Galois Groups vs $\mathbb{Q}(\alpha_i)$ Suppose I have $f(x) = (x^2-3)(x^2-2)$. The roots are $\pm\sqrt{3},\pm\sqrt{2}$. So the splitting field of $f$ over $\mathbb{Q}$, which is  a Galois extension, is the smallest subfield of $\mathbb{C}$ containing all the roots. Specifically, $\mathbb{Q}(\pm\sqrt{3},\pm\sqrt{2}) = \mathbb{Q}(\sqrt{3},\sqrt{2}).$
But, a group is Galois if all homomorphism of the roots are also roots of the polynomial. For my $f(x)= (x^2-3)(x^2-2)$, there are clearly automorphisms of the roots that do not work. So, if $\sigma(\pm \sqrt{2})= \pm \sqrt{3}$, then $\sigma(\alpha)$ are not roots of the equation, since we get $((\sqrt{2})^2-3)((-\sqrt{2})^2-3)(3^2-2)((-\sqrt{3})^2-2) \neq 0$.
Does this mean $f$ doesn't have a Galois group? 
 A: If $f(x)$ is a polynomial in $F[x]$ (assume separable), and $K$ is the splitting field of the polynomial, then it is true that every element of $\mathrm{Gal}(f)$ (the Galois group of the splitting field $K$ over $F$) must permute the roots of $f$. In fact, if $g$ is a factor of $f$ over $F$, then the roots of $f$ that are roots of $g$ must be permuted amongst themselves. 
So if $\deg(f)=n$, then $\mathrm{Gal}(f)$ is isomorphic to a subgroup of $S_n$.
But it not true that every permutation of the roots necessarily defines an element of $\mathrm{Gal}(f)$. Not even if $f$ is irreducible: there are cubics whose Galois group is cyclic of order $3$, so $\mathrm{Gal}(f)$ is not isomorphic to $S_3$.
Your error here is thinking that every permutation of the set $\{\sqrt{2},-\sqrt{2},\sqrt{3},-\sqrt{3}\}$ must correspond to an element of $\mathrm{Gal}(f)$. It does not. In fact, the Galois group is isomorphic to the Klein $4$-group, generated by the map that sends $\sqrt{2}\mapsto-\sqrt{2}$ and fixed $\sqrt{3}$; and the map that sends $\sqrt{3}\mapsto-\sqrt{3}$ and fixes $\sqrt{2}$.
