TITLE: Embedding of Baumslag-Solitar group into a certain group
QUESTION [9 upvotes]: 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

REPLY [3 votes]: 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.