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Let $F_1$ and $F_2$-non-trivial groups.

  1. Is it correct that the number of ends of the free product $F_1\ast F_2$ is infinite?

My thoughts about this: Since $e(G)=\infty$ then $G=F_1\ast F_2$, a non-trivial free product using Stalling's theorem. Applying Grushko theorem, $F_1$, $F_2$ are generated by strictly fewer elements. The intersection of $F_1$ with the free subgroup is a free subgroup. So from induction $F_1$ is a free group. The same is true for $F_2$. And $G=F_1 \ast F_2$ is free. Is it correct?

  1. If it is true. Is it possible to find examples of groups $G$ with infinitely many ends which are not represented as a free product of groups $G=F_1\ast F_2$?

If you don't mind can you explain this in more details?

Thank you!

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    $\begingroup$ What you have written seems incoherent. Stallings' theorem says that a group has more than one end iff it admits a nontrivial decomposition as an amalgamated free product or as an HNN extension. Note also that $C_2*C_2$ has two ends. $\endgroup$
    – Derek Holt
    Commented Sep 12, 2023 at 19:33
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    $\begingroup$ In general, you should also allow amalgamation/HNN extension over finite subgroups. Then it is a classical result due to Stallings and Bergman. $\endgroup$
    – Moishe Kohan
    Commented Sep 12, 2023 at 19:55
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    $\begingroup$ en.wikipedia.org/wiki/Stallings_theorem_about_ends_of_groups $\endgroup$
    – Moishe Kohan
    Commented Sep 12, 2023 at 20:03
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    $\begingroup$ Amalgamate two groups isomorphic to $S_3$ over order 2 subgroups. $\endgroup$
    – Moishe Kohan
    Commented Sep 12, 2023 at 20:21
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    $\begingroup$ @UserIn You can take $\operatorname{SL}_2(\mathbf{Z})$, which is isomorphic to the amalgamated free product of two cyclic groups, one of order 6 and one of order 4, over an order 2 cyclic subgroup. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Sep 12, 2023 at 22:57

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