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For a given group presentation of a group(finitely presented), I want to check whether it is residually solvable or not. Is there any good way to do it?

Actually, I'm curious whether the finitely presented group $G = < x_i, 1\leq i \leq m | w_i x_i w_i^{-1}x_{i+1}, 1\leq i \leq n-1 > $, where $w_i=x_j^{\pm 1}$ for some $j$.

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@HJ: There are misprints in the definition of $G$. Could you please correct them? Is it supposed to be an LOG presentation? – Mark Sapir Dec 8 '10 at 21:36
HJ - as stated, this is just a free group. (Killing a conjugate of $x_i$ is the same as killing $x_i$.) I presume you're aware of Magnus's Theorem that free groups are residually nilpotent (in particular, residually solvable). – HJRW Dec 8 '10 at 21:42
@Henry, it is a clear misprint. He probably meant an LOG presentation. – Mark Sapir Dec 8 '10 at 21:53
@Mark, Yes, I corrected the presentation. There were missing $x_[i+1}$. – hjjang Dec 9 '10 at 20:29

This problem is undecidable. If it were not, you could use it to construct an algorithm for testing if a given f.p. group is trivial or not, which is well known to be undecidable:

Input: f.p. group G

  • Test if G is residually solvable.
  • If it is not, output "non-trivial".
  • If it is, find the abelianization of G.
  • If the abelianization is trivial, output "trivial".
  • Otherwise output "non-trivial".
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Let me add that decidability of this problem is still open for one-relator groups: – Jon Bannon Dec 8 '10 at 13:14

It is not clear whether the set of finite residually solvable group presentations is even recursively enumerable. Unlike for the word or triviality problem where there exists an algorithm which says "yes" iff the answer is "yes", I do not think there exists such an algorithm in this case. But I do not think anybody proved that the algorithm does not exist. Same for residually finite groups.

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