# Torsion in free products of groups

While reading a paper about the modular group $\Gamma = PSL_{2}(\mathbb{Z})$, I read that $PSL_{2}(\mathbb{Z}) \cong C_{2} * C_{3}$ and consequently, all the torsion elements in $\Gamma$ are of order 2 or 3. While I understand the isomorphism, I don't know how to prove the second statement.

More in general, is it true that if $G \cong C_{n_{1}} * \ldots * C_{n_{k}}$, then all torsion elements in $G$ have order $n_{1}, \ldots, n_{k}$ ?

• I edited the title for clarification. Mar 27 '11 at 10:01
• Well in C4 * C15 the torsion elements have orders 1,2,3,4,5,15, so "have order" should be "have order dividing one of". Textbook versions of en.wikipedia.org/wiki/Kurosh_subgroup_theorem often explain user8268's answer. Your question is the KST restricted to cyclic subgroups. Mar 27 '11 at 15:07

More generally, if $A$ and $B$ are groups, then the only torsion elements of $A*B$ are conjugates of elements of $A$ and elements of $B$; this was proven by Schreier. In particular, the order of any element of finite order must be the order of an element of $A$ or of an element of $B$.

Added. Schreier proved it as part of his construction of free products and free products with one amalgamated subgroup. The result can also be obtained as a consequence of the much stronger theorem of Kurosh; this is done, for example, in Rotman's Introduction to the Theory of Groups, Chapter 11.

Theorem. (Kurosh, 1934). If $H$ is a subgroup of a free product $\mathop{*}\limits_{i\in I} A_i$, then $H = F*\left(\mathop{*}\limits_{\lambda\in\Lambda}H_{\lambda}\right)$, for some possibly empty index set $\Lambda$, where $F$ is a free group and each $H_{\lambda}$ is a conjugate of a subgroup of some $A_i$.

As a corollary, you get

Corollary. If $G = \mathop{*}\limits_{i\in I}A_i$, then every finite subgroup of $G$ is conjugate to a subgroup of some $A_i$. In particular, every element of finite order in $G$ is conjugate to an element of finite order in some $A_i$.

• I'd like to clarify some details of the Kurosh subgroup theorem. $F$ is a free group on what set? A subset of $G$? Wikipedia claims so, when $|\Lambda|=2$. Is my formulation of KST correct in every detail: $\forall H\!\leq\!\ast_{i\in I}G_i\!=\!G$ $\exists H_i\!\leq\!G_i,\exists x_i\!\in\!G, i\!\in\!I$ $\exists X\!\subseteq\!G:$ $H=F_X\ast\big({\Large\ast}_{i\in I}\,x_iH_ix_i^{-1}\big)$?
– Leo
Aug 12 '11 at 14:12
• @Leon: free groups can have many different free basis. $F$ is a free group, possibly trivial (on the empty set). Of course, you can identify $F$ with a subgroup of the free product $G$, so it would be free on some subset of the free product. Looks about right. Aug 12 '11 at 15:54
• Thank you. Just one more thing: Wiki's formulation of KST is really strange, as there are only two groups $H\leq A\ast B$, yet it claims $H=(\ast_{i\in I}a_i A_ia_i^{-1})\ast (\ast_{j\in J}b_jB_jb_j^{-1})\ast F(X)$. Could you confirm this is wrong so I can correct it. thanks
– Leo
Aug 12 '11 at 18:45
• @Leon: No, it's not wrong. You can have more subgroups in the product than just one per factor (I did not read your previous comment carefully). For instance, in $\mathbb{Z}*\mathbb{Z}/2\mathbb{Z}$, with the first copy generated by $x$ and the second by $y$, the group generated by $y$ and $xyx^{-1}$ is the free product of two copies of $\mathbb{Z}/2\mathbb{Z}$, and you cannot get that with only one subgroup per factor. Aug 12 '11 at 18:48
• Ah, I knew I was missing something, namely $I\neq\Lambda$ in general (I didn't even notice there were different index sets). thanks
– Leo
Aug 12 '11 at 19:39

Indeed, any torsion element of $G=C_{n_1}*\dots C_{n_k}$ is actually conjugate to an element of one of those $C_{n_i}$. To see it, take a reduced word $a_1*\dots *a_m\in G$ (each $a_j$ is in one of $C_{n_i}$) and suppose moreover that if $m>1$ then $a_1$ and $a_{m}$ are from different $C_{n_i}$'s. Any element of $G$ has a conjugate of this form. Clearly if $m>1$ then such a word is not torsion.