# Problem defining root automorphisms on Galois Group

I'm trying find the Galois Group of $$f(x)=x^4+5x^2+5$$.

I've finded the roots:

$$\alpha_1=i \sqrt{\frac{5-\sqrt5}{2}}$$; $$\alpha_2=-i \sqrt{\frac{5-\sqrt5}{2}}$$; $$\alpha_3=i \sqrt{\frac{5+\sqrt5}{2}}$$; $$\alpha_4=-i \sqrt{\frac{5+\sqrt5}{2}}$$;

And i've finded that:

$$\alpha_1=i \sqrt{\frac{5-\sqrt5}{2}}$$; $$\alpha_2=- \alpha_1$$; $$\alpha_3=\frac{\sqrt5}{\alpha_1}$$; $$\alpha_4=-\frac{\sqrt5}{\alpha_1}$$;

But, the problem is defining the automorphisms. If i define the automorphisms like this:

$$\sigma_1(\alpha_1)=\alpha_1$$;

$$\sigma_2(\alpha_1)=-\alpha_1$$;

$$\sigma_3(\alpha_1)=\frac{\sqrt5}{\alpha_1}$$;

$$\sigma_4(\alpha_1)=-\frac{\sqrt5}{\alpha_1}$$;

The solution of the exercise says that I should get $$\mathbb{Z_4}$$, but none of the automorphisms give me a generator of all group. Am I defining automorphisms well?

The automorphisms is well defined, but i ignored that$$\sqrt5 \in \mathbb{Q}(\alpha_1)$$ and $$\sqrt5=2 \alpha_1^2+5$$, then the automorphisms $$\sigma_3, \sigma_4$$ have order $$4$$