Consider the Galois extension $K:=\mathbb{Q}(\sqrt 2, \sqrt 3, \sqrt{(2+\sqrt 2)(3+\sqrt 3)})/\mathbb{Q}$. As it is clear, it is of degree $8$ extension. So the Galois group is a group of order $8$. I want to identify the Galois group and want to draw its lattice diagram.
For the first glance, it seems the following is the lattice diagram.
But I think I am missing some $i=\sqrt{-1}$ in the whole picture. We note the Galois group of the subextension $\mathbb{Q}(\sqrt 2, \sqrt 3)/\mathbb{Q}$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$ with the following lattice diagram:
I think the Galois group of $K/\mathbb{Q}$ is not abelian. For, the conjugates of $\alpha=\sqrt{(2+\sqrt 2)(3+\sqrt 3)}$ are $\pm \sqrt{(2\pm\sqrt 2)(3\pm\sqrt 3)}$. We have the following $8$ automorphisms: $$\sqrt{(2+\sqrt 2)(3+\sqrt 3)} \to \pm \sqrt{(2\pm\sqrt 2)(3\pm\sqrt 3)}$$ Denote $\beta=\sqrt{(2+\sqrt 2)(3-\sqrt 3)}$ , $\gamma=\sqrt{(2-\sqrt 2)(3+\sqrt 3)}$. Let $\tau, \eta$ be two automorphisms in the Galois group with \begin{align*} &\tau(\alpha)=\beta,~~ \sqrt 2 \mapsto \sqrt 2,~~ \sqrt 3 \mapsto -\sqrt 3 \\ &\eta(\alpha)=\gamma, ~~ \sqrt 2 \mapsto \sqrt -2,~~ \sqrt 3 \mapsto \sqrt 3 \end{align*} Then $\tau \circ \eta \neq \eta \circ \tau$.
So either the Galois group is $D_4$ or $Q_8$. According to the first diagram shows one subgroup of order $4$ corresponding to $\mathbb{Q}(\sqrt 2, \sqrt 3)$ while $D_4$ has $3$ subgroups of order $4$. So the Galois group is $Q_8$, I think.
But I am missing something.
How to find help the Galois group and its lattice diagram ?
Edit: Though in my question, I presumed that $K/\mathbb{Q}$ was Galois and so didn't think much about whether $\alpha \notin \mathbb{Q}(\sqrt 2, \sqrt 3)$. I primarily thought that the Newton polygon of $x^2-\alpha^2$ has a single segment and hence irreducible. But it is not so obvious.
However, I believe $(2+\sqrt 2)(3+\sqrt 3)$ is not square in $\mathbb{Q}(\sqrt 2, \sqrt 3)$. For if, let $(2+\sqrt 2)(3+\sqrt 3)=k^2$ for some $k \in \mathbb{Q}(\sqrt 2, \sqrt 3)$, then $\alpha^2=k^2$. We know $\mathbb{Q}(\sqrt 2, \sqrt 3)/\mathbb{Q}$ is Galois, hence the automorphism $\eta \in \text{Gal}\left(\mathbb{Q}(\sqrt 2, \sqrt 3)/\mathbb{Q} \right)$ mentioned above. Therefore $\eta(\sqrt 2)=-\sqrt 2$ and $\eta(\sqrt 3)=\sqrt 3$.
Now,
\begin{align}
\alpha^2 \eta(\alpha^2)=(2+\sqrt 2)(3+\sqrt 3) (\eta(\alpha))^2=2(3+\sqrt 3)^2, \ \cdots \cdots \cdots (1)
\end{align}
On the other hand, since $k \in \mathbb{Q}(\sqrt 2, \sqrt 3)$ it has the basis expansion $k=a+b \sqrt 2+c \sqrt 3+d \sqrt 6$ for $a,b,c,d \in \mathbb{Q}$. Then
\begin{align}
k~ \eta(k)&=(a+b \sqrt 2+c \sqrt 3+d \sqrt d)(a-b \sqrt 2+c \sqrt 3-d \sqrt 6) \\&=(a^2-2b^2)+(3c^2-6d^2)+2 \sqrt 3 ac-4 \sqrt 3 bd \in \mathbb{Q}(\sqrt 3) \\ \Rightarrow k~\eta(k)
& \in \mathbb{Q}(\sqrt 3), \ \ \cdots \cdots (2)
\end{align}
From $(1)$ and $(2)$, we have $\sqrt 2 \in \mathbb{Q}(\sqrt 3)$, which is a contradiction.
Thus $(2+\sqrt 2)(3+\sqrt 3)$ is not square in $\mathbb{Q}(\sqrt 2, \sqrt 3)$.
