Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $(G/Z(G))'\cap Z(G/Z(G))\neq 1$?

share|improve this question
    
(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
1  
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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.