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Is there an example of a finitely generated (infinite) residually finite group $\Gamma$ for which every linear representation of $\Gamma$ has finite image?

flag	asked Oct 10 2010 at 6:17 anon 41●1

	By linear representation, do you mean finite-dimensional linear representation? – Yemon Choi Oct 10 2010 at 6:20
	Yes, and over a field of characteristic zero. – anon Oct 10 2010 at 6:31
	I erased my answer, since it had the same properties as the Grigorchuk group (and is based on unpublished work), but I wanted to point out that Mark Sapir commented that the original construction of an infinite torsion residually finite group was due to Golod. en.wikipedia.org/wiki/Periodic_group – Agol Oct 11 2010 at 17:42

Pete L. Clark · Answer 1 · 2010-10-10 19:51:42Z UTC

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A group $G$ is just infinite if it is infinite but every proper quotient is finite.

Clearly a just infinite group which is not linear has the property that its image under any linear representation is finite. Thus any group which is finitely generated, residually finite, not linear and just infinite is an example of what you want: for instance, the Grigorchuk group.

edited Oct 10 2010 at 19:51

Pete L. Clark

answered Oct 10 2010 at 7:02

Mustafa Gokhan Benli
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It's not hard to prove that the Grigorchuk grp is not linear. For instance, the Tits alternative says that every fg linear grp G either contains a solvable subgroup of finite index or contains a nonabelian free subgroup. This implies that the "growth function" f(n) of G (here f(n) is the number of elements of G of length at most n in a fixed genset) grows either polynomially (if G has a solvable subgrp of finite index) or exponentially (if G contains a nonabelian free subgrp). However, the first major thm about the Grigorchuk grp is that its growth fcn is superpolynomial but subexponential. – Andy Putman Oct 10 2010 at 20:54

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By the way, I highly recommend reading the final chapter in Pierre de la Harpe's book "Topics in Geometric Group Theory", which is entirely devoted to the Grigorchuk group. It serves as a sort of "universal counterexample" to conjectures in geometric group theory. – Andy Putman Oct 10 2010 at 21:09

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@Andy: solvable groups can have exponential growth. For example ${\mathbb Z}\wr {\mathbb Z}$ is solvable of class 2 and has exponential growth because it contains a free non-cyclic subsemigroup. – Mark Sapir Oct 10 2010 at 22:54

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@Pete, thanks for the far better rewriting of my answer. – Mustafa Gokhan Benli Oct 10 2010 at 23:59

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Andy - with a 'virtually' in front of 'nilpotent', that's the Milnor--Wolf Theorem. – HW Oct 11 2010 at 2:37

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A special residually finite group

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