Galois group of a subextension is a subgroup of the Galois group of the extension? Let $\overline{\mathbb{F}}_{5}$ be the separable closure of $\mathbb{F}_{5}$ and let $G=G(\overline{\mathbb{F}}_{5}/\mathbb{F}_{5})$ be its Galois group.
Say we pick a finite subextension of $\overline{\mathbb{F}}_{5}/\mathbb{F}_{5}$ such as $\mathbb{F}_{5}(\sqrt{2})/\mathbb{F}_{5}$, which has Galois group $\{\varphi_{1},\varphi_{2}\}$, where $\varphi_{1}$ is identity and $\varphi_{2}$ is conjugation.
How can I see this Galois group inside the big Galois group $G$? 
I'm supposed to see it as a quotient of $G$ by some subgroup or simply as a subgroup of $G$? This is still very confusing. Sorry if the question is so easy.

Do you know that the Galois group $G(\overline{\Bbb F}_5/\Bbb F_5)$ is topologically generated by the Frobenius automorphism of $\overline{\Bbb F}_5$, which sends any element $\alpha$ to $\alpha^5$? You can think of it as an infinite cyclic group, but completed with respect to a certain topology.
Calling the Frobenius automorphism $\mathbf f$, its square $\mathbf f^2$ (defined by $\alpha\mapsto\alpha^{25}$) generates a subgroup of index two. That’s the subgroup you’re looking for.
