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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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The preview and the actual parser treat html tags differently; I've substituted with \langle and \rangle. –  Qiaochu Yuan Nov 7 '10 at 0:48
    
Please check, I've edited the question –  Kate Juschenko Nov 7 '10 at 0:49
    
Qiaochu Yuan@ Thanks! –  Kate Juschenko Nov 7 '10 at 0:53
    
In the standard Higman group (the one without finite quotients), there are 4 generators. You really want to have only 3? –  Mark Sapir Nov 7 '10 at 0:58
    
@Mark: yes, I really want to have 3 relations. For 4 the answer is "yes". –  Kate Juschenko Nov 7 '10 at 1:03

1 Answer 1

up vote 6 down vote accepted

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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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 –  Kate Juschenko Nov 7 '10 at 1:52
    
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. –  Mark Sapir Nov 7 '10 at 1:56
    
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. –  Mark Sapir Nov 7 '10 at 1:58
    
@Mark: thank you so much for the paper! –  Kate Juschenko Nov 7 '10 at 4:31

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