On the Structure of the Automorphism Group of Some Finite ... - SJU

On the Structure of the Automorphism Group of Some Finite Groups Bill Semus∗

Sam Smith†

Abstract Given a group G, let Aut(G) denote the group of isomorphisms φ : G → G with multiplication given by composition of functions. The structure of the automorphism groups of finite groups is intimately connected to the structure of the finite groups themselves. In this paper, we describe this connection and determine the structure of Aut(G) for many of the groups encountered in a first course in group theory.

1. Introduction. Historically, the first general structure result for the automorphism group of a finite group follows from a classical result of Gauss in number theory. Let ZZn denotes the additive group of integers mod n and U (n) the multiplicative group of integers mod n. That is, U (n) = {k|1 ≤ k < n and (k, n) = 1}. Gauss analyzed the orders of elements in U (pn ) for p prime. His results can be summarized as Theorem 1.1 (Gauss) Let p be an odd prime and n ≥ 1 or p = 2 and n ≥ 2. Then U (pn ) ∼ = ZZpn−1 (p−1) and U (2n ) ∼ =ZZ2 ×ZZ2n−2 .2 Now note that U (n) is precisely the set of generators of ZZn . Since any automorphism of ZZn sends 1 to a generator, the evaluation map E : Aut(Z Z n) → ∗ †

St. Joseph’s University, Philadelphia PA St. Joseph’s University, Philadelphia PA

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U (n) given by E(φ) = φ(1) is an isomorphism of groups. By Lagrange’s Theorem, if G and H are finite groups of relatively prime order then the funtion P : Aut(G) × Aut(H) → Aut(G × H) given by P (φG , φH ) = φG × φH is an isomorphism of groups. Thus we have Theorem 1.2 Let n be any integer 1. Write the prime factorization of n in the form n = 2n0 pn1 1 pn2 2 · · · pnk k where n0 ≥ 0, each ni ≥ 1 and the pi are odd primes. If n0 ≥ 2 then Aut(Z Z n ) =ZZ2 ⊕ZZ2n0 −2 ×ZZ(p1−1)pn1 −1 × · · · ×ZZ(p 1

nk −1 k −1)pk

.

Otherwise, Aut(Z Z n ) =ZZ(p1 −1)pn1 −1 × · · · ×ZZ(p 1

nk −1 k −1)pk

.2

The goal of this paper is to extend this calculation to other familiar groups. We will consider the following classes: • Cn – the multiplicative cyclic group of order n. We’ll write Cn = hai where a ∈ Cn has order n. • Sn – the nth permutation group is the group of permutations of the set {1, 2, . . . n}. • An – the nth alternating group is the subgroup of Sn consisting of even permutations • Dn – the nth dihedral group consisting of all symmetries of the regular n-gon. Recall Dn has order 2n and presentation Dn = ha, b | a2 = bn = e, aba = b−1 i. • Q – the quaternions §2. Conjugation and the Semi-Direct Product. The structure of the automorphism groups has direct implications for the classification of finite groups. The connection is made by considering the action by automorphisms of a group on itself (or on a normal subgroup) by conjugation. We consider this topic briefly here. Given a group G and an element g ∈ G define ϕg : G → G, the conjugation automorphism, by ϕg (x) = gxg −1 for x ∈ G. Define ϕ : G → Aut(G) by ϕ(g) = ϕg . Then ϕ is a homomorphism with kernel Z(G) – the center of 2

