Order of generators in subgroup of Free Product of Cyclic Groups In this question, it is shown that a subgroup of a free product of cyclic groups is still a free product of cyclic groups. (Subgroup of free product of cyclic group is still a free product of cyclic groups?)
Suppose $G=\langle g_1,g_2,\dots\mid g_1^{k_1}=1, g_2^{k_2}=1\dots\rangle$, where $k_i$ are integers inclusive of 0. (Power 0 to denote those free generators with no relations)
Let $H$ be a subgroup of $G$. Since $H$ is a free product of cyclic groups, write $H=\langle h_1,h_2,\dots\mid h_1^{m_1}=1, h_2^{m_2}=1\dots\rangle$.
Is there any relation/restrictions at all between the orders $m_1,m_2,\dots$ and $k_1,k_2,\dots$?
(By restrictions I mean any "forbidden" values of $m_i$, based on our knowledge of the $k_i$.)
Thanks.

I did some basic "experimenting" by setting $G=\langle a,b\mid a^2=b^3=1\rangle$.
We can have $H=\langle ab\mid\ (ab)^0=1\rangle\cong \mathbb{Z}$,
Hence, it seems that $m_i=0$ is always possible.

Without using the Kurosh subgroup theorem: In a free product, elements of finite order are conjugate into one of the factor groups. (To see this you could consider the action of $G$ on its Bass-Serre tree, where vertex stabilisers are factor groups.)
Hence, in your example if $h_i\in H$ has finite order then $h_i$ is conjugate to $g_j^p$ for some $g_j$ and some $p< k_j$. So $h_i$ has order $m_j=k_j/p$. Therefore, either $m_i=0$ or $m_i$ divides $k_j$ for some $j$.
