8 Conjugation, Centers and Automorphisms

Home , Automorphism, Center (group theory), Conjugacy class, Homomorphism, Inner automorphism

In this section we describe another equivalence relation on a group, that of conjugacy. Conjugacy is the group theory abstraction of the idea of similarity from linear algebra: two square matrices A and B are said to be similar if there exists some invertible matrix S for which B = SAS−1. You should recall that such matrices have the same eigenvalues and, essentially, ‘do the same thing’ with respect to different bases.

Deﬁnition 8.1. Let G be a group and g1, g2 ∈ G. We say that g1 is conjugate to g2 if

−1 ∃h ∈ G such that g2 = hg1h

Conjugation is intimately related to the concept of normal subgroup. Recall that a subgroup H ≤ G is normal if and only if

∀g ∈ G, h ∈ H, ghg−1 ∈ H

A normal subgroup can therefore be deﬁned as a subgroup which is closed under conjugation.

Theorem 8.2. Conjugacy is an equivalence relation.

The proof is an exercise.

Deﬁnition 8.3. The equivalence classes of a group G under conjugacy are termed the conjugacy classes of G.

Examples

1. If G is Abelian then every conjugacy class contains only one element. Observe that g1 and g2 are conjugate if and only if

−1 −1 ∃h ∈ G such that g2 = hg1h = g1hh = g1

∼ 2. The smallest non-Abelian group is D3 = S3. Its conjugacy classes are

{e}, {ρ1, ρ2}, {µ1, µ2, µ3}

This can be computed directly, but it follows immediately from. . .

Theorem 8.4. The conjugacy classes of Sn are the cycle types.

The concept of cycle type is straightforward: for example (123)(45) has the same cycle type as (156)(23), but not the same as (12)(34). Just write an element of Sn as a product of disjoint cycles and its cycle type is clear. The proof involves some messy notation, so you may want to avoid on a ﬁrst reading. In the above example, e is a 0-cycle, ρ1, ρ2 are 3-cycles and µ1, µ2, µ3 are 2-cycles: these form the conjugacy classes by the Theorem.

1 Proof. Suppose that (a1 ··· ak) ∈ Sn is a k-cycle and ρ ∈ Sn is any element. It is easy to see that

−1 ρ(a1 ··· ak)ρ = (ρ(a1) ··· ρ(ak)) is also a k-cycle, since the elements ρ(ai) must be distinct. Now let τ ∈ Sn be written as the product of disjoint cycles τ = τ1 ··· τl. Then

−1 −1 −1 −1 ρτρ = (ρτ1ρ )(ρτ2ρ ) ··· (ρτlρ ) which has the same cycle type as τ (again ρ permutes the elements of {1, 2, . . . , n}) Thus conjugation preserves cycle type.

Conversely, let σ = σ1 ··· σl and τ = τ1 ··· τl be elements of Sn with the same cycle type; otherwise said, σi and τi are k-cycles for the same k. Deﬁne a permutation π by writing σ and τ one on top of the other in standard notation: σ σ ··· σ s s ··· π = 1 2 l 1 2 , τ1 τ2 ··· τl t1 t2 ··· where we write each of the elements of the cycle σi in a row and si, ti are the remaining elements of −1 the set {1, 2, . . . , n} missing from each row, written in any order. We claim that τ = πσπ . If σi,j and th τi,j refer to the j element of the orbits σi and τi respectively, then

−1 πσπ (τi,j) = πσ(σi,j) = πσi,j+1 = τi,j+1 = τ(τi,j)

−1 −1 If t = ti for some i, then πσπ (ti) = πσ(si) = π(si) = ti. Either way, πσπ = τ, as required.

Examples

1. If σ = (13)(24) and ρ = (124) in S4, then

σρσ−1 = σ(1) σ(2) σ(4) = (342) = (234) and, ρσρ−1 = ρ(1) ρ(3)ρ(2) ρ(4)(23)(41) = (14)(23)

2. According to the Theorem, the permutations (145)(627) and (165)(234) in S7 are conjugate by the permutation

1 4 5 6 2 7 3 1 2 3 4 5 6 7 π = = = (23746) 1 6 5 2 3 4 7 1 3 7 6 5 2 4

Indeed

π(145)(627)π−1 = (23746)(145)(627)(26473) = (165)(234)

There are other possible choice of π, for example by writing the orbits in different orders.

