Finding the middle group in a SES.

[Talith]

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.

[jestingrabbit]

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.

[Talith]

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]

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.