Isomorphism between subgroups by preserving index

Question

Let $C$ be a subgroup of a group $A$ such that $C\cong A/\{\pm1\}$ and $D$ be a subgroup of $B$ such that $D\cong B/\{\pm1\}$. Let $\pi:A\to B$ be a group homomorphism, and let $\pi$ induce a surjective group homomorphism (say) $\pi^*$ on the quotient groups. Now, clearly $C$ has index $2$ in $A$, $D$ has index $2$ in $B$, and $C$ is a subgroup of $\pi^{-1}(D)$. Then can we say that $\pi^{-1}(D)$ has index $2$ in $A$ as well? Eventually, I am trying to show that $C\cong\pi^{-1}(D)$. So, if my above argument is not correct then how can I show the desired isomorphism?

Edit: Let me mention the original groups and group homomorphism that I am working with. In the original problem $A=(\mathcal{O}/3^m)^{\times}$, $B=(\mathcal{O}/3)^{\times}$, where $\mathcal{O}=\mathbb{Z}[\sqrt{-3}]$ and $\pi:A\to B$ is defined as $a+b\sqrt{-3}+3^m\mathcal{O}\mapsto a+b\sqrt{-3}+3\mathcal{O}$.

Answer 1

Partial answer and remark

I assume that $[A:C] = [B:D] = 2$, since this may not follow from the isomorphisms $C \simeq A/\{-1,1\}$, $D \simeq B/\{-1,1\}$.

The morphism $\pi$ is surjective, like the canonical projection on $B/D$, denoted by $p_{B/D}$ in what follows, so $\mathrm{Im}(p_{B/D} \circ \pi) = B/D$. But $\mathrm{Ker}(p_{B/D} \circ \pi) = \pi^{-1}(D)$. Hence, $A/\pi^{-1}(D) \simeq B/D$, so $[A:\pi^{-1}(D)] = [B:D] = 2$.

Let $s : \mathcal{O} \to \mathcal{O}$ defined by $s(x)=x^2$. Then $s(A)$ and $s(B)$ are subgroups of $A$ and $B$, and they are contained in every subgroup of $A$ and $B$ with index $2$. I wonder whether $[A:s(A)]$ and $[B:s(B)]$ are larger than $2$ or not.