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From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Subject: Re: Square like groups

With respect to the square group, I wrote:

> 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.

I should really have waited until I got back home to Singmaster's
book, which has a marvelous explanation of why the squares group has
index 6 in the pseudosquare group.

First, the edges are permuted in in all possible ways consistent with
    1. remaining in their "equators" of four edges,
    2. not being flipped, and
    3. having a permutation parity equal to that of the corners.
so we need only consider the 2x2x2 cube, and then we fix the BLD
corner in place.  Corners don't get twisted, so we consider only the
permutation.

We express the generators as permutations of the seven movable
corners, expressed as follows:

            2-------A
           /       / \
          /   T   /   \      F^2 = B^2 = (1,4)(B,C),
         /       /     \     R^2 = L^2 = (1,3)(A,C),
        B-------1   R   3    T^2 = D^2 = (1,2)(A,B).
         \       \     /
          \   F   \   /
           \       \ /
            4-------C

The neat part is to notice that the permutation on {A,B,C} is
determined by the permutation on {1,2,3,4}.  We do this by
representing these generators as symmetries on a tetrahedron, labelled
as follows.

            1-----------C-----------2
             \`-.               .-'/
              \  `A.         .B'  /
               \    `-.   .-'    /
                \      `4'      /
                 \      :      /
                  B     :     A
                   \    :    /
                    \   C   /
                     \  :  /
                      \ : /
                       \:/
                        3

Notice that the symmetry that permutes the tetrahedron's vertex labels
as (1,4) also permutes the edge labels as (B,C), corresponding to F^2
in the cube's action.  Similarly (1,3) implies (A,C) and (1,2) implies
(A,B).

With respect to Mark Longridge's having noticed that the square group
is mapped to itself under conjugation by an antislice (L R), the proof
turns out to be pretty easy.  First, we notice that we may consider
conjugation by a slice (L R') since that differs by a square (R^2)
from the antislice.  Now we work in the group that includes whole-cube
orientations, and perform the slice in the mechanically easy way, as a
4-cycle of face centers and an equatorial 4-cycle of edges.  Note that
all the edges of the equator are flipped (with respect to the
orientation that is preserved by the psueudo-square and square groups)
by the slice.

So if S is a square-group process that rotates the edges in an equator
E, the process

     Slice' S Slice S'

has the following actions:

1. Identity on the corners and the two equators other than E,
   because they are not moved by the slice,
2. Identity on the face centers, because they are not moved by S,
3. Flips each edge of E twice (once in Slice' and once in Slice), so
   restores the orientation, and
4. Is an even permutation of the edges in E (odd in Slice, odd in
   Slice', and equal in S and S').

The even permutation (4) of the edges in E is a slice group process,
as Mark noted, as for instance the 3-cycle (R^2 F^2 R^2 T^2)^2 F^2.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil