Does $G$ always have a subgroup isomorphic to $G/N$?

Question

Let $G$ be a group and $N$ a normal subgroup of $G$. Must $G$ contain a subgroup isomorphic to $G/N$? My first guess is no, but by the fundamental theorem of abelian groups it is true for finite abelian groups, so finding a counterexample has been a little tough. The finite case is also interesting for me.

Answer 1

For a finite counterexample, take a look at the subgroup structure of the quaternions, a nonabelian group of order $8$.

By inspection, it's straightforward to verify that $Q/Z(Q)\simeq V_4$, the Klein group. However, the only subgroups of $Q$ of order $4$ are $\langle i\rangle\simeq\langle j\rangle\simeq\langle k\rangle\simeq\mathbb{Z}/(4)$. So $Q$ does not have a subgroup isomorphic to $Q/Z(Q)$.

Answer 2

Let $G$ be the reals $\mathbb{R}$under addition. Let $N$ be the integers $\mathbb{Z}$ under addition. $G/N$ is the group of reals modulo $1$. This has a non-zero element whose square is $1$, the equivalence class of $0.5$. So $G/N$ is not isomorphic to any subgroup of $G$.