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From: Dan Hoey
To: ajw01@uow.edu.au
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199805010552.PAA00579@wumpus.its.uow.edu.au> (message from
Andrew John Walker on Fri, 1 May 1998 15:52:34 +1000 (EST))
Subject: Re: Square like groups
Andrew Walker asks:
> Does anyone have any information on patterns where each
> face only contains opposite colours, but are not in the square
> group?
We may call this the "pseudosquare" group P. It consists of
orientation-preserving permutations that operate separately on the
three equatorial quadruples of edge cubies and the two tetrahedra of
corner cubies, and for which the total permutation parity is even. So
Size(P) = 4!^5 / 2 = 3981312.
> L' R U2 L R' may be an example.
No, that's in the square group, says GAP. Also, Mark Longridge
noticed (8 Aug 1993) that the square group is mapped to itself under
conjugation by an antislice (though I don't recall a proof--is there
an easy one?). Your position is (L R)' R2 T2 R2 (L R), so this result
would apply. Does anyone have a square process for it?
> If square moves are applied to such patterns to form new groups, how
> many such groups exist?
Consider the subgroup of P consisting of positions in which the
parity of the corner permutation is even. (The edge permutation will
then also be even, and the parity of the permutations of the two edge
tetrahedrons will be equal). Call it AP, for "alternating P".
Size(AP) = Size(P)/2 = 1990656.
The square group S is a subgroup of index 3 in AP, so
Size(S)=Size(AP)/3=663552. I don't have a very criterion for choosing
elements of AP to be in S, except that it has to do with a correlation
between the permutations of the two tetrahedrons of corners, provided
those permutations are of the same parity (as they must be for the
position to be in AP).
According to GAP, these are the only three possibilities. To be
explicit, let us label the cube's corners
1 D B 3
C 2 4 A
Then we can partition S4 into six cosets:
C1 = { (), (3,4)(1,2), (1,4)(2,3), (2,4)(1,3) }
C3 = { (1,2,3), (1,4,2), (1,3,4), (2,4,3) }
C2 = { (1,3,2), (1,4,3), (2,3,4), (1,2,4) }
C4 = { (1,2), (1,4,2,3), (1,3,2,4), (3,4) }
C5 = { (2,3), (1,4), (1,3,4,2), (1,2,4,3) }
C6 = { (1,3), (2,4), (1,4,3,2), (1,2,3,4) }
and similarly D1,D2,...,D4 for S4 acting on {A,B,C,D}. Now let c be
an arbitrary permutation in P that fixes {A,B,C,D} elementwise, and
let Coset(c) be the coset to which c's operation on {1,2,3,4} belongs.
Let d be an arbitrary permutation in P that fixes {1,2,3,4}
elementwise, and let Coset(d) be the coset to which d's operation on
{A,B,C,D} belongs. Then the group generated by ~~ depends only
on Coset(c) and Coset(d):
Coset(d)
D1 D2 D3 D4 D5 D6
Coset(c)
C1 S AP AP P P P
C2 AP S AP P P P
C3 AP AP S P P P
C4 P P P S AP AP
C5 P P P AP S AP
C6 P P P AP AP S
There may be some wisdom to be gained in seeing that C1 is normal in
S4, so S4/C1 is isomorphic to S3. We can represent the Ci and Di by
their action on {1,2,3,A,B,C}. The above table shows whether the
group ,
has order 6, 18, or 24.
I'd love to hear a more explanatory description of this phenomenon,
especially if it explains the absence of a subgroup of index 3 in P.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
~~