In the free group with two generators $F_2\cong\mathbb Z *\mathbb Z$ ($*$ denotes the free product), if two elements $a$ and $b$ commute, then there exists an element $w\in F_2$ such that $\langle a,b\rangle=\langle w\rangle$, that is, the group generated by $a$ and $b$ is given by the cyclic group generated by $w$, or, in other words, $\langle a,b\rangle\cong\mathbb Z$ (in fact, this is true in any free group).
Question. Suppose that, instead of the free group $F_2$, we consider a free product of not necessarily infinite cyclic groups $G=C_1*C_2$ (for example, $\mathbb Z/n\mathbb Z * \mathbb Z$, or $\mathbb Z/n\mathbb Z*\mathbb Z/m\mathbb Z$). Given two elements $a,b\in G$ that commute, is there a similar result for $\langle a,b\rangle$, i.e. $\langle a,b\rangle\cong\mathbb Z$ or $\langle a,b\rangle\cong\mathbb Z/k\mathbb Z$ for some integer $k$?