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Let $\alpha =\sqrt {(2+\sqrt2)(3+\sqrt3)}$ and consider the extension $\mathbb Q(\alpha)/\mathbb Q$. Prove that $\mathbb Q(\alpha)/\mathbb Q$ is a Galois extension with Gal$ (\mathbb Q(\alpha)/\mathbb Q) \simeq \mathbb Q_8$.

I did the basic calculation to find a polynomial which is satisfied by $\alpha$ and that is of the form: $x^8 -24 x^6+144x^4-288x^2+144 =0$. I am unable to show this is irreducible. Thanks for kind help.

Ok. From this link now it is clear that why this extension is galois. But what about the Galois group?

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Let $L$ be your degree 8 field, $G = \text{Gal}(L/\mathbb{Q})$, and let $K_2 = \mathbb{Q}(\sqrt 2)$, $K_3 = \mathbb{Q}(\sqrt 3)$, $K_6 = \mathbb{Q}(\sqrt 6)$ be its three quadratic subfields. Up to isomorphism, there are exactly two non abelian groups of order $8$ : the dihedral $D_8$ and the quaternionic $H_8$, characterized by the number of their cyclic subgroups of order 4: $D_8$ admits exactly 1 such subgroup; for $H_8$ , all its 3 subgroups of order 4 are cyclic. To decide whether G is $D_8$ or $H_8$, you just have to compute the Galois groups of $\text{Gal}(L/K_2)$, etc. If Lucky, you don't even need to do the complete calculations, because there is a criterion which says that $L/K_i$ is embeddable into a cyclic extension of degree 4 iff $-1$ is a norm in $L/K_i$ . My "guess" : $G$ is $H_8$ .

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