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 ?


1 Answer 1


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 $a$.

  • 1
    $\begingroup$ 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/…). $\endgroup$
    – Friedrich
    Dec 9, 2020 at 19:07

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