A von Dyck group is a group with presentation $< a,b | a^m=b^n=(ab)^p=1 >$ with m,n,p natural numbers. Is it known which of these groups are solvable and which are not? Is there a reference for this? Thanks.

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    $\begingroup$ The groups with $\frac{1}{m} + \frac{1}{n} + \frac{1}{p} > 1$ are all finite, and all solvable except $\Delta(2,3,5) \cong A_5$. The groups with $\frac{1}{m} + \frac{1}{n} + \frac{1}{p} = 1$ are all infinite and solvable: their commutator subgroups are isomorphic to $\mathbb{Z}^2$. The groups with $\frac{1}{m} + \frac{1}{n} + \frac{1}{p} < 1$ are all infinite and nonsolvable: indeed, with only finitely many exceptions, each of these groups has a simple group $PSL_2(\mathbb{F}_q)$ as a quotient. $\endgroup$ Jun 21, 2010 at 2:27
  • $\begingroup$ Thank you for the comment. Do you know where I can find a proof for the case of 1/m + 1/n + 1/p = 1? $\endgroup$
    – dave
    Jun 21, 2010 at 3:41
  • $\begingroup$ These days, these groups are usually called triangle groups. You might like to look at the answers to the following question, which was very similar. mathoverflow.net/questions/22459/x-y-xp-yp-xyp-1/22463#22463 $\endgroup$
    – HJRW
    Jun 22, 2010 at 21:27

1 Answer 1


You might try Generators and Relations for Discrete Groups by Coxeter and Moser.

Specifically for 1/m + 1/n + 1/p = 1 there are only 3 cases up to permutation, (2,3,6), (2,4,4) and (3,3,3). Map a and b to an appropriate root of unity to get a homomorphism onto C_6, C_4, or C_3, respectively. The kernel of the map is in all three cases isomorphic to Z^2.


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