Please delete this (if you add a comment above answering my question. actually this should be a comment on Steve D.'s answer)

I wanted to comment on the example: a=...(-p+1,-p+2,...,-1,0)(1,2,...,p)(p+1,...,2p)... b=...(-p+2,-p+3,...,0,1)(2,...,p+1)(p+2,...,2p+1)... Martin Brandenburg wrote that he doesn't see that (ba)^p is the identity and I don't see it for (ab)^p either. [and surely either both at id or neither is.]

Could someone who believes that Steve D is correct please add a short comment at the appropriate point that makes one see that one gets (ab)^p=id?

I for one get:

(ab)^p applied to [-p+2] goes to [2]

Here is who I understand the notation to apply. Maybe there is a notational problem: (ab) [-p+2] = [-p+4] (ab) [-p+4] = [-p+6] (ab)^{(p-1)/2} [-p+2] = [1] b[1]=[-p+2] a[-p+2]=[-p+3] (ab)^{p-1}[-p+2]=[0] b(ab)^{p-1}[-p+2]=b[0]=[1] (ab)^{p}[-p+2]=a[1]=[2].

For you to understand my notation: My notation results in (123)[3]=[1], (34)[5]=[5], ((234)(345)^{2}[2]=[2] I hope that makes it understandable, I guess using the []-brackets i

Please just add a short comment to make the next reader thinking the same thing as I did, where to start the thoughts that he won't get the same result as I did.