(From Milne) Show that the fixed field of $H$ is $\Bbb Q(\sqrt{p})$ when $p=1 mod 4$ and $\Bbb Q(\sqrt{-p})$ when $p=3 mod 4$

Question

This is an exercise in Milne's notes:

Here is the question with an answer. But I find myself hardly understand it. especially the part that if $i\in H$, $iH=H$ and $i(G\setminus H)=G\setminus H$. And I do not under tand the sentence with the "$\Leftrightarrow$". I think in above argument it has proved $i+j=0,i\in H,j\in G\setminus H\Leftrightarrow -1 \text{ is not a square mod p}$ I am so confused. So what is going on here. May I please ask for an explicit argument? Thanks in advance!

Answer 1

Think carefully about what some of these things are. $G=(\Bbb{Z}/(p))^{\times},$ a cyclic (multiplicative) group. $H$ is the subgroup of index 2 in $G$, so that means $H$ consists of elements which are squares and $G\setminus H$ nonsquares. So when he says that "$-1=i^{-1}j$, which is a nonsquare;" he's appealing to the earlier observation that $i(G\setminus H)=(G\setminus H)$. You're taking an element of $H$ times an element of $G\setminus H$, so you must get an element of $G\setminus H$ which is a non-square. $-1$ is a non-square $\bmod p$ iff $p\equiv 3\bmod 4$