Galois group of order 2^4 Find the galois group the polynomial $f(X)=(X^2-2)(X^2-3)(X^2-5)(X^2-7)$ over $\mathbb{Q}$. A splitting field for $f(X)$ is $K=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7})$. We must have $[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7}):\mathbb{Q}]=2^4$. Moreover $Gal(K/\mathbb{Q})$ is abelian, since if $\sigma\in{Gal(K/\mathbb{Q})}$ then $\sigma^2=1$, because $\sigma(\sqrt{2})=\pm\sqrt{2}$ (similary $\sigma(\sqrt{3}),\sigma(\sqrt{5}),\sigma(\sqrt{7})$). By the classification of finite abelian groups; can I conclude that $Gal(K/\mathbb{Q})\approx Z_2\times{Z_2}\times{Z_2}\times{Z_2}$ ? Thanks you all.

You've explicitly identified the generators of the Galois group! You don't need to invoke the classification of finite(ly-generated) abelian groups here.

You know that the Galois group must have order $16$, and it's easy to show that the group generated by the maps $$\sqrt{2} \mapsto -\sqrt{2}, \quad \sqrt{3} \mapsto -\sqrt{3}, \quad \sqrt{5} \mapsto -\sqrt{5}, \quad \sqrt{7} \mapsto -\sqrt{7}$$ is isomorphic to $Z_2 \times Z_2 \times Z_2 \times Z_2$, which has order $16$. Moreover each of these maps must be in the Galois group, so this must actually _be_ the Galois group of your polynomial over $\mathbb{Q}$.