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 $k$?

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    $\begingroup$ Yes, in a free product of cyclics,the centraliser of any non-trivial element is cyclic. $\endgroup$
    – James
    Jul 17 '14 at 17:00

Read Theorem $4.5$ in page $209$, section $4.2$ of Magnus-Karrass-Solitar's "Combinatorial Group Theory": it is what you want.

Basically, two elements in a free product commute iff either they're in the same cyclic subgroup or else they belong to the same conjugate of the same factor.

Be careful: the above theorem in MKS appears in way more general form for amalgamated products. For you, and using the notation in that book, we have $\;H=K=1\;$


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