List some cosets of infinite finitely presented group

asked 2024-04-17 16:30:45 +0200

karl111 gravatar image

updated 2024-05-02 09:50:10 +0200


F.<a,b,c,d,e> = FreeGroup(); 

H = [a^3,

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?

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The Braid Group with 6 strands is:

B.<a,b,c,d,e> = BraidGroup(6)

and in it we have for instance:

sage: a*c == c*a

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 ( 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.

dan_fulea gravatar imagedan_fulea ( 2024-04-29 15:34:15 +0200 )edit

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

karl111 gravatar imagekarl111 ( 2024-05-02 09:48:36 +0200 )edit