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9 Automorphism Groups 9 Automorphism groups For any group G the set of automorphisms of G forms a group under com- position. This is the automorphism group of G denoted Aut(G). Nonabelian groups have inner automorphisms Ág 2 Aut(G) given by ¡1 Ág(x) = gxg and the assignment g 7! Ág gives a homomorphism G ! Aut(G) with kernel Z(G). The image of this homomorphism is evidently the group of inner automorphisms of G which we denote Inn(G). Thus: Inn(G) »= G=Z(G): Proposition 9.1. Inn(G) E Aut(G). Proof. Suppose that à 2 Aut(G) and Ág 2 Inn(G). Then ¡1 ¡1 ¡1 ¡1 ÃÁgà (x) = Ã(gà (x)g ) = Ã(g)xÃ(g ) = ÁÃ(g)(x) ¡1 so ÃÁgà = ÁÃ(g). The automorphisms of G which are not inner automorphisms are called outer automorphisms. However, the outer automorphism group of G is the quotient group: Aut(G) Out(G) := = coker Á Inn(G) Automorphisms of abelian groups Let A denote an additive group. Since A has no (nontrivial) inner automor- phisms we have: Out(G) = Aut(G). The endomorphisms of A form a ring End(A) with pointwise addition [(® + ¯)(x) = ®(x) + ¯(x)] and multiplication given by composition. Proposition 9.2. The automorphism group of an additive group A is the group of units of the ring End(A): Aut(A) = End(A)£: If A = Z=n is cyclic then an endomorphism ® of Z=n is given by multi- plication by ®(1) 2 Z=n. Thus: Corollary 9.3. End(Z=n) = Z=n is a commutative ring and Aut(Z=n) = (Z=n)£ Everyone knows that for n = p is prime, (Z=p)£ is cyclic of order p ¡ 1. Rotman points out that: Theorem 9.4. For p an odd prime, Aut(Z=pm) = (Z=pm)£ is cyclic of order (p ¡ 1)pm¡1 and Aut(Z=2m) = (Z=2m)£ »= Z=2 £ Z=2m¡2. 1 Proof. The order of Aut(Z=pm) = (Z=pm)£ is given by the Euler Á function Á(pm) = (p ¡ 1)pm¡1. So it is enough to find an element of this order. If p ¸ 3, µ k¶ k p (1 + p)p = 1 + pkp + p2 + ¢ ¢ ¢ ´ 1 + pk+1 mod pk+2 2 Consequently, 1 + p 2 (Z=pm)£ has order pm¡1. Let x 2 Z represent a generator of the cyclic group (Z=p)£. Then xpm is an element of order p¡1 in (Z=pm)£ so xpm (1+p) is a generator of (Z=pm)£. Similarly, 52k = (1 + 4)2k ´ 1 + 2k+2 mod 2k+3 shows that 5 2 (Z=2m)£ has order 2m¡2. This means that (Z=2m)£ is either Z=2m¡1 or Z=2m¡2 £Z=2. The epimorphism (Z=2m)£ ³ (Z=8)£ »= Z=2 £ Z=2 [since 32; 52; 72 ´ 1 mod 8] shows that (Z=2m)£ is not cyclic. Complete groups Suppose that K E G with quotient Q = G=K.[G is an extension of K by Q.] Then G acts on K by conjugation. This gives a homomorphism γ : G ! Aut(K) The kernel of γ is the centralizer CG(K). Since γ(K) = Inn(K), we also get an induced homomorphism: θ : Q ! Out(K) which is the first step to measuring the extent to which the extension K ! G ! Q is nontrivial. This is particularly easy in the case when K is “com- plete.” Definition 9.5. A complete group is a group G with 1. no center (Z(G) = 1). Thus G »= Inn(G). 2. no outer automorphisms (Out(G) = 1). So Inn(G) = Aut(G). Theorem 9.6. If K E G and K is complete then G has a complementary normal subgroup Q (K \ Q = 1 and KQ = G making G = K £ Q). Proof. Let Q = CG(K). Then Q \ K = Z(K) = 1. So just need to show KQ = G. Let g 2 G. Then γg = ÁgjK 2 Aut(K) = Inn(K). So there is an h 2 K so that γg = Áh. In other words gxg¡1 = hxh¡1 ¡1 for all x 2 K, i.e., h g 2 CG(K) = Q so g 2 KQ. Theorem 9.7. The symmetric groups Sn are complete for n 6= 2; 6. 