2
$\begingroup$

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 $k$?

$\endgroup$
  • 1
    $\begingroup$ Yes, in a free product of cyclics,the centraliser of any non-trivial element is cyclic. $\endgroup$ – James Jul 17 '14 at 17:00
4
$\begingroup$

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 $\;H=K=1\;$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.