I'm stuck trying to prove that a particular Dirichlet Character $\chi$ must be the Kronecker symbol $\chi_D = \left ( \frac{D}{\cdot}\right )$. The context is the following.
Let $d$ be a square-free integer and define $D$ to be
$D = \begin{cases} |d| & \text{if $d \equiv 1 \pmod{4}$}\\ 4|d| & \text{if $d \equiv 2, 3 \pmod{4}$} \end{cases} $
Now let $H = \left \{ [i] \in (\mathbb{Z}/D \mathbb{Z})^\times \mid \chi_D (i) = \left ( \frac{D}{i}\right ) = 1 \right \}$ and define X to be the set of Dirichlet characters
$ X = \{ \chi \in \widehat{(\mathbb{Z}/D \mathbb{Z})^\times} \mid \ \chi|_H \equiv 1 \} $
That is, $X$ is the set of Dirichlet characters on $(\mathbb{Z}/D \mathbb{Z})^\times$ that are trivial when restricted to $H$.
My Problem
Then I would like to prove that $X = \{ \chi_0 , \chi_D \}$ where $\chi_0$ is the trivial character and $\chi_D$ is the character defined by the Kronecker symbol.
My Attempt
So what I tried is taking an arbitrary nontrivial character $\chi \in X$ and I want to show that $\chi = \chi_D$.
By definition of $X$ and $H$, we know already that $\chi(i) = 1$ for any $i \in H$ and since for any $i \in H$ we have $\chi_D (i) = 1$ then this shows that $\chi |_H = \chi_D |_H$. Thus it remains to show that the two characters agree in the set $(\mathbb{Z}/D \mathbb{Z})^\times \setminus H$.
Now, if $i \in (\mathbb{Z}/D \mathbb{Z})^\times \setminus H$ then that means that $\chi_D (i) = -1$. Therefore it must be proved that $\chi(i) = -1$ also.
But at this point I'm stuck and I don't know what to do. I realize that I'm not using properties of the Kronecker symbol and that might be the key, but I don't know how to proceed. I would really appreciate any help with this problem.
Thank you for any help.