Can group solvability be detected from identities among the generators? For $n=1$ the answer is "yes." -- A group is abelian iff its generators commute.
Let $G_0=G$ be a group and let it be generated by $X_0=X$.  For each $n>0$ let $G_n=[G_{n-1},G_{n-1}]$ and let $X_n=[X_{n-1},X_{n-1}]$.  If $X_n$ is trivial, is $G_n$ trivial?
 A: If I'm not mistaken, $G_0=S_4$ is generated by $X_0=\{(12),(1234)\}$. $G_1=A_4$, $X_1=\{(123),(132)\}$, $G_2$ is Klein-4, $X_2$ is trivial. 
A: For finite groups, solvability can be detected from Engel-like identities. This was not really the question, but
it is very interesting in this context, I think.
The proof is surprisingly complicated, relying on reduction to J.Thompson's list of minimal non-solvable simple groups, on extensive use of arithmetic geometry and on computer algebra and geometry - see http://arxiv.org/abs/math/0303165.
A: This is not true for $n=2$. Indeed, if $X_2=1$ implies $G_2=1$, then the free metabelian group with, say, 2 generators would be finitely presented since $X_2$ is finite, which is not true (Bieri-Strebel). In fact, I think $X_2=1$ does not imply $G_m=1$ for any $m$ but it seems to be harder to prove. 
A: See Exercise 6 in Page 49 of the following book of  E. I. Khukhro:
$p$-Automorphisms of Finite $p$-Groups, London Mathematical Society Lecture Note Series, 246,  Cambridge University Press, Cambridge, 1998. 


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*Let $G$ be a group generated by two elements $x$ and  $y$. Show that the law $\delta_2$ 
of solubility of derived length 2 holds on the generators $x, y$ (while $G$ may 
not be soluble: for example, $\mathbb{S}_5$ is generated by a cycle of order $5$ and a 
transposition). 


The law $\delta_2$ is $[[x_1,x_2],[x_3,x_4]]$.
