I wrote in my e-mail of 1995/01/03
Only one out of 332640 elements of GE (and of G) centralizes P.
That is to say that the index of the centralizer of P in GE has index
332640 in GE. Since all elements of GC commute with all elements of
GE, the index of the centralizer of P in G also has index 332640 in G.
Z is indeed the center of GE', GE, G, G', and GCE.
Mark Longridge answered in his e-mail of 1995/01/06
I get the fact that only the super-flip (or 12-flip) is the centre of
G and the centre of GE. Another way to look at it would be the centre
of the cube group must effect all the corners & edges in the same
way, and only the super-flip fits these conditions when we allow
all 6 generators < U, D, F, B, L, R > to be used.
This sounds very plausible. But I must admit that I find it notoriously
difficult to turn such plausible arguments into proper proofs.
If you try, you may in fact end up with something similar to my proof.
Because the crucial part in my proof is that a central element
must have all components in the wreath product equal, because one
has the full symmetric group S_12 acting on the 12 components.
In the case of the smaller group < U, R > we can get 6 corners twisted
either clockwise or counter-clockwise, thus effecting all the corners
and edges the same, due to the fact we can have 6 twists the same and
< U, R > only contains 6 corners, and so this is the centre of
< U, R >.
This is the ``odd'' element I referred to in my message on shift
But I don't understand how only one out of 332,640 elements of GE
and G centralizes P. I thought that GE had:(12 ^ 2 / 2 ) * 12! = 980,995,276,800 elements
That is to say that the group on the cube of edges only has
980,995,276,800 elements. To be honest I'm not sure what P
represents! Jerry refers to P as the Pons Asinorum, but I think
the term may have two meanings in the two messages.
Sorry, that is just me wrestling with English. What I meant to say was ``... only one out of *every* 332640 elements of GE ...''. That is, of the total 980995276800 elements in GE only 980995276800/332640 = 2949120 elements centralize P. And I used the definition of P from your e-mail of 1995/01/03, i.e., P = (F2 B2) (U2 D2) (L2 R2) = (F2 B2) (L2 R2) (U2 D2) = ... (one gets the same element independent of the order of the three pairs).
Z is the centre of G right? I need an ANSI standard math dictionary,
but I doubt such a book exists.
I'm going to tackle some more cube terminology in my next message.
Z in this case refers to the subgroup generated by the superflip.
I wrote in my e-mail of 1994/12/30
Namely VE has one normal subgroup of size 2 , generated by the element
(1,1,...,1;<identity>). You may not recognize this element, but it is
in fact the superflip, which flips all twelve edges. I shall call this
I would have preferred to call it C, but C was already taken for the
group of rotations of the entire cube. Thus I took Z instead, because
``Zentrum'' is the german word for center. It is not uncommon to use
Z to denote the center of a group G, e.g., Huppert uses Z(G) for the
center in his ``Theory of Groups''.
-- .- .-. - .. -. .-.. --- ...- . ... .- -. -. .. -.- .- Martin Sch"onert, Martin.Schoenert@Math.RWTH-Aachen.DE, +49 241 804551 Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany