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 $\mathbb{Z}_3$?

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 $G:D=6$. 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. 


The free product $\mathbb Z_2$ and $\mathbb Z_3$ (i.e. PSL(2, $\mathbb Z$) is Gromovhyperbolic (as every virtually free group) and nonvirtually cyclic. Therefore by a result of Olshanskii, "SQuniversality of hyperbolic groups". (Russian) Mat. Sb. 186 (1995), no. 8, 119132; translation in Sb. Math. 186 (1995), no. 8, 1199–1211, it is SQuniversal, 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. 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. 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 CannonFloydParry's survey on Thompson groups, but I do not have it with me. Update 2: I cannot find the reference to the result about $V$. It is not in CannonFloydParry. 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). 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). 


I don't know an explicit example off hand, but I would recommend looking at the generalized triangle groups $\langle a,b \  \ a^2 = b^3 = 1 = w^k \rangle$ 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. See Baumslag, Morgan, Shalen, "Generalized triangle groups" Math. Proc. Camb. Phil. Soc. (1987) 102, page 25 and Fine, Rosenberger, "A note on generalized triangle groups" ABHANDLUNGEN AUS DEM MATHEMATISCHEN SEMINAR DER UNIVERSITÄT HAMBURG Volume 56, Number 1, 233244 


I am hazarding a guess, I believe this should do the job A = { {1, x, 0}, {0, 1, 0}, {0, y, 1} } It really does not matter what x and y are, they can be chosen arbitrarily and can even be two formal symbols B = { {0, 0, $i$}, {$i$, 0, 0}, {0, 1, 0} } Then A.A= B.B.B = Id 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 ? 

