Let $f(x)=x^4-6x^2+4 \in \mathbb{Q}[x]$.
The Galois group is $\mathrm{Gal}(L/K) \simeq \mathbb{(Z/2Z)^2}$, but I don't know how to find it.
I know that $\mathbb{Q}(\sqrt{3+\sqrt{5}})$ is a splitting field of $f(x)$ over $\mathbb{Q}$.
$\mathrm{Gal}(L/K)$ acts transitively on the roots of $f(x)$, so there exist $\sigma_1, \sigma_2, \sigma_3$ and $\sigma_4$ with $\sigma_1(\sqrt{3+\sqrt{5}})=\sqrt{3+\sqrt{5}}=\mathrm{id}, \sigma_2(\sqrt{3+\sqrt{5}})=-\sqrt{3+\sqrt{5}}, \sigma_3(\sqrt{3+\sqrt{5}})=\sqrt{3-\sqrt{5}}$ and $\sigma_4(\sqrt{3+\sqrt{5}})=-\sqrt{3-\sqrt{5}}$
So $\sigma_i^2=\mathrm{id}$ and $\mathrm{Gal}(L/K) \simeq \mathbb{(Z/2Z)^2}$
This is what I don't understand. I see that $\sigma_2^2=\sigma_2 \circ \sigma_2=\sigma_1$, but $\sigma_3^2=\sigma_3 \circ \sigma_3 =0,7639 \neq \mathrm{id}$ aswell as $\sigma_4^2=\sigma_4 \circ \sigma_4 = 0,7639 \neq \mathrm{id}$.
How to compute it to get $\sigma_3^2=\sigma_4^2=\mathrm{id}$?