I'm trying to tackle Exercise 3.11 in Cox's PRIMES OF THE FORM $x^2+ny^2$.
$H$ is the subgroup of $(\mathbb{Z}/D\mathbb{Z})^*$ of values represented by $x^2+ny^2$ where $D=-4n$ with $n=2^{a}m$, gcd(2,m)=1. With Chinese Remainder theorem, we can easily obtain $(\mathbb{Z}/D\mathbb{Z})^* \cong (\mathbb{Z}/2^{a+2}\mathbb{Z})^* \times (\mathbb{Z}/m\mathbb{Z})^*$. Then $H=H_1 \times [(\mathbb{Z}/m\mathbb{Z})^*]^2$ where $H_1=H \cap (\mathbb{Z}/2^{a+2}\mathbb{Z})^* \times \{ 1\}$ and $[(\mathbb{Z}/m\mathbb{Z})^*]^2=\{x^2:x\in (\mathbb{Z}/m\mathbb{Z})^* \}$
To simplify our notation, let $G=(\mathbb{Z}/D\mathbb{Z})^*$, $M=(\mathbb{Z}/2^{a+2}\mathbb{Z})^*$, $N=(\mathbb{Z}/m\mathbb{Z})^*$. I'm guessing that there're two steps:
1) $H=\{H\cap M\} \times\{H\cap N\}$. However, this is not necessarily true when the order of $M$ and $N$ are not coprime.
2) $H\cap N=N^2$. Since the principal genus $H=[(\mathbb{Z}/D\mathbb{Z})^*]^2$ when $D \equiv 1 \ (mod4)$. However, using this property to obtain our result is not very clear.
Can someone enlighten me? Thanks a lot!