If G contains a normal subgroup $N \cong \mathbb{Z}_2$ and $G/N \cong \mathbb{Z}$, then $G\cong \mathbb{Z}\times \mathbb{Z_2}$.
If $G$ contains a normal subgroup $H \cong \mathbb{Z_2}$ such that $G/H$ is infinite cyclic, then $G \cong \mathbb{Z} \times \mathbb{Z_2}$