ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Thu, 02 May 2024 09:48:36 +0200List some cosets of infinite finitely presented grouphttps://ask.sagemath.org/question/76970/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?Wed, 17 Apr 2024 16:30:45 +0200https://ask.sagemath.org/question/76970/list-some-cosets-of-infinite-finitely-presented-group/Comment by karl111 for <p>Example:</p>
<pre><code>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
</code></pre>
<p>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. </p>
<p>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.</p>
<p>In my case this means that B(6)/a^3 = G is infinite. </p>
<p>Is there a way to still get a finite list of representatives of the cosets?</p>
https://ask.sagemath.org/question/76970/list-some-cosets-of-infinite-finitely-presented-group/?comment=77198#post-id-77198I've added some context which hopefully makes the question clearer.Thu, 02 May 2024 09:48:36 +0200https://ask.sagemath.org/question/76970/list-some-cosets-of-infinite-finitely-presented-group/?comment=77198#post-id-77198Comment by dan_fulea for <p>Example:</p>
<pre><code>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
</code></pre>
<p>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. </p>
<p>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.</p>
<p>In my case this means that B(6)/a^3 = G is infinite. </p>
<p>Is there a way to still get a finite list of representatives of the cosets?</p>
https://ask.sagemath.org/question/76970/list-some-cosets-of-infinite-finitely-presented-group/?comment=77139#post-id-77139The 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
True
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_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.Mon, 29 Apr 2024 15:34:15 +0200https://ask.sagemath.org/question/76970/list-some-cosets-of-infinite-finitely-presented-group/?comment=77139#post-id-77139