Consider an element in and assume that are the conjugacy classes of . If
in standard cycle notation, then we associate with its -th cyclic factor the element
of and call it the -th cycleproduct of or the cycleproduct associated to with respect to . In this way we obtain a total of cycleproducts, of them arising from the cyclic factors of which are of length . Now let be the number of these cycleproducts which are associated to a -cycle of and which belong to the conjugacy class of (note that we did not say let be the number of different cycleproducts ). We put these natural numbers together into the matrix
This matrix has columns ( is the column index) and as many rows as there are conjugacy classes in ( is the row index). Its entries satisfy the following conditions:
We call this matrix the type of and we say that is of type .
. Lemma
The conjugacy classes of
complete monomial groups have the following properties:
occurs as the type of an element .
where , is defined to be
, and where
means that the proper partition has to be formed that consists of all the
parts of all the summands .
Proof: A first remark concerns the cycleproducts introduced in . Since in each group the products and of two elements are conjugate, we have that is conjugate to
for each integer .
The second remark is, that for each and every ,
This follows from the fact that both and are of type .
A third remark is that implies the existence of an element which satisfies , and for which the cycleproducts and are conjugate.
It is not difficult to check these remarks and then to derive the statement (exercise ).
Exercises
E . Prove that the conjugacy class (in ) of an even element splits into two -classes if and only if the lengths of the cyclic factors of are pairwise different and odd (hint: use ).
E . Fill in the details of the proof of .