[r-t] Semi-interesting observation
Philip Saddleton
pabs at cantab.net
Tue Oct 25 18:49:03 UTC 2005
Robin Woolley said on 25/10/2005 15:05:
> Let, as by convention, H = 13527486 and J = 12436587. Also, let R = 13456782
> and S = 12346587. (H & J were discussed a year or so ago and I adopt the
> convention of the algebra operating from right to left, e.g. HJ means 'H
> operating on J' - I'm too old to change without the danger of making simple
> slips in notation.
>
> Since the regular lead heads are all powers of H: (H^j, j=0,..,7), then the
> single at the lead end when the tenor is becoming 5ths place bell can be
> written as SJH^2.
>
> A little work with a pencil shows R is generated by having the single at
> 3rds which in this notation is SJH^3 followed by five plain leads (if a
> group A method is being used) thus:
>
> R = H^5.SJH^3.SJH^2.
>
> The semi-interesting observation is that if any of the cycles of rounds are
> required, then
>
> R^n = H^5.SJH^3n.SJH^2 for appropriate choice of n.
>
> It is the 3n term which I find semi-interesting.
This is because
R^2 = (H^5.SJH^3.SJH^2)(H^5.SJH^3.SJH^2)
=H^5.SJH^3.SJH^7.SJH^3.SJH^2
=H^5.SJH^3.SJ.SJH^3.SJH^2
=H^5.SJH^3.H^3.SJH^2
=H^5.SJH^6.SJH^2
since H^7 = (SJ)^2 = 1
etc.
and more generally, if
R = K.H^m.K^-1
then R^n = K.H^(mn).K^-1
I leave it as an exercise to prove that the only possibilities for K are
those that go from rounds into one of the six "cyclic" coursing orders
I gave previously.
(And I haven't mentioned conjugacy once)
--
Regards
Philip
http://www.saddleton.freeuk.com/
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