0

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 :)

Shaun
  • 47,747
  • 2
    See this related question. – Dietrich Burde Nov 07 '17 at 13:27
  • See my questions circa this date for more. – Shaun Nov 07 '17 at 13:43
  • 1
    So you should use the two equations you derived to eliminate the generators $x_b$ and $x_{ab}$ from the presentation. Why have you given up at this point? – Derek Holt Nov 07 '17 at 14:15
  • You're right, @DerekHolt; I don't know why I stopped there. I've edited the question accordingly. – Shaun Nov 07 '17 at 14:40
  • If $G$ is generated by $x_a$ and $x_a^2=x_a^3=1$ then $G$ is trivial! I cannot see how you concluded that $G \cong {\mathbb Z}$. But you made a mistake, after substituting for $x_b$ and $x_{ab}$, you get $x_e^{-2} = x_a$ and $x_a^{-2} = x_e$, which simplify to $x_a^3=1$, so the group is isomorphic to $ \langle x_a | x_a^3 = 1 \rangle$, which has order $3$· – Derek Holt Nov 07 '17 at 14:56
  • Thank you. Would you put that in an answer please, @DerekHolt? – Shaun Nov 07 '17 at 14:58
  • Wait: I get that the group is infinite using `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);` in GAP.

     – Shaun Nov 07 '17 at 15:00
  • Oh, I see where I went wrong there. – Shaun Nov 07 '17 at 15:03

2 Answers2

1

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

Shaun
  • 47,747
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.

Shaun
  • 47,747
ahulpke
  • 20,399
  • Yes, I'm a PhD student. I started about a month ago. Thank you for this answer. It's very useful! $\ddot\smile$ – Shaun Nov 07 '17 at 15:16