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Date: Tue, 3 Jan 95 12:59 WET
From: "Martin Schoenert"
To: Cube-Lovers@ai.mit.edu
In-Reply-To: "Jerry Bryan"'s message of Mon, 2 Jan 1995 23:00:01 EST <9501030407.AA25455@life.ai.mit.edu>
Subject: Re: Re: Normal Subgroups of G
Jerry Bryan wrote in his e-mail message of 1995/01/02
I am still absorbing this article, which exceeds my current
knowledge of group theory. But at the risk of asking a dumb
question, doesn't the center of GE (and of G) in fact consist
of more than just the Superflip and the identity? Does it
not also include the Pons Asinorum and the composition of
the Pons Asinorum and the Superflip? Call the Pons Asinorum P
and the Superflip E. I think you are saying Z={I,E}. But
isn't the center {I,P,E,PE}, with subgroups {I,P}, {I,E},
{I,PE}, and {I}? These should all be central, and hence
also normal, I would think.
This is not a dump question. Clearly ``Pons Asinorum'' P looks very
regular, and it is not farfetched to think that it is central. But it is
not. 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.
It is in fact not too difficult to find the centers.
Recall that GE consists of the elements ( c_1, c_2, ..., c_12; p ),
where c_1 + c_2 + ... c_12 = 0 (mod 2) and p in S_12.
Since we can permute the components c_i in any way
by conjugation with an appropriate element (0,0,...,0;p),
it follows that any central element must have c_1 = c_2 = ... = c_12.
Furthermore any central elemement must have a permutation p that is
central in S_12. So we see that we have exactely two elements in the
center of GE, namely (0,0,...,0;) and (1,1,...,1;).
An easy argument shows that this is also the center of GE'.
The same argumentation works for GC, but the element
(1,1,...,1;) is not in GC, since 1 + 1 + ... + 1 <> 0 (mod 3)
(since we have 8 summands). So GC has trivial center.
Again an easy argument shows that this is also the center of GC'.
The center of the direct product GCE is of course the direct product of
the centers of GC and GE. So we see that the center of GCE is again Z.
And again an easy argument shows that this is also the center of G.
If you have more questions, please do ask. I have tried to make my
article selfcontained. I think the only result that I used without
proof is that S_8 and S_12 have only one proper normal subgroup.
The problem is that in order to keep the article reasonably short,
I had to be rather terse at several places.
Have a nice day.
Martin.
-- .- .-. - .. -. .-.. --- ...- . ... .- -. -. .. -.- .-
Martin Sch"onert, Martin.Schoenert@Math.RWTH-Aachen.DE, +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany