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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.

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    $\begingroup$ In a free product $A*B$, the only torsion elements are conjugates of torsion elements of $A$ and torsion elements of $B$. That's why you are getting $0$ in your example ($ab$ is not a conjugate of an element of $A=\langle a\rangle$ or an element of $B=\langle b\rangle$. So you can certainly always get $0$. What you want is Kurosh's Theorem; see math.stackexchange.com/questions/29264/… $\endgroup$ – Arturo Magidin Jul 5 '18 at 16:26
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    $\begingroup$ So the $m_i$ must be either $0$ or divisors of a $k_j$ (possibly different $j$s for different $i$s, and different $i$s may correspond to the same $j$. $\endgroup$ – Arturo Magidin Jul 5 '18 at 16:27
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This is just a repetition of Arturo Magidin's comments. By the Kurosh subgroup theorem, a subgroup of a free product $G$ of groups $G_i$ is itself isomorphic to a free product of a free group and some conjugates of subgroups of some of the groups $G_i$ (where the same $G_i$ can occur arbitrarily many times).

So in your example, for each $m_i$, either $m_i=0$ or $m_i$ divides one of the $k_i$.

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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 $j$.

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    $\begingroup$ Although I can't complain about any proof that uses Bass-Serre trees, nonetheless it's not exactly fair to say that this does not use the Kurosh subgroup theorem, because basically what you are doing is repeating a relevant chunk of the proof of that theorem.. $\endgroup$ – Lee Mosher Jul 5 '18 at 20:15
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    $\begingroup$ @LeeMosher I can't remember how Kurosh is proven! Okay, reason to care about this proof: it still holds for free products of cyclic groups with amalgamation. (While subgroup theorems in this setting are unwieldy!) $\endgroup$ – user1729 Jul 5 '18 at 20:31
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    $\begingroup$ I'm thinking of the proof of Kurosh which starts with the Bass-Serre tree for the given decomposition of $G$, and then restricts to the action of the subgroup $H$ on its minimal subtree. $\endgroup$ – Lee Mosher Jul 5 '18 at 20:32
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    $\begingroup$ (Reason to disregard my proof: what the OP wants to prove is that subgroups of free products of finite cyclic groups have a specific form, and the first step in this classification, which is hidden in the linked-to question, uses Kurosh...) $\endgroup$ – user1729 Jul 5 '18 at 20:32
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    $\begingroup$ @LeeMosher I think I know which proof you are talking about. But I am curious: do you know this proof off the top of your head because you teach it a lot, or for another reason? $\endgroup$ – user1729 Jul 5 '18 at 20:36

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