## Finding the middle group in a SES.

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Talith
Proved the Goldbach Conjecture
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### Finding the middle group in a SES.

I'm in the situation that I have a short exact sequence of groups 1->A->B->C->1. I know that A and C are Z and Z2 respectively and I know that B is torsion-free. Is this enough information to find B? From what I can find on wikipedia, it seems that, in general, you need the SES to split in order to have a nice presentation of B in terms of A and C, however this would give B being equal to Z+Z2 which has torsion, so I guess my sequence doesn't split. I have some heuristic evidence which suggests that B is isomorphic to Z, which obviously fits in to the SES. For those interested, I'm trying to show that the braid group on two strings (B) is isomorphic to the integers, given that I've calculated the pure 2-braid group (A). If this isn't enough information, I guess I'll have to appeal to the homomorphisms in the sequence and chase some elements.
Last edited by Talith on Sun Feb 26, 2012 2:50 am UTC, edited 1 time in total.

jestingrabbit
Factoids are just Datas that haven't grown up yet
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### Re: Finding the middle group in a SES.

Lets make f:A->B be the first homomorphism. If b is in B\f(A), and f(1) = a then B = {a^k * b, a^k | k in Z}. If you specify that b * a = a^k * b and b^2 = a^l for some k and l, then that defines the group operation.

Torsion free implies (a^n * b)^2 = a^(n(k+1) + l), so k and l must be such that n(k+1) + l is never 0 for any integer n, so k+1 does not divide l.

You also need inverses. So, for all n, there exists an m such that e = (a^n * b)(a^m * b) = a^(n+mk + l) ie for all n there is an m such that n+mk + l = 0. In particular, if b^(-1) = a^n * b, then n + l = 0 and nk + l = 0, so k = 1, and that makes B commutative. From before, l= 2i + 1, and b' = a^{-i}*b is such that (b')^2 = a, and <b'> = B.

B is the integers.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

Talith
Proved the Goldbach Conjecture
Posts: 848
Joined: Sat Nov 29, 2008 1:28 am UTC
Location: Manchester - UK

### Re: Finding the middle group in a SES.

Need to apologise for a mistake. I know that A and C(not B) are Z and Z2 respectively and I know that B is torsion-free (that part isn't a mistake). I've corrected the OP now.

It looks like you overlooked that though, because your proof seems to work very well jr. I'll read it again in the morning just to make sure I understand the argument and that there are no holes. I might be able to make the proof a bit cleaner with the extra information that I have about the homomorphisms, but it's nice to know that it's not necessary. Thanks a lot for the help.

jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5959
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

### Re: Finding the middle group in a SES.

Talith wrote:I might be able to make the proof a bit cleaner with the extra information that I have about the homomorphisms

I should hope so, its pretty hacky stuff. I was really just banging stuff together until something shook out. But anyway, np.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.