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Date: Tue, 27 Dec 1994 20:00:04 -0500
From: mreid@ptc.com (michael reid)
Message-Id: <9412280100.AA16989@ducie.ptc.com>
To: Cube-Lovers%ai.mit.edu@ptc.com
Subject: Normal Subgroups of G
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jerry writes

[ ... ]

>                                                             My
> first question is that Frey and Singmaster do not state that At and
> Af are normal subgroups of G.  It seems obvious that they are.

indeed.

> However, is the formal argument that (for example) At is a normal
> subgroup of Ac and Ac is a normal subgroup of G; hence, At is a
> normal subgroup of G?

but this argument is not valid.  your question might be rephrased:

if  H  is a normal subgroup of  G  and  K  is a normal subgroup of  H,
does it follow that  K  is a normal subgroup of  G ??

the answer is no.  here's an easy counterexample:

let  G  be the alternating group  A_4,  H  the subgroup of order 4
{e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)},  and  K  the subgroup
{e, (1 2)(3 4)}.  it is easy to see that  H  is normal  in  G  and
K  is normal in  H.  however, if  x  is any three cycle (for example),
xK != Kx.

[ ... ]

>                         "Sane" is a term used by Frey and Singmaster
> in their proof of conservation of twist and flip.  In general, it
> is easy to see if a cubie is twisted or flipped when it is home,
> but it is not so easy to see if it is twisted or flipped when it
> is not home.  Their proof (and the others I have seen) define a
> frame of reference so that you can tell if a cube is twisted or
> flipped when it is not home.  A cubie which is not twisted or
> flipped in this frame of reference is sane.

here's a completely different proof of "conservation" which doesn't
use any frame of reference.

instead of thinking of permutations of edge cubies, think of
permutations of the facelets of the edges.  any quarter turn
induces two four cycles of these edge facelets, which is an even
permutation.  thus, any legal position has an even permutation
of the edge facelets.  however, a single flipped edge is just
a two cycle of edge facelets, an odd permutation, and therefore
is not a legal position.

my proof for conservation of twist is slightly more sophisticated,
but i think it's worthwhile.

the group of legal corner states may be viewed as a subgroup of the
wreath product  S_8 wr C_3.  we have a natural homomorphism

               S_8 wr C_3  --->  C_3              (*)
defined  by
               (s, c_1, ... , c_8) |-->  c_1 + ... + c_8

(the cyclic group  C_3  is written additively).  it is easy to see
that this is a homomorphism, but it uses the fact that  C_3  is abelian.
(in general, we have a natural homomorphism

               G wr H  --->  H^ab  ( =  H / [H, H] )

defined in the same way.)

conservation of corner twist is equivalent to saying that all legal
corner states are in the kernel of the map given in  (*).  however,
any quarter turn has order 4, so its image in  C_3  must be the
identity.  thus all quarter turns lie in the kernel, and therefore
the same is true of all legal positions.

(actually, i've cheated slightly here.  we actually need a frame of
reference in order to view the group of corner states as a subgroup of
S_8 wr C_3.)

mike