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.

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. 


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. 

