DOCSLIB.ORG

9 Automorphism Groups

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
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.
Load more

Recommended publications
  • Complete Objects in Categories
    Complete objects in categories James Richard Andrew Gray February 22, 2021 Abstract We introduce the notions of proto-complete, complete, complete˚ and strong-complete objects in pointed categories. We show under mild condi- tions on a pointed exact protomodular category that every proto-complete (respectively complete) object is the product of an abelian proto-complete (respectively complete) object and a strong-complete object. This to- gether with the observation that the trivial group is the only abelian complete group recovers a theorem of Baer classifying complete groups. In addition we generalize several theorems about groups (subgroups) with trivial center (respectively, centralizer), and provide a categorical explana- tion behind why the derivation algebra of a perfect Lie algebra with trivial center and the automorphism group of a non-abelian (characteristically) simple group are strong-complete. 1 Introduction Recall that Carmichael [19] called a group G complete if it has trivial cen- ter and each automorphism is inner. For each group G there is a canonical homomorphism cG from G to AutpGq, the automorphism group of G. This ho- momorphism assigns to each g in G the inner automorphism which sends each x in G to gxg´1. It can be readily seen that a group G is complete if and only if cG is an isomorphism. Baer [1] showed that a group G is complete if and only if every normal monomorphism with domain G is a split monomorphism. We call an object in a pointed category complete if it satisfies this latter condi- arXiv:2102.09834v1 [math.CT] 19 Feb 2021 tion.
    [Show full text]
  • The General Linear Group
    18.704 Gabe Cunningham 2/18/05 gcasey@mit.edu The General Linear Group Definition: Let F be a field. Then the general linear group GLn(F ) is the group of invert- ible n × n matrices with entries in F under matrix multiplication. It is easy to see that GLn(F ) is, in fact, a group: matrix multiplication is associative; the identity element is In, the n × n matrix with 1’s along the main diagonal and 0’s everywhere else; and the matrices are invertible by choice. It’s not immediately clear whether GLn(F ) has infinitely many elements when F does. However, such is the case. Let a ∈ F , a 6= 0. −1 Then a · In is an invertible n × n matrix with inverse a · In. In fact, the set of all such × matrices forms a subgroup of GLn(F ) that is isomorphic to F = F \{0}. It is clear that if F is a finite field, then GLn(F ) has only finitely many elements. An interesting question to ask is how many elements it has. Before addressing that question fully, let’s look at some examples. ∼ × Example 1: Let n = 1. Then GLn(Fq) = Fq , which has q − 1 elements. a b Example 2: Let n = 2; let M = ( c d ). Then for M to be invertible, it is necessary and sufficient that ad 6= bc. If a, b, c, and d are all nonzero, then we can fix a, b, and c arbitrarily, and d can be anything but a−1bc. This gives us (q − 1)3(q − 2) matrices.
    [Show full text]
  • Group Theory
    Appendix A Group Theory This appendix is a survey of only those topics in group theory that are needed to understand the composition of symmetry transformations and its consequences for fundamental physics. It is intended to be self-contained and covers those topics that are needed to follow the main text. Although in the end this appendix became quite long, a thorough understanding of group theory is possible only by consulting the appropriate literature in addition to this appendix. In order that this book not become too lengthy, proofs of theorems were largely omitted; again I refer to other monographs. From its very title, the book by H. Georgi [211] is the most appropriate if particle physics is the primary focus of interest. The book by G. Costa and G. Fogli [102] is written in the same spirit. Both books also cover the necessary group theory for grand unification ideas. A very comprehensive but also rather dense treatment is given by [428]. Still a classic is [254]; it contains more about the treatment of dynamical symmetries in quantum mechanics. A.1 Basics A.1.1 Definitions: Algebraic Structures From the structureless notion of a set, one can successively generate more and more algebraic structures. Those that play a prominent role in physics are defined in the following. Group A group G is a set with elements gi and an operation ◦ (called group multiplication) with the properties that (i) the operation is closed: gi ◦ g j ∈ G, (ii) a neutral element g0 ∈ G exists such that gi ◦ g0 = g0 ◦ gi = gi , (iii) for every gi exists an −1 ∈ ◦ −1 = = −1 ◦ inverse element gi G such that gi gi g0 gi gi , (iv) the operation is associative: gi ◦ (g j ◦ gk) = (gi ◦ g j ) ◦ gk.
