From mschoene@math.rwth-aachen.de Tue Jun 20 09:29:59 1995
Return-Path:
Received: from samson.math.rwth-aachen.de by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA23574; Tue, 20 Jun 95 09:29:59 EDT
Received: from hobbes.math.rwth-aachen.de by samson.math.rwth-aachen.de with smtp
(Smail3.1.28.1 #11) id m0sO2bG-000MPBC; Tue, 20 Jun 95 14:40 MET DST
Received: by hobbes.math.rwth-aachen.de (Smail3.1.28.1 #19)
id m0sO2bF-00026zC; Tue, 20 Jun 95 14:40 WET DST
Message-Id:
Date: Tue, 20 Jun 95 14:40 WET DST
From: "Martin Schoenert"
To: Cube-Lovers@ai.mit.edu
Cc: BRYAN@wvnvm.wvnet.edu
In-Reply-To: "Jerry Bryan"'s message of Sun, 18 Jun 1995 15:55:22 -0400 (EDT)
Subject: Re: Re: A Third Way to Calculate the Real Size of Cube Space?
Looking through the old messages about the real size of the cube group,
it appeared to me that no one has shown a proof for the Polya-Burnside
theorem. Since it is not difficult to prove, I decided to write one up.
In the following I will use TeX notation for formulae, i.e., formulae
are included in '$' signs, '{}' are used to group terms, '^' is used
for superscripts, and '_' for subscripts.
If $g \in G$, then I denote the set of elements that are really
equivalent to $g$ by $g^M$. Jerry denotes this set by {m'gm},
but $g^M$ is the more common notation in group theory.
The sum $\sum_{g \in h^M}{1/|g^M|}$ is simply 1, since it is the sum over
all elements in one M-conjugacy class (h^M) of 1 over the length of
that M-conjugacy class. Thus the sum $\sum_{g \in G}{1/|g^M|}$ is the
number of M-conjugacy classes.
Now we need a standard lemma from group theory, which tells us that the
length of a class $g^M$ of an element $g$ under the action of a group $M$
is equal to the size of the group $M$ divided by the size of the subgroup
of those elements of $M$ that fix $g$ (more precisely the index of that
subgroup in $M$, since the lemma is true, even if $M$ is infinite).
So using Jerry's notation this lemma gives $1/|g^M| = |Symm(g)|/|M|$.
Applying that to the above formula we see that the number of
M-conjugacy classes in $G$ is $\sum_{g \in G} {|Symm(g)|/|M|}$
Or, after a trivial change, $1/|M| \sum_{g \in G} {|Symm(g)|}$.
Assume that $(g^m == g)$ is 1 if $g^m$ is equal to $g$ and 0 otherwise.
Then we have $|Symm(g)| = \sum_{m \in M}{(g^m == g)}$.
Thus the number of M-conjugacy classes is
$1/|M| \sum_{g \in G} \sum_{m \in M} {(g^m == g)}$.
Now we can simply change the order of the two summations, so we get
$1/|M| \sum_{m \in M} \sum_{g \in G} {(g^m == g)}$.
But of course $\sum_{g \in G} {(g^m == g)}$ is obviously the number of
fixpoints of $m$. So we obtain the Polya-Burnside lemma: ``The number of
M-conjugacy classes is the average number of fixpoints of the elements
of $M$ w.r.t. their operation on $G$''.
However, here the operation is special, so we can simplify even further.
$g^m$ here means $m^{-1} g m$, so $(g^m == g)$ means $(m^{-1} g m == g)$,
which is equivalent to $(m == g^{-1} m g)$ (multiply the equation first
by $m$ and then by $g^{-1}$ from the left), which is $(m == m^g)$.
So the number of M-conjugacy classes is
$1/|M| \sum_{m \in M} \sum_{g \in G} {(m == m^g)}$.
But $\sum_{g \in G} {(m == m^g)}$ is simply the size of the subgroup of
those elements in $G$ that fix $m$. This is the centralizer of
$m$ in $G$. So the number of M-conjugacy classes is finally
$1/|M| \sum_{m \in M} |Centralizer(G,m)|$.
This is the formulation that I used to compute the real size of the cube
group with GAP.
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