2 If n ¸ 3, Sn has no center. So it suffices to show that every automorphism of Sn is inner for n 6= 6. Lemma 9.8. An automorphism of Sn is inner if and only if it takes trans- positions to transpositions. [If it preserves transpositions.] Proof. Inner automorphisms preserve cycle type so they preserve transposi- tions. Conversely, suppose that Á 2 Aut(Sn) preserves transpositions. Take the sequence of transpositions: (12); (23); (34); ¢ ¢ ¢ ; (n ¡ 1; n) (1) This sequence has the property that each transposition commutes with all the others except for its neighbors (the ones next to it). By assumption Á(12) = (ab);Á(23) = (cd). Since these do not commute, the letters overlap, say b = d. So Á(12) = (ab);Á(23) = (bc). The transposition Á(34) must commute with (ab) and not commute with (bc) so it contains c and a new letter d. Similarly, Á(45) = (de) for some new letter e. Let σ 2 Sn be given by σ(1) = a; σ(2) = b; σ(3) = c, etc. Then Á agrees with Áσ on the transpositions (1). Since these generate Sn we have Á = Áσ. Proof of Theorem 9.7. The proof is a simple counting argument. An au- tomorphism Á of Sn will take conjugate elements to conjugate elements. Therefore, transpositions will go to elements of order 2 of the same cycle type. Since Á¡1 is also an automorphism, we must get all of the elements of this cycle type. Elements of order 2 have cycle type: Tk :(a1b1)(a2b2) ¢ ¢ ¢ (akbk) The number of elements of cycle type Tk is: µ ¶µ ¶ µ ¶ 1 n n ¡ 2 n ¡ 2(k ¡ 1) jT j = ¢ ¢ ¢ k k! 2 2 2 µ ¶ n (n ¡ 2)(n ¡ 3) (n ¡ 4)(n ¡ 5) (n ¡ 2k + 2)(n ¡ 2k + 1) = ¢ ¢ ¢ 2 | 2{zk } | 2k{z¡ 2 } | {z4 } 15 >1 1 ¸ 4 could be 2 Each of the underbraced terms have the same form since 2k · n, 2k ¡ 2 · n ¡ 2 etc. This means (proof by induction) that they are all > 1 with the exception of the last terms which could be 1=2 if n = 2k. For n = 4, 1 15 jT2j = 2 jT1j. For n = 2k ¸ 8, jTkj ¸ 8 jT1j since the first underbraced term 15 is =geq 4 . Thus jT1j 6= jTkj for k > 1 except in the case n = 6; k = 3. The outer automorphisms of S6 Theorem 9.9. Out(S6) = Z=2 3 Looking at the proof of Theorem 9.7 we see that any outer automorphism of S6 interchanges the conjugacy classes T1 and T3. Consequently, the product of any two outer automorphisms is an inner automorphism by Lemma 9.8. Thus Out(S6) has at most two elements and it suffices to find one outer automorphism. I will use an icosahedron to construct the outer automorphisms of S6. The icosahedron has 12 vertices. Take an icosahedron and label the vertices 1; 2; ¢ ¢ ¢ ; 6 using each label twice on opposite vertices (or equivalently, label the faces of a dodecahedron). Turn a corner 6 towards you. The five adjacent vertices will be labelled 1-5 in some order. Turn the 1 up. Then there are four numbers on the bottom, say 3; 5; 2; 4 (counterclockwise). If you turn the icosahedron around you should get the mirror image 4; 2; 5; 3. Denote this labelling of the icosahedron by I(3524) = I(4253) There are 4! = 24 labellings which give 12 possible icosahedra. Now define one more equivalence relation: I(abcd) » I(bdac) (2) I’ll explain where this came from in a minute. This gives 6 equivalence classes of labellings of the icosahedron which can be written uniquely as I(2xyz). Permutation of the numbers 1 to 6 will act on the set of labelled icosahedra. For example the permutation (12) changes I(2xyz) to I(2zyx). A more interesting example is (26) which changes I(2xyz) to I(2xzy). By doing one example we see that each transposition will change two adjacent labels and the resulting icosahedron is not equivalent to the original. Thus transpositions act on the 6 labellings without fixed points, i.e., with cycle type (ab)(cd)(ef). If we number the 6 labellings we get an outer autormophism of S6. Where does this funny equivalence (2) come from? Each labelling of the icosahedron determines a pentad which is a collection of five elements of cycle type T3 which do not commute with each other. Each of these is given in a different color in the figure. For example, in black we have (16)(25)(34) for I(2345) The others are obtained by cyclically permuting the numbers 1 to 5 (corresponds to rotating the picture). The equivalence relation (2) preserves the pentad! 4.
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