Let $H$ be the subgroup $\{i,\alpha\}$ of $\text{Gal}_{\mathbb{Q}}\mathbb{Q}(\sqrt{3},\sqrt{5}),$ where $i$ is the identity map and $\alpha$ is defined as $\alpha(\sqrt{3})=\sqrt{3}$,$\alpha(\sqrt{5})=-\sqrt{5}$. Show that the fixed field of $H$, denoted by $E_H$, is $\mathbb{Q}(\sqrt{3})$.
If someone could verify my proof that would be great. Any "better" proof is more than welcome as well.
My proof:
Since $\alpha(\sqrt{3})=\sqrt{3}$, $\mathbb{Q}(\sqrt{3})\subset E_H$. Since we have $\mathbb{Q}(\sqrt{3})\subset E_H\subset \mathbb{Q}(\sqrt{3},\sqrt{5})$, $[\mathbb{Q}(\sqrt{3},\sqrt{5}):E_H][E_H:\mathbb{Q}(\sqrt{3})]= [\mathbb{Q}(\sqrt{3},\sqrt{5}):\mathbb{Q}(\sqrt{3})]=2.$
If $[\mathbb{Q}(\sqrt{3},\sqrt{5}):E_H]=2$, then we are good. If not, then $\mathbb{Q}(\sqrt{3},\sqrt{5})=E_H$. Then $\alpha(\sqrt{5})\ne \sqrt{5}$. This shows that $E_H = \mathbb{Q}(\sqrt{3})$.
Here is a slightly quicker way that generalizes to more situations.
Recall that if an extension $F$ of $K$ has prime degree then $F=K(\alpha)$ for any $\alpha\in F-K$.
Now you know that the order of your Galois group is 4 and the order of your subgroup is 2 and so you know (by the Galois correspondence) that $[\mbox{ Fix }H:\mathbb{Q}]=2$ (a prime). Clearly $\sqrt{3}$ is fixed by every element in $H$ and $\sqrt{3}$ does NOT belong to $\mathbb{Q}$. Thus $\mbox{ Fix }H = \mathbb{Q}(\sqrt{3})$.