the braid group over the 2 sphere Bn(S2) is infinite if n greather than 3 but this group contain elements of finite order so this element generate a finite cyclic subgroup H in Bn(S2)
my question is : how to calculate the indexe of H in Bn(S2)? if possible
braid groups over the 2 sphere is a quotion of the braid groups.
The index |G:H| of a finite subgroup H in an infinite group G is always infinite.
Note that |G|=|G:H|.|H|
The index |G:H| of a finite subgroup H in an infinite group G is always infinite.
Note that |G|=|G:H|.|H|
Dear Degaichi,
1) The Index is infinite, it is easy to prove it.
2) The subgroup of finite order elements is not necessarily cyclic, only if there is an element (at least) has the order of that subgroup. But, I will mention that the order of that group is equal to what we call in French "exposant".
a braid group over 2 sphere:
http://homepages.warwick.ac.uk/~maseay/doc/braids.pdf
I agree with Yuri S. Semenov. The main fact is :
|G|=|G:H|.|H|
This should be understand as (possibly infinite) Cardinal numbers.
Obviously G infinite and H finite implies that |G:H| is infinite, as Yuri S. Semenov said. But you may obtain a better conclusion :
|G|=|G:H|
In other words, the index of H in G equals the Cardinal of G.
I agree with Yuri S. Semenov. The main fact is :
|G|=|G:H|.|H|
This should be understand as (possibly infinite) Cardinal numbers.
Obviously G infinite and H finite implies that |G:H| is infinite, as Yuri S. Semenov said. But you may obtain a better conclusion :
|G|=|G:H|
In other words, the index of H in G equals the Cardinal of G.
I agree with Yuri S. Semenov. The main fact is :
|G|=|G:H|.|H|
This should be understand as (possibly infinite) Cardinal numbers.
Obviously G infinite and H finite implies that |G:H| is infinite, as Yuri S. Semenov said. But you may obtain a better conclusion :
|G|=|G:H|
In other words, the index of H in G equals the Cardinal of G.
I agree with Yuri S. Semenov. The main fact is :
|G|=|G:H|.|H|
This should be understand as (possibly infinite) Cardinal numbers.
Obviously G infinite and H finite implies that |G:H| is infinite, as Yuri S. Semenov said. But you may obtain a better conclusion :
|G|=|G:H|
In other words, the index of H in G equals the Cardinal of G.
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