7
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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

5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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