# Do you know this finitely presented group on two generators?

I computed using Sage the fundamental group of some topological space and got the infinite group $$\langle a, b\mid aba^{-1}ba\rangle.$$ 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 ?

Note that the abelianized presentation is $$\langle a,b\mid ab^2\rangle$$. 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 $$G=\langle t,x\mid x(t^{-2}xt^2)(t^{-1}xt)^{-1}\rangle.$$ Write $$y=t^{-1}xt$$. Then this yields: $$G=\langle t,x,y\mid t^{-1}yt=x^{-1}y,\; t^{-1}xt=y\rangle.$$ 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 $$\langle t\rangle$$ 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 $$F(x,y)\times\langle t^2y^{-1}\rangle$$. Note that the element $$t^2y^{-1}$$, whose centralizer is this subgroup of index 2, is just equal to $$a$$.
• That's interesting, thank you. Now I found that, letting $c=b^{-1}a$ we get $$\langle a, c \mid a^3=c^2\rangle,$$which is a presentation of the Braid group on three strands (see math.stackexchange.com/questions/101720/…). Commented Dec 9, 2020 at 19:07