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 $(G/Z(G))'\cap Z(G/Z(G))\neq 1$?
A question about intersection of center and commutator subgroup
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abstract-algebra
group-theory
finite-groups
examples-counterexamples
p-groups
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0(1) The center of a group is Z(G), capital "z"; (2) Background, insights...about this question? – 2012-12-21
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0I 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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0"He" is Reynolds Bear, who generally explicit all, every step. – 2012-12-21
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0This smells like character theory, is this for a representation theory course? – 2012-12-21
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0Is G' a subgroup of G – 2012-12-21
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0No... The argument is "Topics in Finite Groups: Minimal Classes" – 2012-12-21
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0$G'$ is the commutator subgroup of $G$ – 2012-12-21
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0Another 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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1Neat 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