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?

  • $\begingroup$ What is $n$ in your first sentence? $\endgroup$ – Mariano Suárez-Álvarez May 21 '13 at 0:21
  • $\begingroup$ I put it in quotes to duck the definition. I guess it is the derived length. $\endgroup$ – Matt Brin May 21 '13 at 0:24
  • $\begingroup$ All answers that have appeared so far have contributed greatly to my understanding. Thanks very much. Sorry if I don't pick one as "the" answer. $\endgroup$ – Matt Brin May 22 '13 at 6:02

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.

  • $\begingroup$ I get X_1={(123), (142)}, and X_2={(14)(23), (12)(34)} which seems to generate Klein-4. I should recheck. $\endgroup$ – Matt Brin May 21 '13 at 0:22
  • 1
    $\begingroup$ Your example is right if you go to a bigger group. For S_n, the set X_1 is {(123), (12n)}. Then X_2 still generates a Klein-4 group. Then X_3 is trivial, but S_n is not solvable. $\endgroup$ – Matt Brin May 21 '13 at 1:32
  • $\begingroup$ Write $a=(12)$, $b=(1234)$. Then $X_1$ just has the two elements $a^{-1}b^{-1}ab$ and $b^{-1}a^{-1}ba$ (and the identity), right? And I think these come to $(132)$ and $(123)$, respectively. $\endgroup$ – Gerry Myerson May 21 '13 at 6:25
  • $\begingroup$ one commutator is the product of (12) and (23) and the other is the product of (12) and (14). That is (12) multiplied (12) conjugated by the rotation (1234) "one way" and (12) multiplied by (12) conjugated by the rotation (1234) "the other way." I still get (123) and (124). $\endgroup$ – Matt Brin May 22 '13 at 5:59
  • $\begingroup$ oops, see my comment to Alireza's answer. $\endgroup$ – Matt Brin May 22 '13 at 17:01

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.


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.


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.

  1. 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 $[[x_1,x_2],[x_3,x_4]]$.

  • $\begingroup$ @Alireza: Isn't it what Matt wrote in his comment to Gerry's answer. I think the only non-clear remaining part of the question is whether $X_2=1$ implies $G_m=1$ for some $m$. $\endgroup$ – Mark Sapir May 22 '13 at 8:04
  • $\begingroup$ I now understand my confusion and my second error. Gerry is right since he interpreted my notation literally and I did not. In [X,X], I erred in letting the elements run over X and also the inverses of elements in X. $\endgroup$ – Matt Brin May 22 '13 at 17:01

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