Which group does not satisfy the Tits alternative? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T20:11:41Z http://mathoverflow.net/feeds/question/38701 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38701/which-group-does-not-satisfy-the-tits-alternative Which group does not satisfy the Tits alternative? Guntram 2010-09-14T15:54:56Z 2010-09-27T09:06:58Z <p>A group is said to satisfy the <strong>Tits alternative</strong> if every finitely generated subgroup of $G$ is either virtually solvable or contains a nonabelian free subgroup.</p> <p>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.</p> <p>What then is an example of a group which does not satisfy the Tits alternative?</p> http://mathoverflow.net/questions/38701/which-group-does-not-satisfy-the-tits-alternative/38705#38705 Answer by Igor Pak for Which group does not satisfy the Tits alternative? Igor Pak 2010-09-14T16:10:38Z 2010-09-14T16:10:38Z <p>Though this is not the only example of this kind, I think you might like to study the <a href="http://en.wikipedia.org/wiki/Grigorchuk_group" rel="nofollow">Grigorchuk group</a>. This Wiki page has lots of information, so there is little point of repeating it. Enjoy! -- IP</p> <p>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... :) </p> http://mathoverflow.net/questions/38701/which-group-does-not-satisfy-the-tits-alternative/38706#38706 Answer by Maxime Bourrigan for Which group does not satisfy the Tits alternative? Maxime Bourrigan 2010-09-14T16:12:27Z 2010-09-14T16:29:39Z <p><a href="http://en.wikipedia.org/wiki/Thompson_groups" rel="nofollow">Thompson's group F</a> 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))).</p> <p>You can find a survey about this group (and two cousins of his) written by Cannon, Floyd and Parry on Brin's webpage at <a href="http://www.math.binghamton.edu/matt/thompson/cfp.pdf" rel="nofollow">http://www.math.binghamton.edu/matt/thompson/cfp.pdf</a></p> http://mathoverflow.net/questions/38701/which-group-does-not-satisfy-the-tits-alternative/38708#38708 Answer by Michele Triestino for Which group does not satisfy the Tits alternative? Michele Triestino 2010-09-14T16:22:18Z 2010-09-27T09:06:58Z <p>There are also the <a href="http://en.wikipedia.org/wiki/Burnside_group" rel="nofollow">Burnside's groups</a> $B(m,n)$ 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:</p> <p>"A f.g. solvable group is of exponential growth if and only if it contains a free sub-semigroup on two generators."</p> <p>You can find details on paragraphs VII.C.27/28 of Pierre de la Harpe's book <em>"Topics in Geometric Group Theory"</em> (Chicago Lectures in Mathematics, 2000)</p> http://mathoverflow.net/questions/38701/which-group-does-not-satisfy-the-tits-alternative/38709#38709 Answer by Francesco Polizzi for Which group does not satisfy the Tits alternative? Francesco Polizzi 2010-09-14T16:26:00Z 2010-09-14T16:26:00Z <p>The paper of Hartley</p> <p>A conjecture of Bachmuth and Mochizuki on automorphisms of soluble groups. Canad. J. Math. 28 (1976), no. 6, 1302--1310 </p> <p>provides many counterexamples.</p> <p>Let me quote from MathSciNet review:</p> <p>"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".</p> http://mathoverflow.net/questions/38701/which-group-does-not-satisfy-the-tits-alternative/38711#38711 Answer by Richard Borcherds for Which group does not satisfy the Tits alternative? Richard Borcherds 2010-09-14T16:41:13Z 2010-09-14T16:41:13Z <p><a href="http://en.wikipedia.org/wiki/Tarski_monster_group" rel="nofollow"> Tarski monsters</a> are counterexamples to almost anything.</p> http://mathoverflow.net/questions/38701/which-group-does-not-satisfy-the-tits-alternative/38722#38722 Answer by Andreas Thom for Which group does not satisfy the Tits alternative? Andreas Thom 2010-09-14T18:08:17Z 2010-09-15T05:46:47Z <p>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 (<a href="http://arxiv.org/abs/0711.1643" rel="nofollow">see here</a>).</p> <p>The notion of randomorphism is due to Nicolas Monod, see his ICM talk from 2006 (<a href="http://egg.epfl.ch/~nmonod/articles/icm.pdf" rel="nofollow">see here</a>).</p> <p><strong>EDIT:</strong> 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 <em>atomic</em> randomorphism, but there are others.</p> http://mathoverflow.net/questions/38701/which-group-does-not-satisfy-the-tits-alternative/38775#38775 Answer by Agol for Which group does not satisfy the Tits alternative? Agol 2010-09-15T04:20:17Z 2010-09-16T03:37:54Z <p><a href="http://en.wikipedia.org/wiki/Wreath_product" rel="nofollow">$A_5\wr \mathbb{Z}$</a> </p>