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I am wondering if $BS(1,4) = \langle a,b\mid bab^{-1}=a^4 \rangle$ can be embedded into the following group generated by $a_0, a_1, a_2$ with relations:

$a_0 a_1 a_0^{-1}=a_1^4$

$a_1 a_2 a_1^{-1}=a_2^4$

$a_2 a_0 a_2^{-1}=a_0^4$

Remark: the group is constructed in analogy to Higman 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

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 $BS(1,4)=$<$a,b:bab^{-1}=a^4$> is embedded into G via $a\mapsto a_1$, $b\mapsto a_0$

Remark: the group is constructed in analogy to Higman group

I am wondering if $BS(1,4) = \langle a,b\mid bab^{-1}=a^4 \rangle$ can be embedded into the following group generated by $a_0, a_1, a_2$ with relations:

$a_0 a_1 a_0^{-1}=a_1^4$

$a_1 a_2 a_1^{-1}=a_2^4$

$a_2 a_0 a_2^{-1}=a_0^4$

Remark: the group is constructed in analogy to Higman group

I am wondering if $BS(1,4)= \langle a,b:bab^{-1}=a^4 \rangle$ can be embedded into the following group generated by $a_0, a_1, a_2$ with relations:

$a_0 a_1 a_0^{-1}=a_1^4$

$a_1 a_2 a_1^{-1}=a_2^4$

$a_2 a_0 a_2^{-1}=a_0^4$

Remark: the group is constructed in analogy to Higman 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 $BS(1,4)=$<$a,b:bab^{-1}=a^4$> is embedded into G via $a\mapsto a_1$, $b\mapsto a_0$

Remark: the group is constructed in analogy to Higman group