Conjugacy classes in complete monomial groups |
Consider an element ( y, p) in H wr S n and assume that C1,C2, ... are the conjugacy classes of H. If
p= Õ nÎ c( p) (j n ...pl n-1j n),
in standard cycle notation, then we associate with its n-th cyclic factor (j n ...pl n-1j n) the element
h n( y, p) := y(j n) y( p-1j n) ...y( p-l n+1j n) = yy p ...y pl n-1(j n)
of H and call it the n-th cycleproduct of ( y, p) or the cycleproduct associated to (j n ...pl n-1j n) with respect to ( y, p). In this way we obtain a total of c( p) cycleproducts, ak( p) of them arising from the cyclic factors of p which are of length k. Now let aik( y, p) be the number of these cycleproducts which are associated to a k-cycle of p and which belong to the conjugacy class Ci of H (note that we did not say ''let aik ( y, p) be the number of different cycleproducts ``). We put these natural numbers together into the matrix
a( y, p):=(aik( y, p)),
This matrix has n columns (k is the column index) and as many rows as there are conjugacy classes in H (i is the row index). Its entries satisfy the following conditions:
aik( y, p) Î N, åi aik( y, p)=ak ( p), åi,k k ·aik( y, p)=n.
We call this matrix a( y, p) the type of ( y, p) and we say that ( y, p) is of type a( y, p).
Lemma: The conjugacy classes of complete monomial groups H wr S n have the following properties:
- CH wr S n ( y', p')=CH wr S n ( y, p) if and only if a( y', p')=a( y, p).
- The order of the conjugacy class of elements of type (aik) in H wr S n , H finite, is equal to
| H | nn!/ Õi,kaik!(k | H | / | Ci | )aik.- Each matrix (bik) with n columns and as many rows as H has conjugacy classes, the elements of which satisfy
bik Î N, åi,kk ·bik=n,occurs as the type of an element ( y, p) ÎH wr S n .- If H is a permutation group and a:= a(hn ( y, p)), then the cycle partition a( d( y, p)), where d denotes the permutation representation of the formula, is equal to
ån ln ·a(hn( y, p)),where ln ·a, a:= a(hn( y, p)), is defined to be (ln ·a1,ln ·a2, ...), and where ån ... means that the proper partition has to be formed that consists of all the parts of all the summands ln ·a(hn( y, p)).
Proof: A first remark concerns the cycleproducts introduced in the formula. Since in each group G the products xy and yx of two elements are conjugate, we have that hn( y, p) is conjugate to
yyp ...y pln-1( pzjn),
for each integer z.
The second remark is, that for each p' ÎS n and every y' ÎHn,
a( y, p)=a(( i, p')( y, p)( i, p')-1)=a(( y',1)( y, p)( y',1)-1).
This follows from the fact that both ( y p', p' pp'-1) and ( y' y y'p-1, p) are of type a( y, p).
A third remark is that a( y, p)=a( y', p') implies the existence of an element p" ÎSn which satisfies p= p" p' p"-1, and for which the cycleproducts hn( y, p) and hn( y' p", p) are conjugate.
It is not difficult to check these remarks and then to derive the statement (exercise).
harald.fripertinger "at" uni-graz.at | http://www-ang.kfunigraz.ac.at/~fripert/ | UNI-Graz | Institut für Mathematik | UNI-Bayreuth | Lehrstuhl II für Mathematik |
Conjugacy classes in complete monomial groups |