On the commutator subgroup of a group Prove that there is no group $G$ such that the commutator subgroup of $G$ to be $S_{3}$.

Denote the commutator subgroup of a group $G$ by $G'$.

Since $S_3$ is complete, you could proceed as in this question. Alternatively, you could use the following idea:

> If $G'/G''$ and $G''$ are cyclic, then $G'' = 1$.

**Hints for proof:** Show that $G''$ is in the center of $G'$. Conclude that $G'/Z(G')$ is cyclic, which implies that $G'$ is abelian.