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 $\mathbb{Z}\_3$? 



Group which "resembles" the free product of a cyclic group of order two and a cyclic group of order three, but isn't. 
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.


