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

2,484

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.

Share:
2,484

Related videos on Youtube

Becky Peters
Author by

Becky Peters

Updated on July 23, 2022

Comments

  • Becky Peters
    Becky Peters over 1 year

    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?

    • Alex Wertheim
      Alex Wertheim almost 7 years
      It's false without stronger assumptions, e.g. $N \subset Z(G)$. For instance, any finite group $G$ with a normal subgroup $N$ of index $p$ prime has cyclic quotient $G/N$, but there are many such groups with this property which aren't abelian.