   Date: Tue, 27 Dec 94 20:00:04 -0500   From: michael reid <mreid@ptc.com >   Subject: Normal Subgroups of G

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     