G. The image of ϕ is denoted Inn(G), and called the group of inner automorphisms of G. It is easy to check that Inn(G) is a normal subgroup of Aut(G). More generally, if H and N are subgroups of G with N normal, then ϕ induces a map ϕ : H → Aut(N ) where ϕ(h) is the restriction of ϕh to N . Here kerϕ = {h ∈ H|hn = nh for all n ∈ N } = ZH (N ), the centralizer of N in H. Conversely, given groups H and N and a homomorphism φ : H → Aut(N ) we can form the semi-direct product N ×φ H. As set, N ×φ H is just the cartesian product N × H. The multiplication is given by (n1 , h1 ) · (n2 , h2) = (n1 · φh1 (n2 ), h1 · h2) where φh1 = φ(h1 ) : N → N. It is easy to check that N ×φ H is indeed a group, that N = {(n, eH )|n ∈ N } is a normal subgroup, that H = {(eN , h)|h ∈ H} is a subgroup and that the conjugation action of H on N is, up to these identifications, just φ. Note that N ×φ H ∼ = N × H if and only if φ is the trivial homomorphism. Thus N ×φ H is abelian if and only if N and H are abelian and φ is trivial. When the homomorphism φ is clear from context, we will denote the semi-direct product by N
H. 2.1 Examples. (1) Let A be any abelian group. We can construct a semi-direct product of the form A
C2 as follows. Write C2 = hai. Define a homomorphism φ : C2 → Aut(A) by φa (x) = x−1 for x ∈ A. We remark that C
C n 2 is isomorphic to Dn . The isomorphism corresponds to the presentation given above. (2) Write C2 × C2 = hai × hbi and C3 = hci. Define φ : C3 → C2 × C2 by φc (a) = b, φc (b) = ab. The group (C2 × C2) ×φ C3 is isomorphic to A4. To see, this consider the subgroup T = {e, (12)(34), (14)(23), (13)(24)}. It is easy to check T ∼ = C2 × C2 is normal in A4. Let c = (123). Then A4 = hT, ci and the conjugation action of c on T is given by φ. (3) Given a group G of order 2 we obtain a nonabelian group of the form G ×φ Aut(G) by taking φ : Aut(G) → Aut(G) to be the identity. We denote this group by G
Aut(G). In particular, we have a new class of nonabelian (4) ∼ groups ZZ
U (n) for n2. Note ZZ
U (3) ∼ = D4 . = S3 and ZZ
U 4 n 3 As the preceding examples suggest, the semi-direct product construction and the structure of the automorphsim groups play critical roles in determining the occurrences of finite groups. As an example, we can “easily” prove the following result (with substantial help from Gauss and Sylow) 3

Theorem 2.2 Let p and q be primes with q < p. If q 6 | p − 1 then Cpq is, up to isomorphism, the unique group of order pq. If q | p − 1, there are, up to isomorphism, two groups of order pq: Cpq and a nonabelian group of the form C
C p q. Proof. Let G have order pq. Let P and Q be p and q-Sylow subgroups, respectively. By Sylow’s Theorem, P is normal in G. Thus G ∼ = P ×φ Q where φ : Q → Aut(P ) is the conjugation homomorphism. By Theorem 1.2, Aut(P ) is cyclic of order p − 1. Thus if q 6 | p − 1, φ must be trivial, as there are no elements of order q in Aut(P ). If q | p − 1 there is a unique subgroup of order q in Aut(P ). Thus there exists a q − 1 nontrivial homomorphims φ : Q → Aut(P ) corresponding to the q − 1 generators of this subgroup. It is easy to check that these φ all give rise to isomorphic semi-direct products. 2 3. Structure of Automorphism Groups. In this section, we study the structure of Aut(G) for various classes of finite groups G. Finite Abelian Groups. Theorem 1.2 determines Aut(G) for G cyclic. We observe that Aut(C2 × C2 ) ∼ = S3 since an automorphism φ of C2 × C2 corresponds precisely to a permutation of the three nonidentity elements. As we show below, the group Aut(C23) already has 168 elements. (Here C23 = C2 × C2 × C2.) In fact, given any prime p and any integer n the group Aut(Cpn) is isomorphic to Aut(Z Z np ), the group of n by n invertible matrices over the field ZZp. These and related “matrix” groups play leading roles in the classification of finite simple groups [1]. They are easy to count using linear algebra. We have Theorem 3.1 Given a prime p and an integer n, the group Aut(Cpn) has order (pn − 1)(pn − p)(pn − p2 ) · · · (pn − pn−1 ). Proof. View Cpn additively asZZnp , an n-dimensional vector space over the field ZZp. An automorphism φ :ZZnp → ZZnp is just a linear isomorphism of this vector space. Fix an ordered basis α = {v1, . . . , vn } for ZZnp over ZZp . Then φ is a linear isomorphism if and only if φ carries α to another ordered basis β. We conclude that Aut(Cpn) is in one-to-one correspondence with the set of ordered bases β of ZZnp overZZp. Now note that an ordered basis β = {w1 , . . . , wn } may be built by first choosing any nonzero vector w1 (pn −1 choices), then choosing 4

any vector w2 not in the span of w1 (pn − p choices), then any vector w3 not in the span of {w1, w2} (pn − p2 choices), etc. 2 Another indication of the size and complexity of Aut(G) for G finite abelian is the following version of Cayley’s Theorem: Theorem 3.2 Let k be any integer ≥ 2. Then every finite group G is isomorphic to a subgroup of Aut(Ckn) for some n. Proof. Write Ckn = ha1 i × · · · han i. Consider the subgroup S of Aut(Ckn) consisting of automorphisms which permute the ai amongst themselves. Then S∼ = Sn . The result now follows from Cayley’s Theorem [2, Theorem 6.1]. 2 Theorems 3.1 and 3.2 indicate the difficulty of obtaining a general structure theorem for the automorphism group of a noncyclic abelian group G. We content ourselves here with describing the groups Aut(C2 × Cn ) for all n. Write n = 2k q for q odd. Then Cn ∼ = C2k × Cq and so Aut(C2 × Cn ) ∼ = Aut(C2 × C2k ) × Aut(Cq). Since the structure of Aut(Cq) is given by Theorem 1.2, the following result completes the calculation: Theorem 3.3 Let k1. Then Aut(C2 × C2k ) ∼ ) 2. = (C2k−1 × C2k−1
C Proof. Write C2 = hai and C2k = hbi. The elements of order 2 in C2 × C2k are k−1 k−1 then a, b2 and ab2 . The elements of order 2k are bj and abj for j odd. k−1 Let φ be an automorphism of C2 × C2n . Then φ(a) = a or φ(a) = ab2 . To see this, observe that φ must send b to bj or abj for some j odd. In either k−1 case, b2 is in φ(hbi, since even powers of a are the identity. Define automorphisms φ1, φ2 and ψ of C2 × C2k on generators as follows: k−1 φ1 (a) = a and φ1 (b) = b3; φ2(a) = a and φ2 (b) = ab3; ψ(a) = ab2 and k−1 ψ(b) = b. Note ψ has order 2. Since 3 has multiplicative order 2 mod 2k , k−1 we have |φ1 |, |φ2| = 2 . Moreover, by the preceding paragraph, φ1 , φ2, and ψ generate Aut(C2 × C2k ). Since a3 = a, φ1 ◦ φ2 = φ2 ◦ φ1. Checking the conjugations, we find ψ ◦ φi ◦ ψ −1 = φ−1 for i = 1, 2. Thus Aut(C2 × C2k ) ∼ = i (hφ1 i × hφ2 i)
hψi where ψ induces the usual C2 -action on C2k−1 × C2k−1 , (Example 2.2 (1)). 2. It is easy to check that Aut(C2 × C4 ) = (C2 × C2
C ) 2 is just the dihedral group D4 . We consider the automorphisms of this class next. The Dihedral Groups. We show 5

Theorem 3.4 Let n2. Then Aut(Dn ) ∼ (n). =ZZ
U n Proof. Let a, b ∈ Dn be the generators as in Example 2.2 (2). That is, |a| = 2, |b| = n and aba = b−1 . The elements of order n in Dn are then the bi for (i, n) = 1. The elements of order 2 are of the form bj a for j = 0, . . . , n − 1, and bn/2 if n is even. Define automorphisms φi,j of Dn for i ∈ U (n) and j ∈ ZZn by specifying φi,j (a) = abj and φi,j (b) = bi . By the preceding paragraph, Aut(Dn ) = {φi,j |i ∈ U (n), j ∈ ZZn }. Composition obeys the equation φi1 ,j1 ◦ φi2 ,j2 = φi1 i2 ,j1 +i1 j2 .

(1)

Consider the subgroups N = {φ1,j |j ∈ ZZn } and U = {φi,0 |i ∈ U (n)}. By (1), the obvious maps give isomorphisms N ∼ = ZZn and U ∼ = U (n). Computing −1 −1 φi,j = φi−1 ,−ji−1 (where i denotes the multiplicative inverse of i in U (n)) we see N is normal in Aut(Dn). Thus Aut(Dn ) ∼ = N ×Φ U for some Φ : U → Aut(N ). To determine Φ we compute using (1) Φφi,0 (φ1,j ) = φi,0 ◦ φ1,j ◦ φ−1 i,0 = φi,0 ◦ φ1,j ◦ φi−1 ,0 = φ1,ij . The result follows. 2 The Symmetric and Alternating Groups. We consider Aut(Sn) and Aut(An) for small values of n. Since Z(Sn ) = {e} for n ≥ 3, Sn ∼ = Inn(Sn ) is a subgroup of Aut(Sn ). For n ≥ 4, observe that ZSn (An ) = {e}. Since An is normal in Sn , Sn is isomorphic to a subgroup of Aut(An), as well. We prove Theorem 3.5 Aut(Sn ) ∼ = Sn for n = 3, 4, and Aut(An) ∼ = Sn for n = 4, 5. Proof. Since Sn is isomorphic to a subgroup of Aut(Sn) and of Aut(An) in all four cases considered, it suffices to show that the order of these automorphism groups is ≤ n!. For Aut(S3) we simply observe that any automorphism φ must permute the elements of order 2, namely (12), (13) and (23), which are also generators of S3. For S4 and A4, we consider the normal subgroup T = {e, (12)(34), (14)(23), (13)(24)}, as in Example 2.2 (2). We claim T is actually a characteristic subgroup of A4 and S4 ; i.e., φ(T ) = T for all automorphisms φ of 6