    [Show full text]
  • Arxiv:Math/0511624V1
    Automorphism groups of polycyclic-by-finite groups and arithmetic groups Oliver Baues ∗ Fritz Grunewald † Mathematisches Institut II Mathematisches Institut Universit¨at Karlsruhe Heinrich-Heine-Universit¨at D-76128 Karlsruhe D-40225 D¨usseldorf May 18, 2005 Abstract We show that the outer automorphism group of a polycyclic-by-finite group is an arithmetic group. This result follows from a detailed structural analysis of the automorphism groups of such groups. We use an extended version of the theory of the algebraic hull functor initiated by Mostow. We thus make applicable refined methods from the theory of algebraic and arithmetic groups. We also construct examples of polycyclic-by-finite groups which have an automorphism group which does not contain an arithmetic group of finite index. Finally we discuss applications of our results to the groups of homotopy self-equivalences of K(Γ, 1)-spaces and obtain an extension of arithmeticity results of Sullivan in rational homo- topy theory. 2000 Mathematics Subject Classification: Primary 20F28, 20G30; Secondary 11F06, 14L27, 20F16, 20F34, 22E40, 55P10 arXiv:math/0511624v1 [math.GR] 25 Nov 2005 ∗e-mail: baues@math.uni-karlsruhe.de †e-mail: fritz@math.uni-duesseldorf.de 1 Contents 1 Introduction 4 1.1 Themainresults ........................... 4 1.2 Outlineoftheproofsandmoreresults . 6 1.3 Cohomologicalrepresentations . 8 1.4 Applications to the groups of homotopy self-equivalences of spaces 9 2 Prerequisites on linear algebraic groups and arithmetic groups 12 2.1 Thegeneraltheory .......................... 12 2.2 Algebraicgroupsofautomorphisms. 15 3 The group of automorphisms of a solvable-by-finite linear algebraic group 17 3.1 The algebraic structure of Auta(H)................
    [Show full text]
  • MATH 412 PROBLEM SET 11, SOLUTIONS Reading: • Hungerford
    MATH 412 PROBLEM SET 11, SOLUTIONS Reading: • Hungerford 8.4, 9.1, 9.2 Practice Problems: These are “easier” problems critical for your understanding, but not to be turned in. Be sure you can do these! You may see them or something very similar on Quizzes and Exams. Section 8.4: 9, 19, 21, Section 9.1: 1, 2, 3, 4, 5 Section 9.2: 1, 3 Problems due Friday. Write out carefully, staple solutions to A, B and C in one pack, and stable solutions to D and E in another. Write your name and section number on each. (A) Let F be a field. Consider the group SL2(F) of 2 × 2 matrices with entries in F and determinant one. 1 (1) Show that the center of SL2(F) is Z = {±Ig, where I is the 2 × 2 identity matrix. (2) Let PSL2(F) be the quotient group SL2(F)=Z. Compute the order of PSL2(Z7) as a group. 2 (3) For each of the following, compute its order as an element of PSL2(Z7). 0 2 • A = ± 3 0 1 1 • B = ± 6 0 1 3 • C = ± 0 1 Note: PSL2(Z7) is the smallest simple group that is neither cyclic nor isomorphic to An for some n. Solution. a b 1 1 1 0 (1) Let A = be an arbitrary element of SL ( ), B = , and C = . We c d 2 F 0 1 1 1 −c a − d b 0 compute AB − BA = and AC − CA = . If A 2 Z, then AB − BA = 0 c d − a −b AC − CA = 0, so b = c = a − d = 0.
    [Show full text]
  • Normal Subgroups of the General Linear Groups Over Von Neumann Regular Rings L
    PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 96, Number 2, February 1986 NORMAL SUBGROUPS OF THE GENERAL LINEAR GROUPS OVER VON NEUMANN REGULAR RINGS L. N. VASERSTEIN1 ABSTRACT. Let A be a von Neumann regular ring or, more generally, let A be an associative ring with 1 whose reduction modulo its Jacobson radical is von Neumann regular. We obtain a complete description of all subgroups of GLn A, n > 3, which are normalized by elementary matrices. 1. Introduction. For any associative ring A with 1 and any natural number n, let GLn A be the group of invertible n by n matrices over A and EnA the subgroup generated by all elementary matrices x1'3, where 1 < i / j < n and x E A. In this paper we describe all subgroups of GLn A normalized by EnA for any von Neumann regular A, provided n > 3. Our description is standard (see Bass [1] and Vaserstein [14, 16]): a subgroup H of GL„ A is normalized by EnA if and only if H is of level B for an ideal B of A, i.e. E„(A, B) C H C Gn(A, B). Here Gn(A, B) is the inverse image of the center of GL„(,4/S) (when n > 2, this center consists of scalar invertible matrices over the center of the ring A/B) under the canonical homomorphism GL„ A —►GLn(A/B) and En(A, B) is the normal subgroup of EnA generated by all elementary matrices in Gn(A, B) (when n > 3, the group En(A, B) is generated by matrices of the form (—y)J'lx1'Jy:i''1 with x € B,y £ A,l < i ^ j < n, see [14]).
