Symmetric group, Braid Groups, and related groups
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The symmetric group, in terms of presentation, is given by a group with generators $x_1,x_2,cdots,x_n$ with following types of relations:
(R1) $x_1x_{i+1}x_i=x_{i+1}x_ix_{i+1}$
(R2) $x_ix_j=x_jx_i$ for $|i-j|geq 2$
(R3) $x_i^2=1$
There is a class of groups which are closely related with this group. The Braid group $B_n$ has presentation:
$$B_n=langle x_1,x_2,cdots,x_n : R1, R2rangle.$$
In other words, dropping relations (R3) from above presentation of symmetric group, we get Braid group. Then my question is
Question: If we drop relations (R1) or (R2) from presentation of $S_n$, what kind of groups we get? Are they finite? Are these groups studied well?
finite-groups symmetric-groups group-presentation
edited Dec 3 at 2:08
Shaun
8,100113577
asked Jun 7 '16 at 8:36
p Groups
6,1681028
add a comment |
up vote
1
down vote
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The symmetric group, in terms of presentation, is given by a group with generators $x_1,x_2,cdots,x_n$ with following types of relations:
(R1) $x_1x_{i+1}x_i=x_{i+1}x_ix_{i+1}$
(R2) $x_ix_j=x_jx_i$ for $|i-j|geq 2$
(R3) $x_i^2=1$
There is a class of groups which are closely related with this group. The Braid group $B_n$ has presentation:
$$B_n=langle x_1,x_2,cdots,x_n : R1, R2rangle.$$
In other words, dropping relations (R3) from above presentation of symmetric group, we get Braid group. Then my question is
Question: If we drop relations (R1) or (R2) from presentation of $S_n$, what kind of groups we get? Are they finite? Are these groups studied well?
finite-groups symmetric-groups group-presentation
edited Dec 3 at 2:08
Shaun
8,100113577
asked Jun 7 '16 at 8:36
p Groups
6,1681028
Well, if you replace (R1) by something suitable you get Coxeter groups of other types than $A$ (with possibly some small adjustments to (R2) for some types).
– Tobias Kildetoft
Jun 7 '16 at 8:44
You mean: the relations (R1) (using (R3)) can be written as $(x_ix_{i+1})^3=1$; instead of this, we write $(x_ix_{i+1})^n_i=1$, then we get Coxeter group.
– p Groups
Jun 7 '16 at 8:46
Yes, precisely (or we can remove the condition by letting the element have infinite order, which obviously gives an infinite group).
– Tobias Kildetoft
Jun 7 '16 at 8:50
Also note that is may be more suggestive to write all the relations here in the form $(xy)^n = 1$ for some $n$.
– Tobias Kildetoft
Jun 7 '16 at 9:11
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The symmetric group, in terms of presentation, is given by a group with generators $x_1,x_2,cdots,x_n$ with following types of relations:
(R1) $x_1x_{i+1}x_i=x_{i+1}x_ix_{i+1}$
(R2) $x_ix_j=x_jx_i$ for $|i-j|geq 2$
(R3) $x_i^2=1$
There is a class of groups which are closely related with this group. The Braid group $B_n$ has presentation:
$$B_n=langle x_1,x_2,cdots,x_n : R1, R2rangle.$$
In other words, dropping relations (R3) from above presentation of symmetric group, we get Braid group. Then my question is
Question: If we drop relations (R1) or (R2) from presentation of $S_n$, what kind of groups we get? Are they finite? Are these groups studied well?
finite-groups symmetric-groups group-presentation
edited Dec 3 at 2:08
Shaun
8,100113577
asked Jun 7 '16 at 8:36
p Groups
6,1681028
The symmetric group, in terms of presentation, is given by a group with generators $x_1,x_2,cdots,x_n$ with following types of relations:
(R1) $x_1x_{i+1}x_i=x_{i+1}x_ix_{i+1}$
(R2) $x_ix_j=x_jx_i$ for $|i-j|geq 2$
(R3) $x_i^2=1$
There is a class of groups which are closely related with this group. The Braid group $B_n$ has presentation:
$$B_n=langle x_1,x_2,cdots,x_n : R1, R2rangle.$$
In other words, dropping relations (R3) from above presentation of symmetric group, we get Braid group. Then my question is
Question: If we drop relations (R1) or (R2) from presentation of $S_n$, what kind of groups we get? Are they finite? Are these groups studied well?
