EQUIVALENCE RELATIONS

A relation from a set S to itself is called a relation on S. Any partial order on a set, such as “ ” on the real numbers, or “is a subset of” on a power set P(X), is a relation on that set. “Equals” is a relation on any set S and is defined by the subset {(a, a)|a S} of S × S. An equivalence relation is a relation that has the most important properties of the “equals” relation.

A relation E on a set S is called an equivalence relation if the following

If E is an equivalence relation on S and a S, then [a] = {x S|xEa} is called the equivalence class containing a. The set of all equivalence classes is called the quotient set of S by E and is denoted by S/E. Hence

S/E = {[a]|a S}.

Proposition 4.1. If E is an equivalence relation on a set S, then

Proof. (i) If aEb, let x be any element of [a]. Then xEa and so xEb by transitivity. Hence x [b] and [a] [b]. The symmetry of E implies that bEa, and an argument similar to the above shows that [b] [a]. This proves that [a] = [b].

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(ii) Suppose that aEb. If there was an element x [a] [b], then xEa, xEb, so aEb by symmetry and transitivity. Hence [a] ∩ [b] = Ø.

(iii) Parts (i) and (ii) show that two equivalence classes are either the same or disjoint. The reflexivity of E implies that each element a S is in the equivalence class [a]. Hence S is the disjoint union of all the equivalence classes.

A collection of nonempty subsets is said to partition a set S if the union of the subsets is S and any two subsets are disjoint. The previous proposition shows that any equivalence relation partitions the set into its equivalence classes. Each element of the set belongs to one and only one equivalence class.

It can also be shown that every partition of a set gives rise to an equivalence relation whose classes are precisely the subsets in the partition.

Example 4.2. Let n be a fixed positive integer and a and b any two integers. We say that a is congruent to b modulo n if n divides a − b. We denote this by

a ≡ b mod n.

Show that this congruence relation modulo n is an equivalence relation on Z. The set of equivalence classes is called the set of integers modulo n and is denoted by Zn.

Solution. Write “n|m” for “n divides m,” which means that there is some integer k such that m = nk. Hence a ≡ b mod n if and only if n|(a − b).

(iii)If a ≡ b mod n and b ≡ c mod n, then n|(a − b) and n|(b − c), so n|(a − b) + (b − c). Therefore, n|(a − c) and a ≡ c mod n.

In the congruence relation modulo 3, we have the following equivalence classes:

[0]= {. . . , −3, 0, 3, 6, 9, . . .}

[1]= {. . . , −2, 1, 4, 7, 10, . . .}

[2]= {. . . , −1, 2, 5, 8, 11, . . .}

[3]= {. . . , 0, 3, 6, 9, 12, . . .} = [0]

Any equivalence class must be one of [0], [1], or [2], so Z3 = {[0], [1], [2]}.

In general, Zn = {[0], [1], [2], . . . , [n − 1]}, since any integer is congruent modulo n to its remainder when divided by n.

One set of equivalence classes that is introduced in elementary school is the set of rational numbers. Students soon become used to the fact that 12 and 36 represent the same rational number. We need to use the concept of equivalence

class to define a rational number precisely. Define the relation E on Z × Z (where Z = Z − {0}) by (a, b) E (c, d) if and only if ad = bc. This is an

equivalence relation on Z × Z , and the equivalence classes are called rational numbers. We denote the equivalence class [(a, b)] by ab . Therefore, since (1, 2) E(3, 6), it follows that 12 = 36 .

Two series-parallel circuits involving the switches A1, A2, . . . , An are said to be equivalent if they both are open or both are closed for any position of the n switches. This is an equivalence relation, and the equivalence classes are the 22n distinct types of circuits controlled by n switches.

Any permutation π on a set S induces an equivalence relation, , on S where a b if and only if b = πr (a), for some r Z. The equivalence classes are the orbits of π. In the decomposition of the permutation π into disjoint cycles, the elements in each cycle constitute one orbit.

COSETS AND LAGRANGE’S THEOREM

The congruence relation modulo n on Z can be defined by a ≡ b mod n if and only if a − b nZ, where nZ is the subgroup of Z consisting of all multiples

of n. We now generalize this notion and define congruence in any group modulo one of its subgroups. We are interested in the equivalence classes, which we call cosets.

Let (G, ·) be a group with subgroup H . For a, b G, we say that a is congruent to b modulo H , and write a ≡ b mod H if and only if ab−1 H .

Proposition 4.3. The relation a ≡ b mod H is an equivalence relation on G. The equivalence class containing a can be written in the form H a = {ha|h H }, and it is called a right coset of H in G. The element a is called a representative of the coset Ha.

Proof. (i) For all a G, aa−1 = e H ; thus the relation is reflexive.

(ii)If a ≡ b mod H , then ab−1 H ; thus ba−1 = (ab−1)−1 H . Hence b ≡ a mod H , and the relation is symmetric.

(iii)If a ≡ b and b ≡ c mod H , then ab−1 and bc−1 H . Hence ac1 = (ab−1)(bc−1) H and a ≡ c mod H . The relation is transitive. Hence ≡ is an equivalence relation. The equivalence class containing a is

{x G|x ≡ a mod H } = {x G|xa−1 = h H }

={x G|x = ha, where h H }

={ha|h H },

Example 4.4. Find the right cosets of A3 in S3.

Solution. One coset is the subgroup itself A3 = {(1), (123), (132)}. Take any element not in the subgroup, say (12). Then another coset is

A3(12) = {(12), (123) Ž (12), (132) Ž (12)} = {(12), (13), (23)}.

Since the right cosets form a partition of S3 and the two cosets above contain all

Example 4.5. Find the right cosets of H = {e, g4, g8} in C12 = {e, g, g2, . . . , g11}.

As the examples above suggest, every coset contains the same number of elements. We use this result to prove the famous theorem of Joseph Lagrange (1736–1813).

80 4 QUOTIENT GROUPS

Lemma 4.6. There is a bijection between any two right cosets of H in G.

Proof. Let Ha be a right coset of H in G. We produce a bijection between Ha and H , from which it follows that there is a bijection between any two right cosets.

Define ψ: H → H a by ψ(h) = ha. Then ψ is clearly surjective. Now suppose

Theorem 4.7. Lagrange’s Theorem. If G is a finite group and H is a subgroup of G, then |H | divides |G|.

Proof. The right cosets of H in G form a partition of G, so G can be written as a disjoint union

G = H a1 H a2 · · · H ak for a finite set of elements a1, a2, . . . , ak G.

By Lemma 4.6, the number of elements in each coset is |H |. Hence, counting all the elements in the disjoint union above, we see that |G| = k|H |. Therefore, |H | divides |G|.

If H is a subgroup of G, the number of distinct right cosets of H in G is called the index of H in G and is written |G : H |. The following is a direct consequence of the proof of Lagrange’s theorem.

Corollary 4.8. If G is a finite group with subgroup H , then

|G : H | = |G|/|H |.

Corollary 4.9. If a is an element of a finite group G, then the order of a divides the order of G.

Proof. Let H = {ar |r Z} be the cyclic subgroup generated by a. By Proposition 3.13, the order of the subgroup H is the same as the order of the element

Corollary 4.11. If G is a group of prime order, then G is cyclic.

Proof. Let |G| = p, a prime number. By Corollary 4.9, every element has order 1 or p. But the only element of order 1 is the identity. Therefore, all the