Cyclicness of a quotient of subgroups of infinite cyclic group

Let $A$ be an abelian group and let $G(A) = A \times S_3$ be the direct product of $A$ and the symmetric group of $3$ elements $S_3$. We define the following operation on $G(A)$:

For each $(x,\sigma),(y,\tau) \in G(A)$ we define $(x,\sigma)(y, \tau)=(xy^{sgn(\sigma)}, \sigma\tau)$. This makes $G(A)$ a group and I don't need to prove it.

For each $B \leq A$ subgroup of $A$, let $N(B)=\{(b,1)\mid b\in B\}$ and $K(B)=\{(b,\sigma)\mid b \in B, \sigma \in A_3\}$, where $A_3 \leq S_3$ is the alternating group, that is the subgroup of $S_3$ whose elements are $\sigma \in S_3$ such that $sgn(\sigma)=1$.

I was asked to prove that both $N(B)$ and $K(B)$ are normal subgroups of $G(A)$, which I managed to do.

Next I was asked to prove that $G(A)/N(B)$ is isomorphic to $G(A/B)$ and I did that too.

The last thing this exercise is asking is to prove the following:

Suppose that $A$ is an infinite cyclic group. For which $B \leq A$ is the quotient $K(A)/N(B)$ a cyclic group?

What I tried is the following:

Since $A$ is infinite cyclic, suppose that $A = \langle z\rangle$, for some $z \in A$.
We know that every subgroup $B \leq A$ must be cyclic too, then we can say that $B=\langle z^b\rangle$, for some $b \in \mathbb Z$.
As $|A_3|=3$, we have that $A_3$ is cyclic, too. So $A_3=\langle \gamma\rangle$, for some $\gamma \in A_3$.

Let's suppose that $B$ is a subgroup of $A$ which makes $K(A)/N(B)$ a cyclic subgroup.
This means that there exists $(z^t,\gamma^s)N(B) \in K(A)/N(B)$ such that, for every $(z^n,\gamma^m)N(B) \in K(A)/N(B)$ there is $l \in \mathbb Z$ such that:
$(z^n, \gamma^m)N(B) = (z^t, \gamma^s)^lN(B)$.

We can prove by induction, since every $\sigma \in A_3$ is such that $sgn(\sigma)=1$, that $(z^t, \gamma^s)^l = (z^{tl}, \gamma^{sl})$.

So we must have:
$(z^n, \gamma^m)N(B) = (z^{tl}, \gamma^{sl})N(B)$.

This happens if and only if:
$(z^n, \gamma^m)^{-1}(z^{tl}, \gamma^{sl})$ is in $N(B)$. $(\star)$

As before, because we have even permutaions, we can write:
$(z^n, \gamma^m)^{-1}(z^{tl}, \gamma^{sl}) = (z^{-n}, \gamma^{-m})(z^{tl}, \gamma^{sl})=(z^{-n}z^{tl},\gamma^{-m}\gamma^{sl})=(z^{-n+tl},\gamma^{-m+sl})$.

Then the condition $(\star)$ is equivalent to:
$(z^{-n+tl},\gamma^{-m+sl})$ is in $N(B)$.

By definiton of $N(B)$ this happens when:
$z^{-n+tl} \in B$ and $\gamma^{-m+sl} = 1$

Since $B=\langle z^b\rangle$ and since $A_3$ is cyclic of order 3, we have:
$z^{-n+tl}=z^{bk}$ for some $k \in \mathbb Z$ and $-m+sl\equiv 0\pmod 3$

Since $A$ is cyclic infinite, we must have, for $z\neq 1$:
$-n+tl=bk$ and $sl\equiv m\pmod 3$

This is where I don't know how to continue. I was thinking about Bézout's identity, since we can write:
$tl-bk=n$
and this means that $n$ is the greatest common divisor of $b$ and $l$, but I don't really know how or whether to use it. I hope someone can help me understand this problem better.

Solution 1:

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Note first that $K(A) \cong N(A) \times Z$, where $Z = \Set{ (1, \sigma) : \sigma \in A_{3}}$ is a cyclic group of order $3$.

Then $K(A) / N(B) \cong N(A)/N(B) \times Z \cong A/B \times Z$.

This should keep you going.