    [Show full text]
  • Generalized Quaternions
    GENERALIZED QUATERNIONS KEITH CONRAD 1. introduction The quaternion group Q8 is one of the two non-abelian groups of size 8 (up to isomor- phism). The other one, D4, can be constructed as a semi-direct product: ∼ ∼ × ∼ D4 = Aff(Z=(4)) = Z=(4) o (Z=(4)) = Z=(4) o Z=(2); where the elements of Z=(2) act on Z=(4) as the identity and negation. While Q8 is not a semi-direct product, it can be constructed as the quotient group of a semi-direct product. We will see how this is done in Section2 and then jazz up the construction in Section3 to make an infinite family of similar groups with Q8 as the simplest member. In Section4 we will compare this family with the dihedral groups and see how it fits into a bigger picture. 2. The quaternion group from a semi-direct product The group Q8 is built out of its subgroups hii and hji with the overlapping condition i2 = j2 = −1 and the conjugacy relation jij−1 = −i = i−1. More generally, for odd a we have jaij−a = −i = i−1, while for even a we have jaij−a = i. We can combine these into the single formula a (2.1) jaij−a = i(−1) for all a 2 Z. These relations suggest the following way to construct the group Q8. Theorem 2.1. Let H = Z=(4) o Z=(4), where (a; b)(c; d) = (a + (−1)bc; b + d); ∼ The element (2; 2) in H has order 2, lies in the center, and H=h(2; 2)i = Q8.
    [Show full text]
  • 14.2 Covers of Symmetric and Alternating Groups
    Remark 206 In terms of Galois cohomology, there an exact sequence of alge- braic groups (over an algebrically closed field) 1 → GL1 → ΓV → OV → 1 We do not necessarily get an exact sequence when taking values in some subfield. If 1 → A → B → C → 1 is exact, 1 → A(K) → B(K) → C(K) is exact, but the map on the right need not be surjective. Instead what we get is 1 → H0(Gal(K¯ /K), A) → H0(Gal(K¯ /K), B) → H0(Gal(K¯ /K), C) → → H1(Gal(K¯ /K), A) → ··· 1 1 It turns out that H (Gal(K¯ /K), GL1) = 1. However, H (Gal(K¯ /K), ±1) = K×/(K×)2. So from 1 → GL1 → ΓV → OV → 1 we get × 1 1 → K → ΓV (K) → OV (K) → 1 = H (Gal(K¯ /K), GL1) However, taking 1 → µ2 → SpinV → SOV → 1 we get N × × 2 1 ¯ 1 → ±1 → SpinV (K) → SOV (K) −→ K /(K ) = H (K/K, µ2) so the non-surjectivity of N is some kind of higher Galois cohomology. Warning 207 SpinV → SOV is onto as a map of ALGEBRAIC GROUPS, but SpinV (K) → SOV (K) need NOT be onto. Example 208 Take O3(R) =∼ SO3(R)×{±1} as 3 is odd (in general O2n+1(R) =∼ SO2n+1(R) × {±1}). So we have a sequence 1 → ±1 → Spin3(R) → SO3(R) → 1. 0 × Notice that Spin3(R) ⊆ C3 (R) =∼ H, so Spin3(R) ⊆ H , and in fact we saw that it is S3. 14.2 Covers of symmetric and alternating groups The symmetric group on n letter can be embedded in the obvious way in On(R) as permutations of coordinates.
    [Show full text]
  • The Number 6
    The number 6 Gabriel Coutinho 2013 Abstract The number 6 has a unique and distinguished property. It is the only natural number n for which there is a construction of n isomor- phic objects on a set with n elements, invariant under all permutations of the set, but not in natural bijection with the set. I will talk about this fact and a closely related result. This is based on Chapter 6 of Cameron and Van Lint [1]. 1 A distinguished property of 6 Theorem 1. The number 6 is the only natural number n that satisfies the following property: If E is an n-set, then there is a construction of exactly n isomor- phic objects on E whose union is invariant under the permuta- tions of E, but none correspond in natural way to an element of E. Some clarification is needed here. Neither the meaning of object nor that of natural is clear. I will make it even worse, and provide a bad definition for one, and an incomplete for the other. An object on a set E is a construction that can be made using the elements of E and set-theoretical operations, such as taking the union. An object on E corresponds naturally to a point of E if the automorphism group of the object is the stabilizer of that point in Sn. Despite such abstract formulation, we will see that this theorem has a very concrete interpretation. In particular, the \shape" of such objects will arise naturally throughout the proof. The idea of the proof is to analyse the automorphism group the objects defined by 1 the automorphisms of the ground set which do not change the object.
