If $G$ is non-abelian group and $Z(G)$ is it's center, what is the least property for $G$ such that $\frac{G}{Z(G)}$ is abelian?
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4$G/H$ is abelian if and only if $H$ contains the commutator subgroup. – Daniel Fischer Aug 16 '15 at 12:59
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Specializing @DanielFischer , you need for all $a,b\in G$, $aba^{-1}b^{-1}\in Z(G)$. – Michael Burr Aug 16 '15 at 13:12
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1There are some interesting special cases that come up some times, such as if $G/Z(G)$ is cyclic, then $G$ is abelian. – Michael Burr Aug 16 '15 at 13:13
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1@MichaelBurr It definitely comes up here more frequently than "some times", e.g., see here. – Dietrich Burde Aug 16 '15 at 14:57