Group which "resembles" the free product of a cyclic group of order two and a cyclic group of order three, but isn't. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:40:07Z http://mathoverflow.net/feeds/question/41922 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41922/group-which-resembles-the-free-product-of-a-cyclic-group-of-order-two-and-a-cyc Group which "resembles" the free product of a cyclic group of order two and a cyclic group of order three, but isn't. Todd Trimble 2010-10-12T17:30:00Z 2010-10-12T22:12:37Z <p>Can someone give an explicit example of a group with two generators $a$, $b$, such that $a^2 = b^3 = 1$ and $a b$ has infinite order, but which is not isomorphic to the free product of $\mathbb{Z}_2$ and $\mathbb{Z}_3$? </p> http://mathoverflow.net/questions/41922/group-which-resembles-the-free-product-of-a-cyclic-group-of-order-two-and-a-cyc/41926#41926 Answer by Richard Kent for Group which "resembles" the free product of a cyclic group of order two and a cyclic group of order three, but isn't. Richard Kent 2010-10-12T18:23:54Z 2010-10-12T18:23:54Z <p>I don't know an explicit example off hand, but I would recommend looking at the generalized triangle groups</p> <p>$\langle a,b \ | \ a^2 = b^3 = 1 = w^k \rangle$</p> <p>where $w$ is a word in $a$ and $b$. Baumslag, Morgan, and Shalen given conditions on when this virtually surjects $\mathbb{Z}$ or a free group of rank two. I would suspect that it wouldn't be too tough to find an explicit example where $ab$ has infinite order.</p> <p>See</p> <p>Baumslag, Morgan, Shalen, "Generalized triangle groups" Math. Proc. Camb. Phil. Soc. (1987) 102, page 25</p> <p>and </p> <p>Fine, Rosenberger, "A note on generalized triangle groups" ABHANDLUNGEN AUS DEM MATHEMATISCHEN SEMINAR DER UNIVERSITÄT HAMBURG Volume 56, Number 1, 233-244</p> http://mathoverflow.net/questions/41922/group-which-resembles-the-free-product-of-a-cyclic-group-of-order-two-and-a-cyc/41927#41927 Answer by Mark Sapir for Group which "resembles" the free product of a cyclic group of order two and a cyclic group of order three, but isn't. Mark Sapir 2010-10-12T18:24:21Z 2010-10-12T22:12:37Z <p>The free product $\mathbb Z_2$ and $\mathbb Z_3$ (i.e. PSL(2, $\mathbb Z$) is Gromov-hyperbolic (as every virtually free group) and non-virtually cyclic. Therefore by a result of Olshanskii, "SQ-universality of hyperbolic groups". (Russian) Mat. Sb. 186 (1995), no. 8, 119--132; translation in Sb. Math. 186 (1995), no. 8, 1199–1211, it is SQ-universal, that is every countable group embeds into a factor group of PSL(2, $\mathbb Z$). In "most" of these groups (by construction) $ab$ will have infinite order. Thus, in particular, there are uncountably many groups of the type you want. </p> <p>Update 1: An explicit example would be this. Take $G=PSL(2,\mathbb Z)$, and any word $w(a,b)$ satisfying very small cancelation (that it no subword of length, say, $\frac{1}{10000}|w|$ occurs twice in $w$ (considered as a cyclic word). Then consider the group $G/\langle\langle w\rangle\rangle$. It is what you want. Geometrically, you just kill the large loop in the standard $K(\pi,1)$ for $PSL(2,\mathbb{Z})$ of course. </p> <p>Another example, as far as I remember, is the R. Thompson group $V$ (it is generated by an element $a$ of order 2 and an element $b$ of order 3 such that $ab$ has infinite order (Mason?). It should be written in the Cannon-Floyd-Parry's survey on Thompson groups, but I do not have it with me. </p> <p>Update 2: I cannot find the reference to the result about $V$. It is not in Cannon-Floyd-Parry. But here is a paper where it is proved that $SL(n,{\mathbb Z})$ is generated by an element of order 2 and an element of order 3, provided $n\ge 13$: Sanchini, Paolo; Tamburini, M. Chiara, Constructive $(2,3)$-generation: a permutational approach. Rend. Sem. Mat. Fis. Milano 64 (1994), 141–158 (1996). </p> <p>Update 3: The paper cited in Update 2 follows this paper: Tamburini, M. Chiara; Wilson, John S.; Gavioli, Norberto On the $(2,3)$-generation of some classical groups. I. J. Algebra 168 (1994), no. 1, 353–370. The result there is quite general (and nice), the generating matrices are explicitly given. To check that $ab$ has infinite order, one just needs to find the characteristic polynomial of $ab$ and show that some roots are not roots of unity. That should be straightforward (using any CAS). </p> http://mathoverflow.net/questions/41922/group-which-resembles-the-free-product-of-a-cyclic-group-of-order-two-and-a-cyc/41929#41929 Answer by Vagabond for Group which "resembles" the free product of a cyclic group of order two and a cyclic group of order three, but isn't. Vagabond 2010-10-12T18:27:59Z 2010-10-12T20:16:05Z <p>I am hazarding a guess, I believe this should do the job</p> <p>A = { {1, x, 0}, {0, -1, 0}, {0, y, 1} }</p> <p>It really does not matter what x and y are, they can be chosen arbitrarily and can even be two formal symbols</p> <p>B = { {0, 0, -$i$}, {$i$, 0, 0}, {0, 1, 0} }</p> <p>Then A.A= B.B.B = Id </p> <p>the order of $A.B$ would be infinite when $x$ and $y$ are suitably chosen, for example one can choose $x$ and $y$ so that the coefficient of the matrices $(A.B)^n$ unbounded ? </p> http://mathoverflow.net/questions/41922/group-which-resembles-the-free-product-of-a-cyclic-group-of-order-two-and-a-cyc/41944#41944 Answer by Derek Holt for Group which "resembles" the free product of a cyclic group of order two and a cyclic group of order three, but isn't. Derek Holt 2010-10-12T21:16:47Z 2010-10-12T21:16:47Z <p>It is straightforward to calculate that the commutator subgroup $G' = D$ of $G = \langle a,b \mid a^2, b^3 \rangle$ is a free group on the generators $x=bab^{-1}a$, $y=b^{-1}aba$, where $|G:D|=6$.</p> <p>Now $(ab)^6$ is equal to the commutator $x^{-1}yxy^{-1}$, which lies in $D'$ but not in $D''$, so if we add any nontrivial element of $D''$ as an extra relator of $G$, then we will get an example with the required property.</p>