Which group does not satisfy the Tits alternative? - MathOverflow

Which group does not satisfy the Tits alternative?

2010-09-14T15:54:56Z

A group is said to satisfy the Tits alternative if every finitely generated subgroup of $G$ is either virtually solvable or contains a nonabelian free subgroup.

Tits proved this for linear groups, and a MathSciNet search gives 38 papers with "Tits alternative" in the title (and 154 papers quoting Tits's original paper), so certainly a lot of groups do enjoy this property.

What then is an example of a group which does not satisfy the Tits alternative?

Answer by Igor Pak for Which group does not satisfy the Tits alternative?

2010-09-14T16:10:38Z

Though this is not the only example of this kind, I think you might like to study the Grigorchuk group. This Wiki page has lots of information, so there is little point of repeating it. Enjoy! -- IP

P.S. I am especially partial to the arXiv preprint mentioned on the bottom of that article, though, apparently, Wikipedia does not realize that it was published awhile ago... :)

Answer by Maxime Bourrigan for Which group does not satisfy the Tits alternative?

2010-09-14T16:12:27Z

Thompson's group F is also an example: it isn't virtually solvable (actually, its commutator subgroup is simple) and does not contain a free subgroup, according to a theorem of Brin and Squier ("Groups of piecewise linear homeomorphisms of the real line.", Invent. Math. 79 (1985))).

You can find a survey about this group (and two cousins of his) written by Cannon, Floyd and Parry on Brin's webpage at http://www.math.binghamton.edu/matt/thompson/cfp.pdf

Answer by Michele Triestino for Which group does not satisfy the Tits alternative?

2010-09-14T16:22:18Z

There are also the Burnside's groups $B(m,n)$ for $n\ge 665$ odd: they are of exponential growth and have the law $x^n=1$ so that they cannot contain any free subgroup on two generators. The fact that they are not solvable follows by the theorem of Rosenblatt:

"A f.g. solvable group is of exponential growth if and only if it contains a free sub-semigroup on two generators."

You can find details on paragraphs VII.C.27/28 of Pierre de la Harpe's book "Topics in Geometric Group Theory" (Chicago Lectures in Mathematics, 2000)

Answer by Francesco Polizzi for Which group does not satisfy the Tits alternative?

2010-09-14T16:26:00Z

The paper of Hartley

A conjecture of Bachmuth and Mochizuki on automorphisms of soluble groups. Canad. J. Math. 28 (1976), no. 6, 1302--1310

provides many counterexamples.

Let me quote from MathSciNet review:

"J. Tits [J. Algebra 20 (1972), 250--270; MR0286898 (44 #4105)] showed that a finitely generated linear group either contains a soluble subgroup of finite index or else contains a nonabelian free subgroup. S. Bachmuth and H. Y. Mochizuki [Bull. Amer. Math. Soc. 81 (1975), 420--422; MR0364452 (51 #706)] conjectured that a finitely generated group of automorphisms of a finitely generated soluble group $G$ satisfies the same conclusion. The conjecture is known to be true when $G$ is nilpotent-by-abelian. This paper shows that the conjecture is false in general and a family of counterexamples is constructed. In particular there are such examples where $G$ has derived length three".

Answer by Richard Borcherds for Which group does not satisfy the Tits alternative?

2010-09-14T16:41:13Z

Tarski monsters are counterexamples to almost anything.

Answer by Andreas Thom for Which group does not satisfy the Tits alternative?

2010-09-14T18:08:17Z

In the generalized sense of measurable group theory, every infinite group satisfies the Tits alternative. Indeed, every group is either amenable and hence orbit equivalent (isomorphic in the category of groups with randomorphisms) to ${\mathbb Z}$ or non-amenable and hence contains ${\mathbb F}_2$ as a random subgroup. The second result is recent and due to Damien Gaboriau and Russell Lyons (see here).

The notion of randomorphism is due to Nicolas Monod, see his ICM talk from 2006 (see here).

EDIT: Answering Henry's comment: $H$ is a random subgroup of $G$ if there is a $H$-equivariant probability measure on the space of maps $\lbrace f: H \to G \mid f(e)=e \rbrace$ endowed with the action $(h.f)(k)= f(kh)f(h)^{-1}$; supported on injective maps. Clearly, every injective homomorphism yields an atomic randomorphism, but there are others.

Answer by Agol for Which group does not satisfy the Tits alternative?

2010-09-15T04:20:17Z

$A_5\wr \mathbb{Z}$