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Do you know this finitely presented group on two generators?

Mathematics Asked on February 9, 2021

I computed using Sage the fundamental group of some topological space and got the infinite group
$$langle a, bmid aba^{-1}barangle.$$
By the change of variables $x=b^{-1}$ and $y=a$, it can also be written as
$$langle x, y mid xy=y^2x^{-1}rangle.$$
Do you know if this group has a name ? Do you know if there exists a table of finitely presented groups of small "complexity" like this one ?

One Answer

Note that the abelianized presentation is $langle a,bmid ab^2rangle$. This suggest changing generators so that $ab^2$ is a generator. Define $x=ab^2$, $t=ab$, so that $a=tx^{-1}t$, $b=t^{-1}x$. This yields the presentation $$G=langle t,xmid x(t^{-2}xt^2)(t^{-1}xt)^{-1}rangle.$$ Write $y=t^{-1}xt$. Then this yields: $$G=langle t,x,ymid t^{-1}yt=x^{-1}y,; t^{-1}xt=yrangle.$$ Since $(x^{-1}y,y)$ is a basis of the free group on $x,y$, we identify a semidirect product of the free group $F(x,y)$ on $x,y$ by a cyclic group $langle trangle$ acting on $F(x,y)$ through powers of the automorphism $(x,y)mapsto (x^{-1}y,y)$. (In particular, $G$ is torsion-free.)

Note that this automorphism is not inner, but its square is $(x,y)mapsto ((x^{-1}y)^{-1}y,y)=(y^{-1}xy,y)$, which is an inner automorphism (right conjugation by $y$). Hence the unique subgroup of index 2 in $G$ is a direct product $F(x,y)timeslangle t^2y^{-1}rangle$. Note that the element $t^2y^{-1}$, whose centralizer is this subgroup of index 2, is just equal to $a$.

Answered by YCor on February 9, 2021

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