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Let $G$ be a finite group such that $G'\cap Z(G)\neq 1$. Suppose also that $G'$ is an elementary abelian $p$-group; $G'\nleq Z(G) $; $(G/Z(G))'$ is a minimal normal subgroup of $G/Z(G)$.
Can we deduce that $(G/Z(G))'\cap Z(G/Z(G))\neq 1$?

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    (1) The center of a group is Z(G), capital "z"; (2) Background, insights...about this question?2012-12-21
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    I have to prove an equivalent assert of "$G$ is also not abelian, but every proper subgroup of $G$ is abelian". At some point in the proof we assume by contradiction $G'\cap Z(G)\neq 1$. Immediately he deduce $ (G/Z(G))′∩Z(G/Z(G))≠1$. Other than the previously results we have also that: The center of $G$ coincide with the Frattini subgroup of $G$; $G$ is finite; $G=G'Z(G)C$ where $C$ is a cyclic subgroup.2012-12-21
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    "He" is Reynolds Bear, who generally explicit all, every step.2012-12-21
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    This smells like character theory, is this for a representation theory course?2012-12-21
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    Is G' a subgroup of G2012-12-21
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    No... The argument is "Topics in Finite Groups: Minimal Classes"2012-12-21
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    $G'$ is the commutator subgroup of $G$2012-12-21
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    Another observation: if you have that $Z(G)=\Phi(G)$, then $G$ can't be nilpotent, or we'd have $(G/Z(G))'=1$.2012-12-21
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    Neat question though,! I would find it very interesting to hear an answer to "When does $G'\cap Z(G))\not= 1$ imply that $(G/Z(G))'\cap Z(G/Z(G))'\not= 1$?"2012-12-21

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