3. Recall earlier when we claimed that V = {e, (12)(34), (13)(24), (14)(23)} is a normal subgroup of A4. While one still has to check to V is a subgroup, the normality is now clear since V contains all the elements in A4 of the cycle type (ab)(cd) which is preserved by conjugation.

2 . Thus A4 is a factor group of order 12 = 3 which must therefore be isomorphic to C . Indeed V 4 3 the cosets of V may be written1 . A4 = {V, (123)V, (132)V} V . An example isomorphism µ : A4 → C from the ﬁrst isomorphism theorem is then V 3  V   e  µ : (123)V = (123) (132)V (132)

where we view C3 in a natural way as a subgroup of S4.

Automorphisms Conjugation by a ﬁxed element of a group is a special case of a general structure-preserving transfor- mation.

Deﬁnition 8.5. An automorphism of a group G is an isomorphism of G with itself. The set of such is labelled Aut G. The inner automorphisms of G are the conjugations

−1 Inn G = {ch : G → G where ch(g) = hgh }

Examples

1. Consider the automorphisms of Z4. To deﬁne a homomorphism φ : Z4 → Z4, it is enough to choose the value of φ(1). Indeed, to obtain a bijection, we need to map 1 (a generator) to a generator. There are therefore two choices:

φ0(x) = x mod 4 φ1(x) = −x mod 4

Observe that the set Aut(Z4) = {φ0, φ1} forms a group (necessarily isomorphic to C2) under composition of functions. There is only one inner automorphism of Z4, since Z4 is abelian.

2. Find the automorphisms of S3 is a little harder. First observe that if φ ∈ Aut(S3), then

(φ(x))n = e ⇐⇒ φ(xn) = e ⇐⇒ xn = e

so that the orders of x and φ(x) are identical. The group S3 is generated by two elements: µ := (12) and ρ := (123), where ρµ = µρ2;

e = ρ3, (132) = ρ2, (13) = ρµ, (23) = ρ2µ

1In fact (123)V = (134)V = (243)V = (142)V and (132)V = (143)V = (234)V = (124)V.

3 We therefore have a maximum of six possible choices of automorphism: φ(µ) is either µ, ρµ or ρ2µ and φ(ρ) = ρ or ρ2. A little thinking shows that all of these functions are in fact automorphisms! Moreover, Aut(G) is a group of order 6 which is easily seen to be non-abelian, whence ∼ Aut(S3) = S3. In fact, each of these functions is conjugation by some element of S3: explicitly

 e   e   e   e   e   e   ρ   ρ   ρ   ρ   ρ   ρ               ρ2   ρ2   ρ2   ρ2   ρ2   ρ2  ce :   7→   c :   7→   c 2 :   7→    µ   µ  ρ  µ  ρ2µ ρ  µ   ρµ               ρµ   ρµ   ρµ   µ   ρµ  ρ2µ ρ2µ ρ2µ ρ2µ ρµ ρ2µ µ  e   e   e   e   e   e   ρ   ρ2   ρ   ρ2   ρ2   ρ               ρ2   ρ   ρ2   ρ   ρ   ρ2  c :   7→   c :   7→   c 2 :   7→   µ  µ   µ  ρµ  µ  ρ2µ ρ µ  µ   ρµ               ρµ  ρ2µ  ρµ   ρµ   ρµ   µ  ρ2µ ρµ ρ2µ µ ρ2µ ρ2µ ∼ It follows that Inn(S3) = Aut(S3) = S3. Indeed we have a general result here:

Theorem 8.6. The sets Aut G and Inn G form groups under composition. Moreover Inn G is a subgroup of Aut G.

Proof. We sketch the argument for Aut G ﬁrst.

Closure It is straightforward to check that composition of two isomorphisms is an isomorphism. Associativity Aut G consists of functions under composition. Identity id : g 7→ g is clearly an automorphism and moreover ∀ψ ∈ Aut(G), ψ ◦ id = id ◦ψ = ψ. Inverse If φ ∈ Aut G it is also staightforward to check that the inverse function φ−1 : G → G is automorphism.

For Inn G, note that each ch is a homomorphism:

−1 −1 ch(g1g2) = h(g1g2)h = (hg1h )(hg2h) = ch(g1)ch(g2)

−1 It also has inverse ch = ch−1 since

−1 −1 ch−1 (ch(g)) = h (hgh )h = g =⇒ ch−1 ◦ ch = id

Thus each ch is an automorphism of G and indeed Inn G is closed under inverses. Finally we check that Inn G is closed under composition:

( ◦ )( ) = ( −1) −1 = ( ) ( )−1 = ( )( ) cg1 cg2 h g1 g2hg2 g1 g1g2 h g1g2 cg1g2 h c

Inn G is therefore a subgroup of Aut G.

