If G/N is cyclic, then G is abelian.

Question

I need to prove if G/N is cyclic, then G is abelian. I was given no other information on the groups. I know that a cyclic group is abelian and that N is a normal subgroup in G. I tried looking up this problem, but the only solutions I could find were answering problems such as "If G/Z(G) is cyclic, then G is abelian" or "Factor group of a center of a abelian group is cyclic" when the person was really asking how to prove "if the factor by the center is cyclic, then the group is abelian." All of these involve the center of the group, but my question prompt has no mention of the center in it. How can I go about solving it without having the use of the center to help me commute?

Answer 1

This is false, let $G=S_3$, the symmetric group of order $6$ and let $N=A_3$ be the alternating group on three symbols. Then $S_3/A_3$ is of order $2$ hence cyclic, but $S_3$ is not abelian.