I have the sequence $0\rightarrow A\xrightarrow{s}\mathbb{Z}_k\xrightarrow{\times \ell}\mathbb{Z}_k\xrightarrow{t}B\rightarrow0$ where $\times\ell$ is multiplication by $\ell\in\mathbb{Z}_{>0}$.
Now, $ker(t)=im(\times\ell)=\ell\mathbb{Z}_k=\mathbb{Z}_{gcd(\ell,k)}$, so by the first isomorphism theorem we have that $B=im(t)\cong\mathbb{Z}_k/\mathbb{Z}_{gcd(\ell,k)}=\mathbb{Z}_{gcd(\ell,k)}$ since $gcd(\ell,k)\vert k$. So we have successfully determined $B$ in this sequence. But I'm unable so far to determine $A$. Would anyone be able to give me some suggestions on how to proceed? Thanks!
I assume that the sequence is exact.
The image of $0 \to A$ is then the kernel of $s$, which is $0$. Hence $s$ is injective and $A \cong \mathrm{im}(s) = \ker(\times \ell) = d\Bbb Z / k \Bbb Z$ for some $d \in \{0,...,k\}$. Can you take it from here?