Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

7 Answers 7

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

share|improve this answer
    
I'm a bit confused by the first parenthetical remark, which seems to suggest that simplicity of the commutator subgroup implies virtual solvability. Certainly this is not literally true -- e.g. $S_n$ for $n \geq 5$. But even supposing that the commutator subgroup $G'$ is infinite, I'm still not quite seeing it: off the top of my head, I would think that you need to know also that $G'$ has finite index in $G$. Please help... –  Pete L. Clark Sep 14 '10 at 18:51
2  
What I meant is : 1. a subgroup of a virtually solvable group is virtually solvable. So F cannot be virtually solvable if F' isn't. 2. An infinite simple group has no finite-index subgroup at all (it is easy to show that every finite-index subgroup contains a finite-index normal subgroup). So, if it were virtually solvable, it would be solvable, but that's absurd. –  Maxime Bourrigan Sep 14 '10 at 20:09
    
@MB: Thanks for the help. For some reason I was vaguely doubtful of 1., but now that you assert it I see how to prove it (a subgroup of a solvable group is solvable!). –  Pete L. Clark Sep 15 '10 at 2:02
add comment

Tarski monsters are counterexamples to almost anything.

share|improve this answer
add comment

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)

share|improve this answer
1  
It seems it could be made simplier, without referring to the theorem of Rosenblatt. The Burnside groups are free groups in the variety of groups determined by the law x^n = 1, so would they be virtually solvable, every infinite group satisfying the law x^n = 1 would be solvable. The latter is obviously not true. Similarly, a free group in any variety of group containing infinite non-solvable groups is not virtually solvable (and, obviously, does not contain a free subgroup). –  Pasha Zusmanovich Sep 18 '10 at 8:28
    
I am taking this back: "obviously" in my previous comment is blatantly wrong (if one replaces solvability by nilpotency, this is what Burnside's problems about, far from being "obvious"). –  Pasha Zusmanovich Sep 18 '10 at 9:40
add comment

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... :)

share|improve this answer
3  
As far as I remember the following is true: the Grigorchuk group has intermidia word growth. Now linear groups have either polynomial or exponential word growth and the Tits alternative played an important role in the original proof of this. So the Grigorchuk group is even more than not satisfying the Tits alternative. –  Yiftach Barnea Sep 14 '10 at 17:04
add comment

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.

share|improve this answer
1  
Could you say a little bit more about what 'contains $\mathbb{F}_2$ as a random subgroup' means? –  HJRW Sep 14 '10 at 18:54
add comment

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".

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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