Conjugacy

There is another equivalence relation defined on a group, which is very important. The concept of conjugate elements and conjugate classes in a group is basic to the study of normal subgroups.

7.5.1 Definition Let G be a group. We say that elements x, y of G are conjugate (written x ∼ y) if y = g⁻¹xg for some g ∈ G.  □

Example. Let σ and σ' two permutations of 𝓢_n. We say that the permutation σ is conjugate to the permutation σ' if there exists a permutation τ in 𝓢_n such that

Fix any x in a group G, then for *every* g in G, the element gxg^-1 is in G, and all of these are "the conjugates" of x. They always exist. Sometimes x itself is the only when, such as when G is abelian.

For example in the permutation group 𝓢₃, the element (1, 2) is conjugate to (3, 1) because (3, 1) = (1, 2, 3) (1, 2) (2 , 3)⁻¹.

Conjugacy is an equivalence relation, i.e. Let (G, *) be a group and a,b ∈ G. If the element a is conjugate to b ∈ G then:

1. Reflexive, each element is conjugate to itself, since ∀a ∈ G, there exists e ∈ G such that a = e * a * e⁻¹, which implies, a ~ a.

2. Symmetric, b ~ a. For a,b ∈ G, a = x * b * x⁻¹, for some x ∈ G ⇒ b = x⁻¹ * a * x = y * a * y⁻¹ where y = x⁻¹ ∈ G.

3. Transitive: For a,b,c ∈ G, if a ~ b and b ~ c, then a ~ c.

   a = x * b * x⁻¹  and   b = y * c * y⁻¹
   ⇒ a = x * (y * c * y⁻¹) * x⁻¹
   ⇒ a = (x * y) * c * (y⁻¹ * x⁻¹)
   ⇒ a = (x * y) * c * (x⁻¹ * y)⁻¹ ⇒ a ~ c.

Hence, G is the disjoint union of the equivalence classes, which are called conjugacy classes. Thus the conjugacy class of x ∈ G is

Note that the conjugacy class of x is {x} if and only if x commutes with each elements of G. We say that a conjugacy class is trivial if it contains only one member.

To find the elements a, b ∈ G which determine the same conjugate of x, notice that axa⁻¹ = bxb⁻¹ if and only if (b⁻¹a) = x(b⁻¹a), it turns out that we need to find the elements that commute with x. We can calculate the size of a conjugacy class in terms of another subgroup of G.

7.5.2 Definition. The centraliser of the element x ∈ G, written C_G(x), is the set of all elements of G which commute with x:

The centralizer of x contains at least x itself as well as the center Z(G) of G, because the elements in the center commute with every element in G. Also, x ∈ Z(G) iff C(x) = G.

There is a procedure by which conjugates of elements in a permutation group may be computed with ease.

7.5.3 Proposition Let σ a permutation, written as a product of disjoint cycles. Then

1. Every conjugate permutation σ' = τστ⁻¹ of σ has the same cyclic structure of σ. Furthermore, the integers appearing in the cycles of σ' are obtained applying the permutation τ to the integers that appear in the cycles of σ i.e. it is obtained from that for σ by substituting τ(k) for k throughout.

2. If σ and σ' are two permutation of 𝓢_n which have the same cyclic structure, then they are conjugate.

Proof. Let a and b two consecutive numbers in the writing of any of the two cycles of σ (the last element and the first are as well consecutives), i.e.

Let’s examine how τστ⁻¹ acts on the different elements. Let τ(a) = s and τ(b) = t, then

that is, b is the consecutive element of a in the cyclic structure of σ, the τ(b) is the consecutive of τ(a) in the cyclic structure of τστ⁻¹. This means that if for example

two permutations with the same cyclic structure. We prove now that they are conjugates. Indeed we've σ' = τστ⁻¹, where τ is a permutation such that

that is, having the behavior we described above on the elements of the cyclic structure of σ and on the elements that do not appear in σ in a compatible way with a permutation.

Remark. It is possible that two permutations of A_n which are conjugated in 𝓢_n are not conjugated in A_n, because there is not τ in A_n such that σ'= τστ⁻¹, that is the τs permutations are in 𝓢_n but not in A_n.

7.5.4 Example. In 𝓢₇ let

We have that

Since we have that (τ(1), τ(5)) (τ(2), τ(3), τ(4)) = (4,2)(6,1,3).

