# List some cosets of infinite finitely presented group

Example:

```
F.<a,b,c,d,e> = FreeGroup();
H = [a^3,
a*b*a*b^(-1)*a^(-1)*b^(-1),
b*c*b*c^(-1)*b^(-1)*c^(-1),
c*d*c*d^(-1)*c^(-1)*d^(-1),
d*e*d*e^(-1)*d^(-1)*e^(-1),
a*c*a^(-1)*c^(-1),
a*d*a^(-1)*d^(-1),
a*e*a^(-1)*e^(-1),
b*d*b^(-1)*d^(-1),
b*e*b^(-1)*e^(-1),
c*e*c^(-1)*e^(-1),];
G = F/H
```

We take the free group on 5 generators and quotient out the braid relations. This yields the braid group on 6 strands. Since we also quotient out a^3, we get that G is a factor group of the braid group on 6 strands.

Geometrically, we are looking at braids up to the t_k move, which untwists k consecutive halftwists. The t_2 move forgets about over and undercrossings and thus B(n) mod t_2 = S(n). In 'Factor groups of the braid group' coxeter studies those groups and proves that the factor group B(n)/a^k is finite iff 1/n + 1/k > 1/2.

In my case this means that B(6)/a^3 = G is infinite.

Is there a way to still get a finite list of representatives of the cosets?

The Braid Group with 6 strands is:

and in it we have for instance:

My generators $a,b,c,d,e$ above are thus not the generators from the posted question. This group $B$ is something quite different. The given group $G$ is a quotient of the Coxeter group (https://en.wikipedia.org/wiki/Coxeter...) where the diagram is connecting any two different bullets. (Else we have at least two generators that commute...)

So please make clear what is $H$, how is $G$ a factor of a braid group $B_6$, which result of Coxeter is involved, so that the relation $\frac 16+\frac 13\le \frac 12$ should ring a bell, and how to get a finite list of representative for which finite group... Describing the mathematical problem can help.

I've added some context which hopefully makes the question clearer.