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If $G$ is a group, $N \trianglelefteq G$ is a normal abelian subgroup, and $\frac{G}{N}$ is abelian, is $G$ necessarily abelian?


I'm working through a proof that if $\frac{G}{Z(G)}$ is cyclic then $G$ is abelian, and I was wondering if "cyclic" could be weakened to "abelian".

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  • $\begingroup$ Consider $S_3$. $\endgroup$
    – lulu
    Sep 13 '17 at 0:08
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Hint $S_3$ has a subgroup $H$ with $3$ elements. Since this subgroup $H$ has index $2$ is normal.

Now, both $H$ and $G/H$ have cardinality a prime number, thus are cyclic, but $S_3$ is not abelian.

Note This is not true even for the case $H=Z(G)$. To see this,use the fact that every group with $p$ or $p^2$ elements is abelian (if $p$ is primes), that there exists a non-abelian group with $p^3$ elements, and that the center of a $p$-group is non-trivial.

It follows that $G=$ any non-abelian group with $p^3$ elements, $H=Z(G)$ is also a counterexample.

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