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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 $n_{1}, \ldots, n_{k}$ ?

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I edited the title for clarification. – Rasmus 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 often explain user8268's answer. Your question is the KST restricted to cyclic subgroups. – Jack Schmidt Mar 27 '11 at 15:07
up vote 6 down vote accepted

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 $A_i$.

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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: $\forall H\!\leq\!\ast_{i\in I}G_i\!=\!G$ $\exists H_i\!\leq\!G_i,\exists x_i\!\in\!G, i\!\in\!I$ $\exists X\!\subseteq\!G:$ $H=F_X\ast\big({\Large\ast}_{i\in I}\,x_iH_ix_i^{-1}\big)$? – Leon 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. – Arturo Magidin 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 – Leon 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. – Arturo Magidin 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 – Leon 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.

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