4 Theorem 8.7. Inn G / Aut G.

−1 −1 Proof. Let τ ∈ Aut G and ch ∈ Inn G. It is enough to see that τchτ ∈ Inn G. Since τchτ is a function, we compute what it does to a general element g ∈ G.

−1 −1 −1 −1 (τchτ )(g) = τ ch τ (g) = τ hτ (g)h = τ(h)ττ−1(g)τ(h−1) (since τ is a homomorphism) = τ(h)gτ(h)−1 (again since τ is an homomorphism)

= cτ(h)(g)

−1 Therefore τchτ = cτ(h), which is an element of Inn G.

Since conjugation by an element g ∈ G is an automorphism of G, conjugation must map subgroups of G to isomorphic subgroups. We also call such subgroups conjugate.

Example Let H = {e, µ1} ≤ D3. Then = −1 = { } = { } cρ1 H ρ1 Hρ1 e, ρ1µ1ρ2 e, µ2

If we compute all six possible conjugations of H, we obtain

{e, µ1} = H = ce H = cµ1 H,

{e, µ2} = cρ1 H = cµ3 H

{e, µ3} = cρ2 H = cµ2 H

These comprise all the two element subgroups of D3.

Centers We say that an element g in a group G commutes with another element a ∈ G if the order of multipli- cation is irrelevant: i.e. if ga = ag. It is a natural question to ask if there are any elements in a group which commute with all other elements. There are two simple cases to consider:

• If G is abelian, then every element commutes with every other element!

• The identity e commutes with everything, regardless of the group.

In general, the set of such elements will fall somewhere between these extremes. This subset will turn out to form another normal subgroup of G.

Deﬁnition 8.8. The center of a group G is the subset of G deﬁned by

Z(G) := {g ∈ G : ∀h ∈ G, gh = hg}

We will prove that Z(G) / G shortly, although a simple proof (check the axioms of a subgroup and that the left and right cosets are equal) is straightforward. Instead, here are a few examples.

5 Examples

1. Z(G) = G ⇐⇒ G is Abelian.

2. Z(D3) = {e}. This is straightforward to check as there are only 6 elements! ◦ 3. In general Z(D2n+1) = {e} and Z(D2n) = {e, ρn/2}, where ρn/2 is rotation by 180 . For example, it is easy to see in D2n+1 that any rotation and reﬂection fail to commute.

4. Z(Sn) = {e} if n ≥ 3. × 5. Z(GLn(R)) = {λIn : λ ∈ R }. The challenge with centers is that they are typically very difﬁcult to compute, at least in part because we need to think about non-abelian groups in order to generate interesting examples. As an example we give an argument for computing Z(GLn(R)). 2 Let A ∈ GLn(R) be a matrix with only one eigenvector v, and let Z ∈ Z(GLn(R)). Then

AZv = ZAv = λZv where λ is the corresponding eigenvalue of A. It follows that Zv is an eigenvector of A, whence Zv is parallel to v. Since our construction depended only on the initial choice of vector v, it follows that Zv must be parallel to v for every possible choice of v. The only such matrices are multiples of the identity.

Theorem 8.9. Z(G) / G.

Proof. We have already done most of the work. Deﬁne φ : G → Inn G by φ(g) = cg. Thus each element φ(g) is the function ‘conjugation by g.’ We quickly see that φ is a homomorphism:

−1 −1 −1 φ(gh)(x) = cgh(x) = (gh)x(gh) = g(hxh )g = cg(ch(x)) = φ(g) ◦ φ(h) (x)

We compute the kernel of φ:

g ∈ ker φ ⇐⇒ cg = id ⇐⇒ ∀h ∈ G, ghg−1 = h ⇐⇒ g ∈ Z(G)

It follows that ker φ = Z(G) / G.

In fact φ : G → Inn G is a surjective homomorphism (every inner automorphism is a conjugation). The 1st isomorphism theorem tells us that . G =∼ Inn G Z(G) for any group G. 2 n T T Such A exist: extend v to an orthonormal basis v, v2,... vn of R and let A = I + v(v2 + ··· + vn ). With respect to this basis, A has 1’s down the main diagonal and the diagonal immediately above the main, and zeros elsewhere: a classic Jordan form from linear algebra.