I saw from a paper which claimed that this is a easy consequence from the definitions, but I can't give a proof of it just by the definitions. So could you give me some ideas? Thanks!
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Well, for one it must be a free product of a finite number of (non-trivial) hyperbolic groups, otherwise the finite number of generators won't work. – DonAntonio Aug 10 '16 at 22:19
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@DonAntonio, thanks I will fix it. – Aug 10 '16 at 22:30
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E.g. use that a group is hyperbolic iff it has a finite presentation satisfying the Dehn algorithm – YCor Aug 10 '16 at 23:01
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@YCor, I saw a paper saying that it's an easy consequences of definitions, I wonder if it could be proved just by definitions? – Aug 10 '16 at 23:06
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Yes, it can be directly proved from the triangle definition, sounds like a nice exercise. – YCor Aug 11 '16 at 06:36
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@Katherine: But what is your definition of hyperbolic groups? There exist several equivalent definitions. – Seirios Aug 11 '16 at 09:04
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@Seirios, the paper I read just gave the definition by $\delta-slim$, so I think that's the definition. – Aug 11 '16 at 19:37
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@YCor, thanks, can you give some hints? – Aug 12 '16 at 05:11
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Hint 1: Let $S_1$ (resp. $S_2$) be a finite generating set of a hyperbolic group $G_1$ (resp. $G_2$). Then $S_1 \cup S_2$ generates $G_1 \ast G_2$. Moreover, given some $g \in G_1 \ast G_2$, every geodesic between $1$ and $g$ can be written as a sequence of elements $$1, \ s_1^1, \ldots, \ s_1^1 \cdots s_{n_1}^1, \ldots, \ s_1^1 \cdots s_{n_1}^1 \cdots s_1^k \cdots s_{n_k}^k,$$ where, if we set $s^i = s_1^i \cdots s_{n_i}^i$, then $g=s^1 \cdots s^k$ is an alternating word in $G_1$ and $G_2$ and, for every $1 \leq i \leq k$, the path $$1, \ s^i_1, \ldots, \ s^i_1 \cdots s^i_{n_i}$$ defines a geodesic in the Cayley graph of $G_1$ or $G_2$ with respect to the generating set $S_1$ or $S_2$ (depending on whether $s^i \in G_1$ or $s^i \in G_2$).
Hint 2: Let $\Delta$ be a geodesic triangle. Say that $1,g,h$ are the three corners of $\Delta$. Written $g$ and $h$ as alternating words, let $p$ denote their maximal common prefix. Then, using the previous description of the geodesics, notice that $p$ belongs to the three sides of $\Delta$.
Seirios
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Thank you, but actually, it's just based on the definition of free product, and I have got the step before the hint. And what I I am stuck on is how to relate it to $\delta-slim$. – Aug 12 '16 at 18:30
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