Conjugacy classes in complete monomial groups |

Consider an element *( y, p)* in *H wr S _{ n} * and assume that

p= Õ_{ nÎ c( p)}(j_{ n}...p^{l n-1}j_{ n}),

in standard cycle notation, then we associate with its * n*-th cyclic factor
*(j _{ n} ...p^{l n-1}j_{ n})* the element

h_{ n}( y, p) := y(j_{ n}) y( p^{-1}j_{ n}) ...y( p^{-l n+1}j_{ 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} ...p^{l n-1}j_{ n})*
with respect to

a( y, p):=(a_{ik}( 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:

a_{ik}( y, p) ÎN, å_{i}a_{ik}( y, p)=a_{k}( p), å_{i,k}k ·a_{ik}( 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 groupsH wr Shave the following properties:_{ n}

C.^{H wr S n }( y', p')=C^{H 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
(ain_{ik})H wr S,_{ n}Hfinite, is equal to| H |^{n}n!/ Õ_{i,k}a_{ik}!(k | H | / | C^{i}| )^{aik}.- Each matrix
(bwith_{ik})ncolumns and as many rows asHhas conjugacy classes, the elements of which satisfyoccurs as the type of an elementb_{ik}ÎN, å_{i,k}k ·b_{ik}=n,( y, p) ÎH wr S._{ n}- If
His a permutation group anda:= a(h, then the cycle partition_{n}( y, p))a( d( y, p)), whereddenotes the permutation representation of the formula, is equal towhereå_{n}l_{n}·a(h_{n}( y, p)),l, is defined to be_{n}·a, a:= a(h_{n}( y, p))(l, and where_{n}·a_{1},l_{n}·a_{2}, ...)åmeans that the proper partition has to be formed that consists of all the parts of all the summands_{n}...l._{n}·a(h_{n}( 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
*h _{n}( y, p)* is conjugate to

yy_{p}...y_{ pln-1}( p^{z}j_{n}),

for each integer *z*.

The second remark is, that for each * p' ÎS _{ n} * and every

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

A third remark is that *a( y, p)=a( y', p')* implies the existence of an
element * p" ÎS _{n} * which satisfies

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 |