Let $E\subset \mathbb{C}$ be the splitting field of $f(x)\in \mathbb{Q}[x]$ such that $\sqrt{-d}\in E$, for a positive integer $d$. Show that $Gal(E/\mathbb{Q}) \cong H\rtimes \mathbb{Z}_2$ for some group $H$. [Hint: Use complex conjugation]
My attempt:
In order to use Galois correspondence, $\mathbb{Q}(\sqrt{-d})\subset E$. If $d$ is not a square, then $[\mathbb{Q}(\sqrt{-d}):\mathbb{Q}] = 2$ and hence $E$ has a subextension of order $4$ and hence $Gal(E/\mathbb{Q})$ must have a subgroup of order 2. I'm not sure how to go further to use complex conjugation and if I need to construct the explicit homomorphism involved in the definition of the semidirect product. Thanks.
Let $H$ be the stabiliser of $\sqrt{-d}$ in $G$, and $\sigma$ denote the operation of complex conjugation. Then $H$ has index $2$ in $G=\text{Gal}(E/\Bbb Q)$ and $\sigma\notin H$. Moreover $\langle\sigma\rangle$ is normal in $G$. All of these combine to prove that $G$ is a semidirect product of $H$ and $\langle\sigma\rangle$.