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Let $G$ be a group generated by $a_0, a_1, a_2$ with relations: $a_0 a_1 a_0^{-1}=a_1^4$ $a_1 a_2 a_1^{-1}=a_2^4$ $a_2 a_0 a_2^{-1}=a_0^4$ I am wondering if $BS(1,4)=\langle a,b:bab^{-1}=a^4\rangle$ is embedded into G via $a\mapsto a_1$, $b\mapsto a_0$ Remark: the group is constructed in analogy to Higman group |
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This is a fundamental group of a graph of groups with B-S groups as vertex groups and cyclic groups as edge groups. So yes, each Baumslag-Solitar group naturally embeds in your group. Edit. It is not a graph of groups but a complex of groups: triangle with B-S groups at vertices, cyclic groups as edges and 1 in the triangle. So you need to check the Haefliger condition to prove that it is developable. One can also (better) use Stallings, John R. Non-positively curved triangles of groups. Group theory from a geometrical viewpoint (Trieste, 1990), 491–503, World Sci. Publ., River Edge, NJ, 1991. Edit 2. No, this method does not work. To compute the Gersten-Stalling angle, one needs to count only the alternating length, not the total length of the word. The alternating length of each relator is not 7 but 4, so each angle is $\pi/2$ and the total angle is $3\pi/2$, too large. Edit 3. In fact the group is finite, so none of the B-S groups embeds. See this paper (the link was sent to me by Dani Wise). |
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