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I'd like to know the Growth Function of the Triangle Groups. From the Gromov's theorem we know that every virtually nilpotent group has a polynomial growth. It seems that T(2,4,4) is virtually nilpotent. How can I prove it? More generally, are Triangle Groups always nilpotent? Why? Thank you!

Deraam
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1 Answers1

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By "triangle group" I mean the group $T(i, j, k)=\langle a, b, c\mid a^i, b^j, c^k, abc\rangle$. If you mean the group where every generator has order two then my group has index two in your group. As this question is about "virtual" properties, this is not an issue!

To see that $T(2, 4, 4)$ is virtually-nilpotent, you can prove that the commutator subgroup is abelian. Can you see why this will work? To prove that the commutator subgroup is abelian you can use the Reidemeister-Schreier method to find a presentation of it; I wrote an answer with some links about this method here.

For your "more generally" question:

  1. If $1/i+1/j+1/k<1$ then the triangle group $T(i, j, k)$ is not virtually nilpotent, and indeed has exponential growth. This is because it is hyperbolic, and so contains a free subgroup of rank two.
  2. If $1/i+1/j+1/k>1$ then the triangle group $T(i, j, k)$ is finite so virtually nilpotent.
  3. If $1/i+1/j+1/k=1$ then there are three groups in this class, $T(2,3,6)$, $T(2,4,4)$, and $T(3,3,3)$. In the comments below, Derek Holt has pointed out that all three are virtually abelian, in fact virtually-$\mathbb{Z}^2$, which can be seen by proving that in each case the derived subgroup is isomorphic to $\mathbb{Z}^2$ (that is, using the same idea as my suggestion above).
user1729
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    In fact $T(2,3,6)$, $T(2,4,4)$ and $T(3,3,3)$ are all virtually (free) abelian. You can prove this directly by calculating a presentation of $[G,G]$, which has finite index and is free abelian of rank $2$ in all three cases. – Derek Holt Jun 10 '19 at 11:03