This is related to Chpt 5, Sec 1(Quadratic Residue) Algebraic Number Theory by Taylor Frohlich
Narrow ideal class group is defined by quotient $C^+=\frac{I}{P^+}$ where $I$ is the set of fractional ideals and $P^+$ is the abelian group generated by all positive principals remaining positive under all real embedding.
Let $p,q$ be primes. Let $p,q^\star\equiv 1(\mod 4)$ where $q^\star=(-1)^{\frac{q-1}{2}}q$. Suppose $(\frac{q^\star}{p})=1$. Then $q^\star$ splits in $Q(\sqrt{p})$. Since $Q(\sqrt{p})$'s discriminant has only $p$ as prime divisor, $h(Q(\sqrt{p})$ is odd where $h(X)$ is the class number of number ring associated to the number field $X$. However $p=SS^\sigma$ where $S$ is the prime ideal lying above $p$ and $\sigma:\sqrt{p}\to-\sqrt{p}$.
$\textbf{Q:}$ Then the book says $S$ in narrow ideal class group is a square by class number odd. I do not see any particular obvious reason this has to be the case. Why is $\bar{S}\in C^+$ square? Probably one needs $(C^+)^2=(C^+)^{1-\sigma}$. I think I should consider $C_K\to C_K$ by $x\to x^2$ automorphism where $C_K$ is the class group. Since $h_O$ odd, kernel is trivial and this yields bijection by finiteness of $C_K$. Hence every $C_K$ representative is a square. Then $C^+\to C_K$ is surjection. Hence $S$ is a square in $C^+$. Is this correct?