Identifying $\langle x_e, x_a, x_b, x_{ab}\mid x_ex_a=x_{ab}, x_ax_e=x_b, x_bx_{ab}=x_a, x_{ab}x_b=x_e\rangle.$

Question

Consider the group presentation $$\langle x_e, x_a, x_b, x_{ab}\mid x_ex_a\stackrel{(1)}=x_{ab}, x_ax_e\stackrel{(2)}=x_b, x_bx_{ab}\stackrel{(3)}=x_a, x_{ab}x_b\stackrel{(4)}=x_e\rangle.$$ What group does it define?

So far I have that the group is infinite using the Size function in GAP:

F:=FreeGroup(4);

rels:=[(F.1)*(F.2)*(F.3)^(-1), (F.2)*(F.1)*(F.3)^(-1), (F.3)*(F.4)*(F.2)^(-1), (F.4)*(F.3)*(F.1)^(-1)];

G:=F/rels;

Size(G);

(This code is in error: the first relator should be (F.1)*(F.2)*(F.4)^(-1).)

. I also have that $(1) \,\&\,(4)$ give $x_b=x_a^{-1}$ and $(2) \,\&\,(3)$ give $x_{ab}=x_e^{-1}$. One can obtain $x_e=x_a^{-2}$ by $(2)$.

I think it's simply $\Bbb Z$ but I'm not sure: I have $x_a^{3}=id.$ by $(1)$ and $x_a^2=id.$ by $(2)$, and each of the generators can be written in terms of $x_a$, so the presentation can be transformed into $$\Bbb Z=\langle x_a\rangle.$$

Is this right? Please help :)

Answer 1

[I seem to remember from a prior question that you are a PhD student] Have you seen Nielsen transformations or Tietze transformations, that are a formal framework to change presentations. They will give you a framework to do such substitutions. You could ask GAP to do some simplification, this is not guaranteed to be optimal, but in this case helps:

gap> f:=FreeGroup("e","a","b","c");
<free group on the generators [ e, a, b, c ]>
gap> rels:=ParseRelators(f,"ea=c,ae=b,bc=a,cb=e");
[ e*a*c^-1, a*e*b^-1, b*c*a^-1, c*b*e^-1 ]

amd use IsomorphismSimplifiedFpGroup to apply a heuristics of Tietze transformations (with g:=f/rels;):

gap> iso:=IsomorphismSimplifiedFpGroup(g);
[ e, a, b, c ] -> [ e, e, e^-1, e^-1 ]
gap> h:=Image(iso);
<fp group on the generators [ e ]>
gap> RelatorsOfFpGroup(h);
[ e^3 ]

So the group is cyclic of order 3. Not sure where the claim of infinity comes from.

In general the automated simplification is not going to give you always a the result you want, and it is neccessary to work by hand.

Answer 2

The group is isomorphic to $\Bbb Z_3$. See Derek Holt's comment.