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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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  • $\begingroup$ The preview and the actual parser treat html tags differently; I've substituted with \langle and \rangle. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 7, 2010 at 0:48
  • $\begingroup$ Please check, I've edited the question $\endgroup$
    – Kate Juschenko
    Commented Nov 7, 2010 at 0:49
  • $\begingroup$ In the standard Higman group (the one without finite quotients), there are 4 generators. You really want to have only 3? $\endgroup$
    – user6976
    Commented Nov 7, 2010 at 0:58
  • $\begingroup$ @Mark: yes, I really want to have 3 relations. For 4 the answer is "yes". $\endgroup$
    – Kate Juschenko
    Commented Nov 7, 2010 at 1:03
  • $\begingroup$ @Jon: the group is a possible candidate for example of non-sofic group $\endgroup$
    – Kate Juschenko
    Commented Nov 7, 2010 at 1:07

2 Answers 2

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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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  • $\begingroup$ Thank you for your suggestions. I have to note that the following: $a_0 a_1 a_0^{-1}=a_1^2$; $a_1 a_2 a_1^{-1}=a_2^2$; $a_2 a_0 a_2^{-1}=a_0^2$ is trivial group, so it is essential that we have power 4. It might be that 3 will also work $\endgroup$
    – Kate Juschenko
    Commented Nov 7, 2010 at 1:52
  • $\begingroup$ You need to compute the Gersten-Stallings angles at each corner of the triangle. For example two sides have $\langle a_0\rangle$ and $\langle a_1 \rangle$ on the edges. Then compute the shortest word in $a_0, a_1$ that is equal to 1. In your case, I am sure, it is 7 (the length of the defining relation involving these two letters). Hence the angle is $2\pi/7$. The sum of all three angles is then $6\pi/7\lt \pi$. Hence by the Gersten-Stallings-Haefliger... the complex of groups is developable, and the B-S group embeds. $\endgroup$
    – user6976
    Commented Nov 7, 2010 at 1:56
  • $\begingroup$ Note that in your other example with $2$ instead of $4$, the angles are $2\pi/5$, the sum is $6\pi/5\gt \pi$ and the theorem does not apply. $\endgroup$
    – user6976
    Commented Nov 7, 2010 at 1:58
  • $\begingroup$ @Mark: thank you so much for the paper! $\endgroup$
    – Kate Juschenko
    Commented Nov 7, 2010 at 4:31
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A few minutes of running GAP shows that the group has order $6751269$. Hence, as the final edit of the accepted answer points out, the group is finite, and no Baumslag-Solitar group embeds in it.

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  • $\begingroup$ $3^97^3$... I wonder if this is an easily characterizable group? $\endgroup$
    – Steven Stadnicki
    Commented May 19, 2021 at 17:38
  • $\begingroup$ @StevenStadnicki All I know for sure (right now) is that it is solvable :-) $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented May 19, 2021 at 18:57

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