structure of extension by quadratic elements

Question

Is it true that if $\sqrt{b}\not\in\mathbb{Q}(\sqrt{a})$ for $a,b$ not squares in $\mathbb{Q}$, then $Gal(\mathbb{Q}(\sqrt{a},\sqrt{b}))\cong Gal(\mathbb{Q}(\sqrt{2},\sqrt{3}))$? Im seeing a bunch of examples where its holding (e.g. $a=-1,b=5$) and I was wondering if its a rule

Answer 1

As abstract groups, both Galois groups are isomorphic to $C_2\times C_2$: explicitly, we have automorphisms $$\sigma:\sqrt a\mapsto-\sqrt a, \sqrt b \mapsto \sqrt b\\\tau:\sqrt a\mapsto\sqrt a, \sqrt b \mapsto- \sqrt b$$ and one can check, using the fact that $\sqrt b\notin \mathbb Q(\sqrt a)$, that $\sigma\tau = \tau\sigma,\ \sigma^2=\tau^2 = 1$.

However, bear in mind that this is an isomorphism of abstract groups, and there is no canonical choice of isomorphism - i.e. any isomorphism requires us to manually and arbitrarily pick where each automorphism maps to.

In particular, these groups do not map to the same subgroup of $\mathrm{Gal}(\mathbb Q(\sqrt a ,\sqrt b,\sqrt2,\sqrt 3)/\mathbb Q)$.