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8 Conjugation, Centers and Automorphisms

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8 Conjugation, Centers and Automorphisms 8 Conjugation, Centers and Automorphisms 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 2 G. We say that g1 is conjugate to g2 if −1 9h 2 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 8g 2 G, h 2 H, ghg−1 2 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 9h 2 G such that g2 = hg1h = g1hh = g1 ∼ 2. The smallest non-Abelian group is D3 = S3. Its conjugacy classes are feg, fr1, r2g, fm1, m2, m3g 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 cy- cles 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, r1, r2 are 3-cycles and m1, m2, m3 are 2-cycles: these form the conjugacy classes by the Theorem. 1 Proof. Suppose that (a1 ··· ak) 2 Sn is a k-cycle and r 2 Sn is any element. It is easy to see that −1 r(a1 ··· ak)r = (r(a1) ··· r(ak)) is also a k-cycle, since the elements r(ai) must be distinct. Now let t 2 Sn be written as the product of disjoint cycles t = t1 ··· tl. Then −1 −1 −1 −1 rtr = (rt1r )(rt2r ) ··· (rtlr ) which has the same cycle type as t (again r permutes the elements of f1, 2, . , ng) Thus conjugation preserves cycle type. Conversely, let s = s1 ··· sl and t = t1 ··· tl be elements of Sn with the same cycle type; otherwise said, si and ti are k-cycles for the same k. Define a permutation p by writing s and t one on top of the other in standard notation: s s ··· s s s ··· p = 1 2 l 1 2 , t1 t2 ··· tl t1 t2 ··· where we write each of the elements of the cycle si in a row and si, ti are the remaining elements of −1 the set f1, 2, . , ng missing from each row, written in any order. We claim that t = psp . If si,j and th ti,j refer to the j element of the orbits si and ti respectively, then −1 psp (ti,j) = ps(si,j) = psi,j+1 = ti,j+1 = t(ti,j) −1 −1 If t = ti for some i, then psp (ti) = ps(si) = p(si) = ti. Either way, psp = t, as required. Examples 1. If s = (13)(24) and r = (124) in S4, then srs−1 = s(1) s(2) s(4) = (342) = (234) and, rsr−1 = r(1) r(3)r(2) r(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 p = = = (23746) 1 6 5 2 3 4 7 1 3 7 6 5 2 4 Indeed p(145)(627)p−1 = (23746)(145)(627)(26473) = (165)(234) There are other possible choice of p, for example by writing the orbits in different orders. 3. Recall earlier when we claimed that V = fe, (12)(34), (13)(24), (14)(23)g 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 = fV, (123)V, (132)Vg V . An example isomorphism m : A4 ! C from the first isomorphism theorem is then V 3 0 V 1 0 e 1 m : @(123)VA = @(123)A (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 = fch : G ! G where ch(g) = hgh g Examples 1. Consider the automorphisms of Z4. To define a homomorphism f : Z4 ! Z4, it is enough to choose the value of f(1). Indeed, to obtain a bijection, we need to map 1 (a generator) to a generator. There are therefore two choices: f0(x) = x mod 4 f1(x) = −x mod 4 Observe that the set Aut(Z4) = ff0, f1g 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 f 2 Aut(S3), then (f(x))n = e () f(xn) = e () xn = e so that the orders of x and f(x) are identical. The group S3 is generated by two elements: m := (12) and r := (123), where rm = mr2; e = r3, (132) = r2, (13) = rm, (23) = r2m 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: f(m) is either m, rm or r2m and f(r) = r or r2. A little thinking shows that all of these functions are in fact automor- phisms! 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 0 e 1 0 e 1 0 e 1 0 e 1 0 e 1 0 e 1 B r C B r C B r C B r C B r C B r C B C B C B C B C B C B C B r2 C B r2 C B r2 C B r2 C B r2 C B r2 C ce : B C 7! B C c : B C 7! B C c 2 : B C 7! B C B m C B m C r B m C Br2mC r B m C B rm C B C B C B C B C B C B C @ rm A @ rm A @ rm A @ m A @ rm A @r2mA r2m r2m r2m rm r2m m 0 e 1 0 e 1 0 e 1 0 e 1 0 e 1 0 e 1 B r C B r2 C B r C B r2 C B r2 C B r C B C B C B C B C B C B C B r2 C B r C B r2 C B r C B r C B r2 C c : B C 7! B C c : B C 7! B C c 2 : B C 7! B C m B m C B m C rm B m C Br2mC r m B m C B rm C B C B C B C B C B C B C @ rm A @r2mA @ rm A @ rm A @ rm A @ m A r2m rm r2m m r2m r2m ∼ 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 8y 2 Aut(G), y ◦ id = id ◦y = y. Inverse If f 2 Aut G it is also staightforward to check that the inverse function f−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.
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