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Suppose $G$ is a finite group and $A$ an abelian subgroup. Suppose for some natural number $n\geq 2$, elements of $\gamma_n(G)$ have the form $[a, x]$ where $a\in A$ and $x\in G$. Then $G$ is solvable.

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A similar statement is true for finite dimensional Lie algebras of characteristic zero. See my short note in Bull. Australian Math. Soc.(2011): A note on derivations of Lie algebras. –  M. Shahryari Dec 5 '12 at 17:03
    
and what is $\gamma_n(G)$? –  Dima Pasechnik Dec 5 '12 at 17:38
    
@Dima: it is the $n$-th term of the lower central series. –  M. Shahryari Dec 5 '12 at 18:34

2 Answers 2

up vote 10 down vote accepted

I claim that the conjecture is false, and there is a counterexample with $G=S_5$, $n=2$, and $A$ cyclic of order 6. In this case $\gamma_2(G)=[G,G]=G'=A_5$ has order $5!/2=60$. Take $A$ to be the abelian group $A=\langle(1,2,3),(4,5)\rangle$, but any conjugate of $A$ will also work. A simple computation shows that the set $C:=\{[a,x]\mid a\in A,\ x \in G\}$ also has 60 elements. Since $C\subseteq A_5$, this proves that $C=A_5$. However, $G$ is clearly not solvable.

What a pity, the motivating idea for the question was attractive.

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I've just recovered my previous account and now my new and old accounts are merged. I am sorry for this long delay of accepting your answer. I hope you will accept my excuse. –  M. Shahryari Feb 26 at 15:54

A remark in a special case.

Let $H$ be a finite group, $a$ a fixed point free automorphism of $H$, and $A = \langle a \rangle$. Let $G$ be the semidirect product of $H$ by $A$. Then it is well known that every element of $H$ is of the form $[a, h]$, for $h \in H$, so that $H = \gamma_2(G)$.

Now it is indeed true that $H$ (and thus $G$) is soluble in this case, but the proof requires the classification of finite simple groups.

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@Andreas Caranti: yes and this point was my motivation in fact: it was proved by F. Ladish may be 4 years ago in Comm. Algebra using CFSG. I proved a version of it for Lie algebras using Cartan criterion (above mentioned paper). The above conjecture was one of my favorite problems during the past 3 years. –  M. Shahryari Dec 13 '12 at 20:46

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