7.5.5 Example. Given

It is well known that an equivalence relation provides a unique way to classify the elements of a set.

7.5.6 Proposition. There are as many conjugate classes in 𝓢_n as different cyclic structures.

In 𝓢₃, 𝓢₄ and 𝓢₅ the different cyclic structures

Where (−) represents the id permutation. Thus in 𝓢₃, 𝓢₄, 𝓢₅ there are respectively 3, 5 and 7 conjugate classes, namely

Let's count now the number of permutations in each conjugate class. It suffices to count the number of distinct r-cycles, r = 1, ..., n and the cyclic products in 𝓢_n.

7.5.7 Proposition. The number of r-cycles in the symmetric group 𝓢_n is given, for 1 ≤ r ≤ n, by the following equivalent formula

Proof. By its cyclic nature, a cycle of length r can be represented in r different equivalent ways by cyclically shuffling its members. For example, the cycle (1, 2, 3) can also be represented as (2, 3, 1) and (3, 2, 1). To obtain the distinct cycles of length r we divide r! by r, that is (r − 1)!. We multiply then, this number by the number of ways we can pick r numbers among n i.e.: C_{n, r} ways (combinations without repetitions); Thus we have

With a similar reasoning (see also the Exercises) we can derive the number of permutations in 𝓢_n whose cycle structure consists of r_ii cycles of a given length.

7.5.8 Proposition. The number of permutations in 𝓢_n whose cycle structure consists of r_i cycles of length k_i, for i = 1, ..., m is

Proof. We write down a template consisting of r_i cycles of length k_i, whit − in place of the elements {1,..., n}. For example, if the cycle lengths are 3,3,2,1, then the template is

We can fill the blanks in n! different ways. However, there are two ways in which the number n! is an overcount of the desiered number. For each r-cycle, you divide by k_i as only the cyclic order within a cycle plays a role, not which element we start with. Second, for each cycle length k_i, the r_i! arrangements of the r_i cycles of length k_i are counted separately, but they give the same permutations, so we must divide also by r₁! ⋅⋅⋅ r_m!.  □

If you try to count permutations of type (-,-)(-,-)(-,-) in 𝓢₆ you select the pairs forming the 2-cycles in three stages. Because you don't care about the ordering of the stages, the answer is (C_6,2 ⋅ C_4,2 ⋅ C_2,2)/3! = 15.

Based on the two previous examples we can state the following proposition.

7.5.9 Definition. Let n an integer, we say that the sequence of positive integers n₁,n₂ ⋅⋅⋅ n_t with n₁ > n₂ > n₃ ⋅⋅⋅ > n_t is a partition of the integer n.

For example the partitions of the integer 4 are:

you will notice that each parittion of the integer 4, corresponds to one of the five cyclic structure of 𝓢₄, (the last corresponds to the identity, the first to a 4-cycle).

7.5.10 Proposition. The number of Conjugacy class 𝓢_n is the same as the partitions of the integer n.

If we want to calculate the Conjugacy classes of A_n we must be careful: permutations which are conjugate in S_n are not necessarily conjugate in A_n. e.g in S₃, (1 2 3) is conjugate to (1 3 2) because (2 3)⁻¹ (1 2 3) (2 3) = (1 3 2); but (2 3) is not in A₃. This tells us that (1 2 3) and (1 3 2) are in separate conjugacy classes.

Let's examine another example.

Example 7.5.11. The elements of the alternating group A₄ of S₄ are given below:

To go from (123) to (243) requires you change 1 into 2 and 2 into 4. So clearly (12)(24)(123) ((12)(24))⁻¹ = (243). It involves two substitutions. And (12)(24) is in A₄. You can get the others in the class the same pattern.

In the full symmetric group S₄ all 3-cycles are conjugate, e.g. going from (123) to (132) takes one substitution, 2 into 3:

But (2 3) does not lie in A₄, thus we split the 3-cycle of S₄ in two conjugacy classes so that (123) and (132) are in separate classes of A₄. The same observation applies to other pairs of 3-cycles that involve the same set of objects. On the other hand (12)(34), (13)(24) and (14)(23) remain conjugate in A₄, for example

Thus is found that hte conjugacy classes of A₄ are as follows