Embedding of Baumslag-Solitar group into a certain group 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 
 A: 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). 
A: 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.
