I have been reading Cox's "Primes of the form $x^2+ny^2$" and I have been a bit confused about a seeming conflict between a statement in the proof of Theorem 3.15(ii) (page 49) and Lemma 3.17 (page 50).
For context, Theorem 3.15(ii) shows that for a negative integer $D \equiv 0, 1 \pmod 4$, the principal genus mod $D$ consists exactly of the square elements in the form class group $\mathcal C(D)$. Lemma 3.17 is an intermediate step which shows that the map $\Psi: (\mathbb Z/D \mathbb Z)^\times \rightarrow \{\pm 1\}^\mu$ given by the tuple of evaluations at the assigned characters, is surjective and its kernel is precisely the subgroup $H$ of values represented by the principal form.
Now, at the end of page 50, it is said that in the case when $D$ is even (i.e., $D=-4n$ for some $n \in \mathbb N$), the subgroup $H$ might be larger than the subgroup of squares in the group $\ker \chi$ (which is an exactly equality for odd $D$). But if we show that the principal genus consists only of square elements (as Theoem 3.15(ii) does), then isn't it automatic that $H$ (which is the image of any class in the principal genus under the map $\Phi: \mathcal C(D) \rightarrow \ker \chi/H$) consists only of squares? So basically my question is the following:
Does the subgroup $H$ of $\ker \chi \subset (\mathbb Z/D \mathbb Z)^\times$ represented by the principal form, always consist of the squares in $\ker \chi$? (In that case, is the statement at the end of page 50 basically saying that we don't know this fact at that stage?) Or am I misunderstanding something and there are explicit counterexamples (if so, what are some of them?) to the expectation that $H$ always consists of squares, irrespective of the parity of $D$?
Thank you.