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} $?

Question

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