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 $n_{1}, \ldots, n_{k}$ ?
 A: 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 $A_i$.
A: 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.
