Isomorphisms between quotient groups Is it true that if $\mathbb{Z} \cong A/B$ with $A, B$ abelian groups then $A \cong \mathbb{Z} \times B$? I think it must be true, but can't show it.

You have a short exact sequence
$$
0\longrightarrow B\longrightarrow A\longrightarrow \mathbb Z\longrightarrow 0
$$
where the map $B\to A$ is the inclusion (since you take the quotient, $B$ is a subgroup of $A$) and the map $A\to\mathbb Z$ is the composition of the projection $A\to A/B$ with the isomorphism $A/B\to\mathbb Z$.
Since $\mathbb Z$ is free, the sequence splits, so that $A\cong B\oplus \mathbb Z$ (here $\oplus$ is just an alternative notation for the product, since we have a finite sum/product). The proof of these general facts is similar to what Bernard did. See for example the page Splitting Lemma.
Note that this is not true if instead of $\mathbb Z$ you have, for example, $\mathbb Z/2\mathbb Z$: for $A=\mathbb Z$ and $B=2\mathbb Z$ you have $A\not\cong \mathbb Z\oplus B$.
