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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