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.
 A: If $A/B\simeq \mathbf Z$, it is isomorphic to a direct summand of $A$. Indeed, consider the commutative triangle:
\begin{align}
p: A \longrightarrow & A/B\\
s\nwarrow\;&\enspace\downarrow \varphi \\
&\enspace\,\mathbf Z
\end{align}
where $p$ is the canonical map, $\varphi$ the given isomorphism and $ s$ is the homomorphism defined by $s(1)=$ an inverse image of $1$ by $\varphi\circ p$. By construction, we have $\;(\varphi\circ p)\circ s=\operatorname{id}_\mathbf Z$ and $s$ is injective.
Note any $x\in A$ can be written as 
$$x=s(\varphi\circ p(x))+\bigl(x-s(\varphi\circ p(x))\bigr),$$
and $\;x-s(\varphi\circ p(x))\in\ker p=B$, since $$\;p\bigl(x-s(\varphi\circ p(x))\bigr)=p(x)-(p\circ s)(\varphi\circ p(x))=p(x)-p(x)=0. $$
Further, $\operatorname{Im} s\cap B=\{0\}$: indeed if $x\in B$ and $x=s(n)$, then $n=(\varphi\circ p)(s(n))=\varphi(p(x))=0$, so that $x=s(0)=0$. Thus
$$A=\operatorname{Im} s\oplus B\simeq\mathbf Z\oplus B. $$
A: 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 $A\not\cong \mathbb Z\oplus B$.