A4 (or S4.) In A4 this is easy, since T contains all the elements of order 2. In S4 , there are also the transpositions (ij). But suppose φ(a) = (ij) where a = (12)(34). Then, since b = (14)(23) commutes with a, φ(b) must commute with φ(a) = (ij). Thus either φ(b) = (kl) or φ(b) = (ij)(kl) where {k, l} ⊂ {1, 2, 3, 4} is disjoint from {i, j}. In either case, the element (ij)(kl) ∈ T is also in φ(T ). Now note that the nonidentity elements of T are precisely a conjugacy class of S4. Since φ preserves the conjugacy equivalence relation, φ must send all of T into T . We now count the elements φ ∈ Aut(A4). There are 8 elements of order 3 in A4 and they are precisely the 8 three cycles. Let x = (123). There are 8 choices for φ(x). Now consider y = (124) : Since yx−1 = (14)(23) ∈ T a characteristic subgroup of A4, φ(yx−1) ∈ T, also. That is, φ(y) = t · φ(x) for some t ∈ T. Since φ is one-to-one, t 6= e and so there are 3 choices for φ(y). Recall that x and y generate A4 . Thus there are at most (8)(3) = 24 = 4! elements of Aut(A4). To count the elements of Aut(S4), we note that any automorphism φ must send transposition to transposition, since the other elements of order 2 are all in T = φ(T ). Thus there are 6 choices for φ((12)), the 6 transpositions. Suppose φ((12)) = (ij). Then φ((34)) = (kl) for {k, l} disjoint from {i, j}, as above. There remain 4 choices for φ((23)). Since (12), (23) and (34) generate S4 , the proof here is complete. Finally, we must count the automorphisms φ of A5. Consider the generators x = (12)(34) and y = (135) of A4. Note that xy = (14352) has order 5. By the argument above, φ(x) must be of the form (ij)(kl) for distinct i, j, k, l ∈ {1, . . . , 5}. Thus there are 15 choices for φ(x). Now φ(y) must be a 3-cycle (abc) such that (ij)(kl)(abc) is a 5- cycle. Up to the order of a, b, c, this forces a ∈ {i, j}, b ∈ {k, l} and c ∈ {1, . . . , 5} − {i, j, k, l}. Since each 3-cycle has two distinct orders, this gives 8 choices for φ(y). We conclude that |Aut(A5)| ≤ 8(15) = 120 = 5!, as needed. 2 We conclude with The Quaternions. Theorem 3.6 Aut(Q) ∼ = S4 . Proof. We first observe that |Aut(Q)| = 4!. For let φ ∈ Aut(Q). Then there are 6 choices for φ(i); namely, ±i, ±j, ±k. After φ(i) is chosen, there remain 4 choices for φ(j) since φ(j) = ±φ(i) implies φ(k) = ±1, a contradiction. 7

Now consider the normal subgroup Inn(Q) = {ϕ1 , ϕi , ϕj , ϕk } ∼ = C2 × C 2 of Aut(Q). Define φout1 by φout1 (i) = j, φout1 (j) = k. Let A = hInn(Q), φout1i. ∼ Computing the conjugates φout1 ϕx φ−1 out1 for x = i, j, k we see that A = (C2 × C2
C ) 3∼ = A4 as in Example 2.2 (2). Finally, define φout2 by setting φout2 (i) = j and φout2 (j) = i. Then, checking the conjugate action of φout2 on A, we see ∼ Aut(Q) = hA, φout2 i ∼ = A
C 4 2 = S4 . 2

References [1] M. Aschbacher, Finite Group Theory, Cambridge University Press, Cambridge, 1986. [2] J. Gallian, Contemporary Abstract Algebra, Fourth Edition HoughtonMifflin Company, New York, 1998

About the Authors: Bill Semus St. Joseph’s University 5600 City Ave. Philadelphia PA 19131 Bill Semus will begin graduate studies in mathematics at the University of Pennsylvannia this fall. This work was completed as part of his year-long Senior Honors Project at St. Joseph’s University.

Sam Smith St. Joseph’s University 5600 City Ave. Philadelphia PA 19131 email: [email protected] Sam Smith is an algebraic topologist and an Associate Professor of Mathematics at St. Joseph’s University.

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