3
$\begingroup$

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?

Improve this question
$\endgroup$
3
  • $\begingroup$ What is $n$ in your first sentence? $\endgroup$
    – Mariano Suárez-Álvarez
    May 21, 2013 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, 2013 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, 2013 at 6:02

4 Answers 4

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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, 2013 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, 2013 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, 2013 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, 2013 at 5:59
  • $\begingroup$ oops, see my comment to Alireza's answer. $\endgroup$
    – Matt Brin
    May 22, 2013 at 17:01
5
$\begingroup$

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.

Improve this answer
$\endgroup$
4
$\begingroup$

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.

Improve this answer
$\endgroup$
2
$\begingroup$

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

Improve this answer
$\endgroup$
2
  • $\begingroup$ @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 $m$. $\endgroup$
    – user6976
    May 22, 2013 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, 2013 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.