ExercisesEnumeration of symmetry classesThe cycle type of the induced action on 2-subsetsSome congruences

Some congruences

According to Corollary we obtain from Theorem the following results:

Corollary: For any subgroups G <= Sn and H <= Sm, the following congruences hold:
å pÎGmc( p) º0 ( | G | ), åh ÎHa1(h)n º0 ( | H | ),
and also
å( r, p) ÎH ´G Õi=1na1( ri)ai( p) º0 ( | H | | G | ),
as well as
å( y, p) ÎH wr G Õ n=1c( p)a1(hn( y, p)) º0 ( | H | n | G | ).

Further congruences show up in the enumeration of group elements with prescribed properties. This theory of enumeration in finite groups is, besides the enumeration of chemical graphs, one of the main sources for the theory of enumeration which we are discussing here. A prominent example taken from this complex of problems is the following one due to Frobenius: The number of solutions of the equation xn=1 in a finite group G is divisible by n, if n divides the order of G. There are many proofs of this result and also many generalizations. Later on we return to this problem, at present we can only discuss a particular case which can be treated with the tools we already have at hand.

Example: Let g denote an element of a finite group which forms its own conjugacy class and consider a prime number p, which divides | G | . We want to show that the number of solutions x ÎG of the equation xp=g is divisible by p. In order to prove this we consider the action of Cp on the set YX:=G p. The orbits are of length 1 or p. An orbit is of length 1 if and only if it consists of a single and therefore of a constant mapping (g', ...,g'), say. We now restrict our attention to the following subset M ÍG p:
M:= {(g1, ...,gp) | g1 ...gp=g }.
As g forms its own conjugacy class, we obtain a subaction of Cp on M (for example g1 ...gp is conjugate to gpg1 ...gp-1). Hence the desired number k of solutions of xp=g is equal to the number of orbits of length 1 in M. Now we consider the number l of orbits of length p in M. It satisfies the equation k+pl= | M | . As each equation g1g'=g has a unique solution g1 in G, we moreover have that | M | = | G | p-1. Thus
k ºk+pl= | M | º0 (p),
which completes the proof. We note in passing that the center of G consists of the elements which form their own conjugacy class, so that we have proved the following:
Corollary: If the prime p divides the order of the group G, then the number of p-th roots of each element in the center of G is divisible by p. In particular the number of p-th roots of the unit element 1 of G has this property (and it is nonzero, since 1 is a p-th root of 1), and hence G contains elements of order p.
This result can be used in order to give an inductive proof of Sylow's Theorem which we proved in example.

harald.fripertinger "at" uni-graz.at http://www-ang.kfunigraz.ac.at/~fripert/
UNI-Graz Institut für Mathematik
UNI-Bayreuth Lehrstuhl II für Mathematik
last changed: January 19, 2005

ExercisesEnumeration of symmetry classesThe cycle type of the induced action on 2-subsetsSome congruences