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I was wondering when is the group $$\langle a,b \mid a^i, b^j, (ab)^k \rangle$$ finite?


Here are some examples:

Tetrahedral, Octahedral and Icosahedral groups: $\langle s,t \mid s^2, t^3, (st)^3 \rangle\,\!$, $\langle s,t \mid s^2, t^3, (st)^4 \rangle\,\!$, $\langle s,t \mid s^2, t^3, (st)^5 \rangle\,\!$

$D_{2n}$: $\langle r, f| r^n, f^2, (rf)^2\rangle$

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  • $\begingroup$ Is there a finite group presented this way with none of i,j,k equal to 2? $\endgroup$ – user58512 Feb 5 '13 at 20:26
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    $\begingroup$ Finite if and only if 1/i + 1/j + 1/k is bigger than 1. Look up triangle groups. $\endgroup$ – user641 Feb 5 '13 at 20:29
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The group presentation you have given is a presentation for a collection of groups called ordinary triangle groups or von Dyck groups. These are subgroups of larger general triangle groups.

The behavior is very different depending on whether $\frac 1i + \frac 1j + \frac 1k <1$, or $=1$, or $> 1$. These yield Hyperbolic, Euclidean, or Spherical triangle groups (respectively). Of these, only the spherical case corresponds to finite triangle groups and (correspondingly) finite von Dyck groups.

So in all, it will be finite iff $\frac 1i + \frac 1j + \frac 1k > 1$.

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