From cube-lovers-errors@mc.lcs.mit.edu Thu Aug 20 18:08:04 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id SAA21659; Thu, 20 Aug 1998 18:08:03 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Aug 20 18:06:49 1998
Date: Thu, 20 Aug 98 18:06:30 EDT
Message-Id: <9808202206.AA13431@sun28.aic.nrl.navy.mil>
From: Dan Hoey
To: jbryan@pstcc.cc.tn.us
Cc: reid@math.brown.edu, cube-lovers@ai.mit.edu
Subject: Group centers (oops)
I wrote:
> So if c is any element of the center of G* -- i.e., c commutes
> with all elements of M and G -- then Symm(x)=Symm(x c). As is
> well known to cube-lovers, the center of the usual cube group consists
> of the identity and the superflip. In the supergroup, we may also
> compose these with Big Ben (all face centers rotated 90 degrees) and
> Noon (Big Ben squared).
In short, I should not have included Big Ben in that paragraph, only Noon.
The long explanation is that both of these are in the center of the
usual supergroup, as is any position that differs from Solved only by
face center orientation. But for the Symm(x)=Symm(x c) argument to
work, c must be in the center of the group generated by the union of
the supergroup with M. This is equivalent to saying that c must be
in the center of the supergroup and be M-symmetric. Big Ben is only
C-symmetric.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil