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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.

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 Cannon-Floyd-Parry'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 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).

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).

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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.

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 Cannon-Floyd-Parry'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 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).

2 added 723 characters in body

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.

Update: 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 Cannon-Floyd-Parry's survey on Thompson groups, but I do not have it with me.

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