Prove or disprove: There exists a group $G$ and a normal subgroup $N$ such that $G$ is non-abelian, but both $N$ and $G/N$ are abelian.

Can anyone give me some hint on this question, please? What theorem(s) in abstract algebra is(are) related to this question, please? Thank you!

  • $\begingroup$ could you explain what are your thoughts on this question... $\endgroup$
    – user87543
    Nov 16, 2013 at 4:52
  • 3
    $\begingroup$ One way is to just start writing down non-abelian groups and checking if this holds. In fact, try writing down $S_3$.... $\endgroup$
    – user61527
    Nov 16, 2013 at 4:53
  • $\begingroup$ Let's take $S_3$ as an example. Clearly $S_3$ is not abelian and its normal subgroup is $A_3$. It is true that $S_3/A_3$ is isomorphic to $\{1, -1\}$ and therefore, $G/N$ is abelian. Also $A_3$ is abelian, I think. So At least I get an example for this. That is we cannot dis-prove this statement. $\endgroup$
    – LaTeXFan
    Nov 16, 2013 at 5:02
  • 1
    $\begingroup$ Another example is $Q_8=\{1,-1,i,-i,j,-j,k,-k\}$ (the elementary quaternions) where all proper subgroups are normal and abelian, and also all quotients by proper subgroups are abelian. $\endgroup$
    – egreg
    Nov 16, 2013 at 11:47
  • $\begingroup$ You could look up the concept of a semidirect product. $\endgroup$
    – Carsten S
    Nov 16, 2013 at 11:50

1 Answer 1


Hint If $H$ has prime order $p$ it is abelian. Moreover, if $G:H=2$ then $H$ is normal and $ G/H$ is also abelian.

So look for a non-abelian group of order $2p$ where $p$ is prime....


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.