# 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.

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!

• could you explain what are your thoughts on this question... – user87543 Nov 16 '13 at 4:52
• One way is to just start writing down non-abelian groups and checking if this holds. In fact, try writing down $S_3$.... – user61527 Nov 16 '13 at 4:53
• 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. – LaTeXFan Nov 16 '13 at 5:02
• 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. – egreg Nov 16 '13 at 11:47
• You could look up the concept of a semidirect product. – Carsten S Nov 16 '13 at 11:50

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....