Ganesh Ujwal
New member
- Joined
- Aug 10, 2014
- Messages
- 32
Let \(\displaystyle B_n\) be the braid group; that is, a group generated by \(\displaystyle \sigma_1,\cdots,\sigma_{n-1}\) with relations
1. \(\displaystyle \sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}\) for \(\displaystyle i=1,\cdots,n-2\);
2. \(\displaystyle \sigma_i\sigma_j=\sigma_j\sigma_i\) if \(\displaystyle i,j\in\{1,\cdots,n-1\}\) and \(\displaystyle |i-j|\geq 2\).
For \(\displaystyle 1\leq i<j\leq n\), let the usual generator \(\displaystyle A_{i,j}\) of pure braid groups be defined as
I need to prove that the following two sets have the same normal closure:
1. The set of all \(\displaystyle [A_{j,k},h^{-1}A_{j,k}h]\), where \(\displaystyle 1\leq j<k\leq n\) and \(\displaystyle h\) is an element of the subgroup generated by \(\displaystyle A_{j,j+1},A_{j,j+2},\cdots,A_{j,n}\).
2. The set of all \(\displaystyle [A_{j,k},g^{-1}A_{j,k}g]\), where \(\displaystyle 1\leq j<k\leq n\) and [/tex]g[/tex] is an element of the subgroup generated by \(\displaystyle A_{1,k},A_{2,k},\cdots,A_{k-1,k}\).
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My attempt: I believe once we proved that \(\displaystyle 1\Rightarrow2\), the other direction should be similar. Then I am planning to prove that assuming 1 is true, for each \(\displaystyle i=1,\cdots,k-1\), \(\displaystyle [A_{j,k},A_{i,k}^{-1}A_{j,k}A_{i,k}]=1\). But the difficulty is :
- I do not see any obvious way to prove this. Should I use the ususal presentation of pure braid group? If yes, how?
- Even if I can prove \(\displaystyle [A_{j,k},A_{i,k}^{-1}A_{j,k}A_{i,k}]=1\), does this imply 2? Anyway, generally in a group \(\displaystyle G\) where \(\displaystyle a,b,g\in G\), \(\displaystyle g\) commutes with \(\displaystyle g^a\) and \(\displaystyle g^b\)
does not imply that \(\displaystyle g\) commutes with \(\displaystyle g^{ab}\), where \(\displaystyle g^a=a^{-1}ga\).
1. \(\displaystyle \sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}\) for \(\displaystyle i=1,\cdots,n-2\);
2. \(\displaystyle \sigma_i\sigma_j=\sigma_j\sigma_i\) if \(\displaystyle i,j\in\{1,\cdots,n-1\}\) and \(\displaystyle |i-j|\geq 2\).
For \(\displaystyle 1\leq i<j\leq n\), let the usual generator \(\displaystyle A_{i,j}\) of pure braid groups be defined as
| \(\displaystyle A_{i,j}=(\sigma_{j-1}\sigma_{j-2}\cdots\sigma_{i+1})\sigma_i^2(\sigma_{i+1}^{-1}\cdots\sigma_{j-2}^{-1}\sigma_{j-1}^{-1})\) |
I need to prove that the following two sets have the same normal closure:
1. The set of all \(\displaystyle [A_{j,k},h^{-1}A_{j,k}h]\), where \(\displaystyle 1\leq j<k\leq n\) and \(\displaystyle h\) is an element of the subgroup generated by \(\displaystyle A_{j,j+1},A_{j,j+2},\cdots,A_{j,n}\).
2. The set of all \(\displaystyle [A_{j,k},g^{-1}A_{j,k}g]\), where \(\displaystyle 1\leq j<k\leq n\) and [/tex]g[/tex] is an element of the subgroup generated by \(\displaystyle A_{1,k},A_{2,k},\cdots,A_{k-1,k}\).
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My attempt: I believe once we proved that \(\displaystyle 1\Rightarrow2\), the other direction should be similar. Then I am planning to prove that assuming 1 is true, for each \(\displaystyle i=1,\cdots,k-1\), \(\displaystyle [A_{j,k},A_{i,k}^{-1}A_{j,k}A_{i,k}]=1\). But the difficulty is :
- I do not see any obvious way to prove this. Should I use the ususal presentation of pure braid group? If yes, how?
- Even if I can prove \(\displaystyle [A_{j,k},A_{i,k}^{-1}A_{j,k}A_{i,k}]=1\), does this imply 2? Anyway, generally in a group \(\displaystyle G\) where \(\displaystyle a,b,g\in G\), \(\displaystyle g\) commutes with \(\displaystyle g^a\) and \(\displaystyle g^b\)
does not imply that \(\displaystyle g\) commutes with \(\displaystyle g^{ab}\), where \(\displaystyle g^a=a^{-1}ga\).
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