Show that there are σ, τ ∈ Gal(L|Q) and isomorphism without Sylow theorem

Consider the extension $$Q ⊂ L = Q(i, \sqrt[4]{2})$$. I wish to first show that there are $$σ, τ ∈ Gal(L|Q)$$ such that

$$σ(i) = i, σ(\sqrt[4]{2}) = i\sqrt[4]{2}τ (i) = −i, τ (\sqrt[4]{2}) = \sqrt[4]{2}.$$

Next I wish to prove that $$σ$$ and $$τ$$ generate $$Gal(L|Q)$$. Lastly, I want to show that $$Gal(L|Q) \cong D_8$$, where $$D_8$$ is the dihedral group of order 8.

By Galois group of $$x^4-2$$, I know how to compute the Galois group of $$x^4-2$$ but I am not sure how to precisely show these automorphisms exist and they generate the Galois group. Or is the proof exactly the same as in how the answer is phrased over there, I think I am missing something simple here.. Also, is it possible to have an answer that proves the isomorphism desired in the last part without the use of Sylow theorem?

• There aren't many groups of order $8$; just exhaust the possibilities. I'm also not sure what's unclear about proving that the automorphisms exist; you wrote them down. Why don't you want to use the Sylow theorems? Oct 15 '18 at 15:32
• Yes, the proof is exactly the same as in the question you have linked. It is the standard proof to show that the Galois group of $x^4-2$ is isomorphic to $D_4$ of order $8$. Oct 15 '18 at 15:32