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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? – DonAntonio Dec 21 '12 at 17:54
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. – W4cc0 Dec 21 '12 at 18:02
"He" is Reynolds Bear, who generally explicit all, every step. – W4cc0 Dec 21 '12 at 18:05
This smells like character theory, is this for a representation theory course? – Alexander Gruber Dec 21 '12 at 18:07
Neat question though,! I would find it very interesting to hear an answer to "When does $G'\cap Z(G))\not= 1$ imply that $(G/Z(G))'\cap Z(G/Z(G))'\not= 1$?" – Alexander Gruber Dec 21 '12 at 18:58

No, we can't. Minimal counterexample: $G=\text{SmallGroup}(96,197)$.

In here, $G'\cong \mathbb{Z}_2\times\mathbb{Z}_2\times \mathbb{Z}_2$ and $Z(G)\cong \mathbb{Z}_2$.

$G/Z(G)\cong\text{SmallGroup}(48,49)$, and $(G/Z(G))'\cong \mathbb{Z}_2\times\mathbb{Z}_2$ is a minimal normal subgroup of $G/Z(G)$. We have that $Z(G/Z(G))\cong \mathbb{Z}_2\times\mathbb{Z}_2$ as well, but the two intersect trivially.

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First of all, Thanks for your answer. Here's my real problem 1 2 3 4 5 6 7 . My question refers to the last picture. I obviously miss a condition that runs everything... Finding it would provide an answer also to your question above (at least in some particular case). – W4cc0 Dec 22 '12 at 12:28

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