Exercise 15, § 6
Algebra, IV, p. 101
Let be a finite set with elements. We denote by the set of elements of such that and . With every equivalence relation R in T we associate an element of as follows: if are the equivalence classes under R, numbered so that , put with if and if . Let () be indeterminates; for each put , where the summation is extended to the set of mappings such that R is the equivalence relation associated with . Put .
a) Prove that if of the equivalence classes under R arc of cardinal i for . (See formula (11) of IV, p. 65 for the definition of .)
b) Prove the relation where the are as in a) and for every integer .
c) Show that , where is defined as in Exercise 14.
d) Hence deduce an expression of the as a polynomial in with rational coefficients, where .
e) For put . Show that , where the integer is defined as follows: suppose that (resp. ) of the components of (resp. ) are equal to i (for ). Then is the product , where N is the number of pairs of equivalence relations R, S on T such that and such that the intersection of an equivalence class under R with an equivalence class under S has at most one element. Deduce the symmetry relation .
f) Show that there is an involution in , characterized by the fact that the relations and are equivalent, for i, j in the range from 1 to k. Show that and . Hence obtain the relation (with the notation of IV, p. 66).
Conventions (cf. page IV, 66).
A symbol such as , where are integers , denotes an element of where having components equal to , components equal to , etc. Thus denotes the element of equal to , while denotes the element of .
On the right of the row of index of the matrix C the expression of as monomial in is given. For example, for and we have
We have used a corresponding convention for the columns of the matrix D. For example, for and we have
We recall that is the sum of all monomials of the form for in the ring .
An empty box indicates a zero element of the matrices C and D.
k = 2 2 1^2 2 1 2 S_1^2 1^2 1 S_2
2 1^2 2 1 -2 1^2 1 S_1^2 S_2
k = 3 3 2.1 1^3 3 1 3 6 S_1^3 2.1 1 3 S_1 S_2 1^3 1 S_3
3 2.1 1^3 3 1 -3 3 2.1 1 -3 1^3 1 S_1^3 S_1 S_2 S_3
k = 4 4 3.1 2^2 2.1^2 1^4 4 1 4 6 12 24 S_1^4 3.1 1 2 5 12 S_1^2 S_2 2^2 1 2 6 S_2^2 2.1^2 1 4 S_1 S_3 1^4 1 S_4
Matrix C
4 3.1 2^2 2.1^2 1^4 4 1 -4 2 4 -4 3.1 1 -2 -1 4 2^2 1 -2 2 2.1^2 1 -4 1^4 1
S_1^4 S_1^2 S_2 S_2^2 S_1 S_3 S_4 Matrix D
TABLE
k = 5 5 4.1 3.2 3.1^2 2^2.1 2.1^3 1^5 5 1 5 10 20 30 60 120 S_1^5 4.1 1 3 7 12 27 60 S_1^3 S_2 3.2 1 2 5 12 30 S_1 S_2^2 3.1^2 1 2 7 20 S_1^2 S_3 2^2.1 1 3 10 S_2 S_3 2.1^3 1 5 S_1 S_4 1^5 1 S_5
Matrix C
5 4.1 3.2 3.1^2 2^2.1 2.1^3 1^5 5 1 -5 5 5 -5 -5 5 4.1 1 -3 -1 5 1 -5 3.2 1 -2 -1 5 -5 3.1^2 1 -2 -1 5 2^2.1 1 -3 5 2.1^3 1 -5 1^5 1
S_1^5 S_1^3 S_2 S_1 S_2^2 S_1^2 S_3 S_2 S_3 S_1 S_4 S_5 Matrix D
Solution
No solution has been written yet.