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 



Embedding of Baumslag-Solitar group into a certain group
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](http://www.ma.utexas.edu/users/allcock/research/bs6.pdf)  (the link was sent to me by Dani Wise). 


