Determine the original group according to its quotient Suppose that $G$ is a group and the group of integers $\mathbb Z$ is its normal subgroup with $G/\mathbb Z\cong\mathbb Z$. Then can I say that $G$ is isomorphic to $\mathbb Z\times\mathbb Z$?

The answer is no. ${\mathbb Z}^2$ is the only abelian group with that property, but there is a nonabelian group defined by the presentation $\langle x,y \mid y^{-1}xy=x^{-1} \rangle$.
Let $x$ generate the normal infinite cyclic subgroup $N$ and choose $y \in G$ such that $yN$ is a generator of $G/N$. Then $G=\langle x,y \rangle$ and, since $y^{-1}xy$ must also generate $N$, we have $y^{-1}xy$= $x$ or $x^{-1}$. So there are just two isomorphism classes of groups with thsi property.