finite-groups symmetric-groups group-presentation
finite-groups symmetric-groups group-presentation
edited Dec 3 at 2:08
Shaun
8,100113577
asked Jun 7 '16 at 8:36
p Groups
6,1681028
edited Dec 3 at 2:08
Shaun
8,100113577
asked Jun 7 '16 at 8:36
p Groups
6,1681028
edited Dec 3 at 2:08
Shaun
8,100113577
edited Dec 3 at 2:08
Shaun
8,100113577
edited Dec 3 at 2:08
Shaun
8,100113577
8,100113577
asked Jun 7 '16 at 8:36
p Groups
6,1681028
asked Jun 7 '16 at 8:36
p Groups
6,1681028
asked Jun 7 '16 at 8:36
p Groups
6,1681028
6,1681028
Well, if you replace (R1) by something suitable you get Coxeter groups of other types than $A$ (with possibly some small adjustments to (R2) for some types).
– Tobias Kildetoft
Jun 7 '16 at 8:44
You mean: the relations (R1) (using (R3)) can be written as $(x_ix_{i+1})^3=1$; instead of this, we write $(x_ix_{i+1})^n_i=1$, then we get Coxeter group.
– p Groups
Jun 7 '16 at 8:46
Yes, precisely (or we can remove the condition by letting the element have infinite order, which obviously gives an infinite group).
– Tobias Kildetoft
Jun 7 '16 at 8:50
Also note that is may be more suggestive to write all the relations here in the form $(xy)^n = 1$ for some $n$.
– Tobias Kildetoft
Jun 7 '16 at 9:11
add a comment |
Well, if you replace (R1) by something suitable you get Coxeter groups of other types than $A$ (with possibly some small adjustments to (R2) for some types).
– Tobias Kildetoft
Jun 7 '16 at 8:44
You mean: the relations (R1) (using (R3)) can be written as $(x_ix_{i+1})^3=1$; instead of this, we write $(x_ix_{i+1})^n_i=1$, then we get Coxeter group.
– p Groups
Jun 7 '16 at 8:46
Yes, precisely (or we can remove the condition by letting the element have infinite order, which obviously gives an infinite group).
– Tobias Kildetoft
Jun 7 '16 at 8:50
Also note that is may be more suggestive to write all the relations here in the form $(xy)^n = 1$ for some $n$.
– Tobias Kildetoft
Jun 7 '16 at 9:11
Well, if you replace (R1) by something suitable you get Coxeter groups of other types than $A$ (with possibly some small adjustments to (R2) for some types).
– Tobias Kildetoft
Jun 7 '16 at 8:44
Well, if you replace (R1) by something suitable you get Coxeter groups of other types than $A$ (with possibly some small adjustments to (R2) for some types).
– Tobias Kildetoft
Jun 7 '16 at 8:44
You mean: the relations (R1) (using (R3)) can be written as $(x_ix_{i+1})^3=1$; instead of this, we write $(x_ix_{i+1})^n_i=1$, then we get Coxeter group.
– p Groups
Jun 7 '16 at 8:46
You mean: the relations (R1) (using (R3)) can be written as $(x_ix_{i+1})^3=1$; instead of this, we write $(x_ix_{i+1})^n_i=1$, then we get Coxeter group.
– p Groups
Jun 7 '16 at 8:46
Yes, precisely (or we can remove the condition by letting the element have infinite order, which obviously gives an infinite group).
– Tobias Kildetoft
Jun 7 '16 at 8:50
Yes, precisely (or we can remove the condition by letting the element have infinite order, which obviously gives an infinite group).
– Tobias Kildetoft
Jun 7 '16 at 8:50
Also note that is may be more suggestive to write all the relations here in the form $(xy)^n = 1$ for some $n$.
– Tobias Kildetoft
Jun 7 '16 at 9:11
Also note that is may be more suggestive to write all the relations here in the form $(xy)^n = 1$ for some $n$.
– Tobias Kildetoft
Jun 7 '16 at 9:11
add a comment |
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Well, if you replace (R1) by something suitable you get Coxeter groups of other types than $A$ (with possibly some small adjustments to (R2) for some types).
– Tobias Kildetoft
Jun 7 '16 at 8:44
You mean: the relations (R1) (using (R3)) can be written as $(x_ix_{i+1})^3=1$; instead of this, we write $(x_ix_{i+1})^n_i=1$, then we get Coxeter group.
– p Groups
Jun 7 '16 at 8:46
Yes, precisely (or we can remove the condition by letting the element have infinite order, which obviously gives an infinite group).
– Tobias Kildetoft
Jun 7 '16 at 8:50
Also note that is may be more suggestive to write all the relations here in the form $(xy)^n = 1$ for some $n$.
– Tobias Kildetoft
Jun 7 '16 at 9:11