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. 



A conjecture on solvablity of finite groups
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.


