COMMON SYSTEMS OF COSET REPRESENTATIVES
ASHAY DHARWADKER DISTINGUISHED PROFESSOR OF
INSTITUTE OF MATHEMATICS
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Using the axiom of choice,
we prove that given any group
G and a finite subgroup H,
there always exists a common system of coset representatives for the left
and right cosets of H in G. This result played a major role in the proof
of the Four Colour Theorem in 2000 and the
Grand Unification of the Standard Model with Quantum Gravity
in 2008. Google Scholar Citations © 2005
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Thanks to Fabrice Larere
for asking under what conditions there exist common systems of coset representatives
and for providing the first example. Thanks to Peter Cameron for pointing
out exactly where in the proof the finiteness of the subgroup is essential
and for providing the second example. The Endowed Chair of the Institute of Mathematics was bestowed upon
Distinguished Professor Ashay Dharwadker in 2012 to honour his fundamental contributions to Mathematics and Natural Sciences.
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We shall prove that given any group G
and a finite subgroup H, there always exists a common system of
coset representatives for the left and right cosets of H in G.
Precise definitions and examples are given below. The proof uses the standard
von Neumann - Bernays - Gödel (NBG) axioms of set theory [1]
together with
The Axiom of Choice. Given any set X of nonempty pairwise disjoint sets, there is a set Y, called a choice set, that contains exactly one element of each set in X. A nonempty set I together with a binary relation ≤ is called a partially ordered set if, for all i, j, k in I
The Well-Ordering Principle. Every set can be well-ordered. Proof. See [1], the proof of proposition 4.37. The axiom of choice implies Zorn's lemma. Zorn's lemma implies the well-ordering principle. ☐ In particular, given any set X, we may index the elements of X by a well-ordered index set I and write X = { xi | i in I }. In this notation we may now state and prove The Transfinite Induction Principle. Let X = { xi | i in I } be any set indexed by a well-ordered set I. If P is a property such that, for any i in I, whenever all xj with j < i have property P, then xi has property P, then all elements of X have property P. Proof. Let Y = { x in X | x has property P }. Suppose X - Y is nonempty, then there is a least element xi in X - Y. By the definition of least element and X - Y we must have, for any xj with j < i, that xj has the property P. But then, by hypothesis, xi has property P, a contradiction. Therefore, X - Y is empty and X = Y. ☐ A set G together with a binary operation (written here in the usual multiplicative notation) is called a group if
Lemma 1. Let G be a group and H a subgroup. Then G is the disjoint union of the set of double cosets { HgH | g in G }. Proof. Suppose x belongs to the double cosets Hg1H and Hg2H. Then x = h'1g1h'2 = h''1g2h''2 for some h'1, h'2, h''1, h''2 in H. Then g1 = h'1-1h''1g2h''2h'2-1 and so, for any h1, h2 in H, we have h1g1h2 = h1h'1-1h''1g2h''2h'2-1h2 showing that Hg1H is contained in Hg2H. Similarly g2 = h''1-1h'1g1h'2h''2-1 and so, for any h1, h2 in H, we have h1g2h2 = h1h''1-1h'1g1h'2h''2-1h2 showing that Hg2H is contained in Hg1H. Thus Hg1H = Hg2H. This proves that distinct double cosets cannot have any elements in common and must be disjoint. Since every g in G can be written as g = 1g1, every g in G belongs to at least one double coset, namely HgH. This proves that the union of the disjoint double cosets is all of G. ☐ Lemma 2. Let G be a group and H a subgroup. Let HgH be a fixed double coset of H in G. Then
Lemma 3. Let G be a group and H a finite subgroup. Let HgH be a fixed double coset of H in G. Then there exists a system of representatives for the left cosets of H in G that are contained in HgH such that distinct representatives belong to distinct right cosets of H in G that are contained in HgH. Proof. There are two cases.
Proof. By lemma 1 and the axiom of choice, select a set { gj in G | j in J } such that { HgjH | j in J } is the set of disjoint double cosets whose union is G. By lemma 3 and the axiom of choice, select a set { Sj | j in J } where Sj is a set of representatives for the left cosets of H in G that are contained in HgjH such that distinct representatives belong to distinct right cosets of H in G that are contained in HgjH. Form the union S of all the sets in { Sj | j in J }. Then S is a system of representatives for all left cosets of H in G such that distinct representatives belong to distinct right cosets of H in G. But, as observed above, there is a bijection between the set of all left cosets of H in G and the set of all right cosets of H in G. Thus each right coset of H in G must have an element in S. It follows that the set S must be a common system of representatives for the left and right cosets of H in G. ☐ Example 1. We first give an example of a group G and subgroup H that satisfy the hypotheses of the proposition. Let G = S3 denote the symmetric group on three letters consisting of all permutations of the set {1, 2, 3}
Consider the subgroup H = { 1, ε }. The double cosets of H in G are { 1, ε } and { α, β, γ, δ }. The left cosets of H in G are { 1, ε }, { α, γ } and { β, δ }. The right cosets of H in G are { 1, ε }, { α, δ } and { β, γ }. We may select ε as a common representative for the left coset { 1, ε } and the right coset { 1, ε } contained in the double coset { 1, ε }. We may select γ and δ as common representatives for the left cosets { α, γ }, { β, δ } and right cosets { α, δ }, { β, γ } respectively, contained in the double coset { α, β, γ, δ }. The union of the selected representatives { ε, γ, δ } is a common system of representatives for the left and right cosets of H in G in this example where H is finite. Example 2. Finally, we give an example of a group G and subgroup H that do not satisfy the hypotheses of the proposition and for which there cannot exist a common system of representatives of the left and right cosets of H in G. Consider the group G generated by x, y subject to the relation xy = y2x. Let H be the subgroup generated by y. Then H = { yn | n is any integer } is an infinite subgroup. Using the relation inductively, it is easy to see that for any integer n, xynx-1 = y2n. Thus the subgroup xHx-1= { y2n | n is any integer } is properly contained in the subgroup H. This implies that the left coset xH is properly contained in the right coset Hx which is equal to the double coset HxH. But then by lemma 2, the double coset HxH contains at least two left cosets and exactly one right coset. Thus, it is impossible to select representatives for the left cosets of H in HxH that belong to distinct right cosets of H in HxH. By lemma 1 and lemma 2, it follows that it is impossible to select representatives for the left cosets of H in G that belong to distinct right cosets of H in G. Thus, there cannot exist a common system of representatives for the left and right cosets of H in G in this example where H is infinite. |
REFERENCES |
[1] Elliott Mendelson, Introduction to Mathematical Logic, Wadsworth Inc., 1987. |
[2] Nathan Jacobson, Basic Algebra I, W. H. Freeman & Co., 1974. |
Copyright © 2005 by Ashay Dharwadker. All rights reserved.