first thing is to do is to check how many conjugates (over $\Bbb Q$) of $\alpha$ are in $K$. Obviously, we have $\alpha, - \alpha \in K$. Let $\beta = \sqrt{(2-\sqrt 2)(3+\sqrt 3)}$. Then $\beta\alpha = \sqrt 2(3+\sqrt 3) \in K$, so that $\beta, - \beta \in K$. Similarly, if $\gamma = \sqrt{(2+\sqrt 2)(3- \sqrt 3)}$ and $\delta = \sqrt{(2-\sqrt 2)(3-\sqrt 3)}$, we have $\gamma\alpha = (2+\sqrt 2)\sqrt 6$ and $\delta\alpha = 2\sqrt 3$.
Therefore, all $8$ conjugates of $\alpha$ are in $K$, so that $K$ is Galois over $\Bbb Q$.
Let $\pi$ be the automorphism sending $(\alpha,\sqrt 2,\sqrt 3)$ to $(\beta, - \sqrt 2, \sqrt 3)$. We have $\pi(\pi(\alpha)) = \pi(\beta) = \pi(\sqrt 2(3+\sqrt 3)/\alpha) = -\sqrt 2(3+\sqrt 3)/\beta = - \alpha$. Hence $\pi^2(\alpha,\sqrt 2,\sqrt 3) = (- \alpha,\sqrt 2,\sqrt 3)$, then $\pi^3(\alpha,\sqrt 2,\sqrt 3) = (- \beta, - \sqrt 2, \sqrt 3)$ and $\pi^4$ is the identity.
Let $\tau$ be the automorphism sending $(\alpha,\sqrt 2,\sqrt 3)$ to $(\gamma, \sqrt 2, -\sqrt 3)$. We have $\tau(\tau(\alpha)) = \tau(\gamma) = \tau((2+\sqrt 2)\sqrt 6/\alpha) = - \alpha$, so $\tau^2 = \pi^2$.
$\tau(\pi(\alpha)) = \tau(\beta) = \tau(\sqrt 2(3+\sqrt 3)/\alpha) = \sqrt 2(3-\sqrt 3)/\gamma = (\sqrt 2(3-\sqrt 3)/(2+\sqrt 2)\sqrt 6)\alpha = (6\sqrt 2/\alpha^2\sqrt 6)\alpha =2\sqrt 3/\alpha = \delta$
$\pi(\tau(\alpha)) = \pi(\gamma) = \pi((2+\sqrt 2)\sqrt 6/\alpha) = -(2-\sqrt 2)\sqrt 6/\beta = (-(2-\sqrt 2)\sqrt 6/\sqrt2(3+\sqrt 3))\alpha = (-2\sqrt 6/\alpha^2\sqrt 2)\alpha = -2\sqrt 3/\alpha = -\delta = \tau(\pi ^3(\alpha))$
So we have $\pi^4 = 1, \tau^2 = \pi^2, \pi \tau = \tau \pi^3$. $\tau^2 = \pi^2 = (\tau\pi)^2 = \tau\pi(\tau\pi)$. The Galois group is isomorphic to the quaternion group.