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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!

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  • $\begingroup$ could you explain what are your thoughts on this question... $\endgroup$
    – user87543
    Nov 16, 2013 at 4:52
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    $\begingroup$ 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
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    $\begingroup$ 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

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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....

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