Let $m$ and $n$ be to positive integers strictly larger than $1$. Is it possible to find a group $G$ in which there are two elements, say $a$ and $b$, such that the order of $a$ is $m$, the order of $b$ is $n$ and their product has infinite order?
If $m$ and $n$ are not coprime I think I have solved using a restricted wreath product (which is then directly summed to some cyclic group); but in the general case I could not achieve anything.
On
You could just give the presentation: $\langle a,b|a^n=b^m=e \rangle$ A word that is equivalent to the zero word satisfies each of its $a$-blocks is a multiple of $m$ and each of its $b$ blocks is a multiple of $n$. but the first block of $a$'s of a word of the form $(ab)^n$ does not have a multiple of $m$ quantity of $a$'s. Then $(ab)^n$ is never the zero word.
If you already talk of wreath product then I think you're ready for the following: in the free product of any two non-trivial groups, the product $\;ab\;$ , when each letter belongs to a different free factor, always has infinite order.
For example, if $\;C_2=\langle \,a\,\rangle\; \;,\;\;C=\langle\, b\,\rangle\;$, then in $\;C_2*C_3\;$ we have that $\;ord(a)=2\;,\;\;ord(b)=3\;$ , but certainly $\;ord(ab)=\infty\;$ .