# Is the Galois group of $\mathbb{Q} \left(\sqrt{5}+\sqrt{7}+i \right)$ isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times\mathbb{Z}_{2}$?

First of all, I proved that $$\mathbb{Q}(\sqrt{5}+\sqrt{7}+i)=\mathbb{Q}(\sqrt{5},\sqrt{7},i)$$.Found that $$|G|=|Gal(\mathbb{Q}(\sqrt{5},\sqrt{7},i)|=[\mathbb{Q}(\sqrt{5},\sqrt{7},i):\mathbb{Q}]=8$$. That's because $$\mathbb{Q}(\sqrt{5},\sqrt{7},i)$$ is a splitting field of $$f(x)=(x^2-5)(x^2-7)(x^2+1)/\mathbb{Q}$$.Finally I showed that every $$\sigma \in G$$ has an order of 2 and that G is abelian.

Do the above really prove that $$G \simeq Z_2 \times Z_2 \times Z_2 ?$$

Edit: here is a link on how i showed that $$\mathbb{Q}(\sqrt{5}+\sqrt{7}+i)=\mathbb{Q}(\sqrt{5},\sqrt{7},i)$$ https://www.overleaf.com/read/bsrbxwnxnkcg

• oops... i meant 1.Thank you for the point out
– GGG
May 23, 2022 at 16:48
• What have you computed for the minimal polynomial of $\sqrt{5}+\sqrt{7}+i$? May 23, 2022 at 16:56
• @DietrichBurde I haven't yet , i really thought it wasn't necessary to be honest.I would probably use the formula $Irr(\sqrt{5}+\sqrt{7}+i,\mathbb{Q})=\prod_{\sigma \in G}(x-\sigma(\sqrt{5}+\sqrt{7}+i))$
– GGG
May 23, 2022 at 17:08
• Just taking the last sentence alone, the structure of finite abelian groups would say there is only one (the one you identify) isomorphism class of abelian groups of size 8 in which every element is its own inverse.
– irh
May 23, 2022 at 17:14
• The minimal polynomial is $x^8-44x^6+542x^4+964x^2+841$, and it has Galois group $C_2\times C_2\times C_2$. May 23, 2022 at 18:20