    [Show full text]
  • SL{2, R)/ ± /; Then Hc = PSL(2, C) = SL(2, C)/ ±
    PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 60, October 1976 ERRATUM TO "ON THE AUTOMORPHISM GROUP OF A LIE GROUP" DAVID WIGNER D. Z. Djokovic and G. P. Hochschild have pointed out that the argument in the third paragraph of the proof of Theorem 1 is insufficient. The fallacy lies in assuming that the fixer of S in K(S) is an algebraic subgroup. This is of course correct for complex groups, since a connected complex semisimple Lie group is algebraic, but false for real Lie groups. By Proposition 1, the theorem is correct for real solvable Lie groups and the given proof is correct for complex groups, but the theorem fails for real semisimple groups, as the following example, worked out jointly with Professor Hochschild, shows. Let G = SL(2, R); then Gc = SL(2, C) is the complexification of G and the complex conjugation t in SL(2, C) fixes exactly SL(2, R). Let H = PSL(2, R) = SL{2, R)/ ± /; then Hc = PSL(2, C) = SL(2, C)/ ± /. The complex conjugation induced by t in PSL{2, C) fixes the image M in FSL(2, C) of the matrix (° ¿,)in SL(2, C). Therefore M is in the real algebraic hull of H = PSL(2, R). Conjugation by M on PSL(2, R) acts on ttx{PSL{2, R)) aZby multiplication by - 1. Now let H be the universal cover of PSLÍ2, R) and D its center. Let K = H X H/{(d, d)\d ED}. Then the inner automorphism group of K is H X H and its algebraic hull contains elements whose conjugation action on the fundamental group of H X H is multiplication by - 1 on the fundamen- tal group of the first factor and the identity on the fundamental group of the second factor.
    [Show full text]
  • THE CONJUGACY CLASS RANKS of M24 Communicated by Robert Turner Curtis 1. Introduction M24 Is the Largest Mathieu Sporadic Simple
    International Journal of Group Theory ISSN (print): 2251-7650, ISSN (on-line): 2251-7669 Vol. 6 No. 4 (2017), pp. 53-58. ⃝c 2017 University of Isfahan ijgt.ui.ac.ir www.ui.ac.ir THE CONJUGACY CLASS RANKS OF M24 ZWELETHEMBA MPONO Dedicated to Professor Jamshid Moori on the occasion of his seventieth birthday Communicated by Robert Turner Curtis 10 3 Abstract. M24 is the largest Mathieu sporadic simple group of order 244823040 = 2 ·3 ·5·7·11·23 and contains all the other Mathieu sporadic simple groups as subgroups. The object in this paper is to study the ranks of M24 with respect to the conjugacy classes of all its nonidentity elements. 1. Introduction 10 3 M24 is the largest Mathieu sporadic simple group of order 244823040 = 2 ·3 ·5·7·11·23 and contains all the other Mathieu sporadic simple groups as subgroups. It is a 5-transitive permutation group on a set of 24 points such that: (i) the stabilizer of a point is M23 which is 4-transitive (ii) the stabilizer of two points is M22 which is 3-transitive (iii) the stabilizer of a dodecad is M12 which is 5-transitive (iv) the stabilizer of a dodecad and a point is M11 which is 4-transitive M24 has a trivial Schur multiplier, a trivial outer automorphism group and it is the automorphism group of the Steiner system of type S(5,8,24) which is used to describe the Leech lattice on which M24 acts. M24 has nine conjugacy classes of maximal subgroups which are listed in [4], its ordinary character table is found in [4], its blocks and its Brauer character tables corresponding to the various primes dividing its order are found in [9] and [10] respectively and its complete prime spectrum is given by 2; 3; 5; 7; 11; 23.
    [Show full text]
  • Automorphisms of Group Extensions
    TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 155, Number t, March 1971 AUTOMORPHISMS OF GROUP EXTENSIONS BY CHARLES WELLS Abstract. If 1 ->■ G -í> E^-Tt -> 1 is a group extension, with i an inclusion, any automorphism <j>of E which takes G onto itself induces automorphisms t on G and a on n. However, for a pair (a, t) of automorphism of n and G, there may not be an automorphism of E inducing the pair. Let à: n —*■Out G be the homomorphism induced by the given extension. A pair (a, t) e Aut n x Aut G is called compatible if a fixes ker á, and the automorphism induced by a on Hü is the same as that induced by the inner automorphism of Out G determined by t. Let C< Aut IT x Aut G be the group of compatible pairs. Let Aut (E; G) denote the group of automorphisms of E fixing G. The main result of this paper is the construction of an exact sequence 1 -» Z&T1,ZG) -* Aut (E; G)-+C^ H*(l~l,ZG). The last map is not surjective in general. It is not even a group homomorphism, but the sequence is nevertheless "exact" at C in the obvious sense. 1. Notation. If G is a group with subgroup H, we write H<G; if H is normal in G, H<¡G. CGH and NGH are the centralizer and normalizer of H in G. Aut G, Inn G, Out G, and ZG are the automorphism group, the inner automorphism group, the outer automorphism group, and the center of G, respectively.
    [Show full text]