A PERMUTATION REPRESENTATION. OF THE AUTOMORPHISMS OF A GROUP. Thesis for the Degree of Ph. D. MICHIGAN STATE UNIVERSTTY MAX GILBERT GRUENDL 1969 This is to certify that the thesis entitled A PERMUTATION REPRESENTATION OF THE AUTOMORPHISMS OF A GROUP presented by Max Gi lbert Gruend 1 has been accepted towards fulfillment of the requirements for Ph . D 0 degree in Mathemat iCS QMM Major professor, Date 7/24/69- I / ' 0-169 , , fl... PMs-4rd“: tun-um :_’_‘_4: 2-1-” . .37 r 1 [firm :\ (—1 i Y Myth n 1.1 “rate bun c153)! I la mm: “m; ‘9' ”NBING BY ‘1 nuns &SDNS' a . : am mm mo. ; - T .- ABSTRACT A PERMUTATION REPRESENTATION OF THE AUTOMORPHISMS OF A GROUP By Max Gilbert Gruendl This thesis investigates the permutation representation of the automorphisms of a group G on the composition and chief series of G denoted by (A(G), CG) and (A(G), DG), respectively. The problem is motivated by the thesis of A. Polimeni, Michigan State University, 1965, in which he considers the group G when (A(G), CC) is a transitive permutation group. We generalize in two directions: first since every chief series is a composition series,restrictions on the group G are found when (A(G), D6) is a transitive permutation group (Chapter II), and second we assume that (A(G), CG) is an intransitive permutation group (Chapter III). In Chapter II the main results are: (1) Let G be a nilpotent group, (A(G), DC) is a transitive permutation group if and only if G is isomorphic to one of the follow- ing groups: (a) cyclic p-group for some prime p, (b) elementary abelian p-group for some prime p, (c) the quaternions, (d) the group generated by a, b, and c where ap = bp = cp = [a,b] = [a,c] = 1 and [b,c] = a for some prime p, p ¢ 2. Max Gilbert Gruendl (2) If G is a supersolvable group but not nilpotent, and (A(G), DC) is a transitive permutation group then G = G'Sq where G' is the normal p-Sylow subgroup of G and Sq is a non-normal q-Sylow subgroup of G where p > q and such that (a) G' = Fit(G) , (b) Sq is elementary abelian or cyclic, (c) 6(G) is the largest characteristic subgroup of G contained in G', (d) if, in fact, G' is abelian then Z(G) = (1). (3) If G is a solvable group and (A(G), DC) is a transitive permutation group then the following are true: (a) Socn+1(G)/Socn(G) is an elementary abelian p-group. (b) If K is a characteristic subgroup of G then there is an integer n such that K = Socn(G). (t)/G(t+1) (c) G is an abelian p-group and, in particular, G/G' is elementary abelian or cyclic. (d) Fit(G) is a p-group, and there is an integer n (n) such that G = Fit(G). (e) There is no characteristic subgroup between @(G) and Fit(G) so that §(Fit(G)) S §(G). (f) If Z(G) # (1) then Z(Fit(G)) 2 Z(G) 2 Soc(G). (g) If L0 > L1 >...> Ln = (l) is the lower nilpotent series of G then Li/Li+1 is a p-group for some prime p, and Lo/L1 is isomorphic to one of the four groups of (1)° Chapter III is divided into three sections. The first section contains basic results which are used in the other two sections. In the second section (A(G), CG) is assumed to be intransitive, that is, Max Gilbert Gruendl n(G) 2 2 where n(G) is the number of orbits of (A(G), CG). The results are restricted to p-groups. (4) If G is a p-group, G/Zm is not the quaternions, and -2 m = d(G) then n(G) 2 d(G) where d(G) is the class of G. (5) If G is a p-group, Qn is not isomorphic to H.< G for n 2 3, and n(G) 2 2 then the exponent of G s pn(G). (6) If G is a p-group such that Z(G) is not cyclic then the exponent of G s pm where m is the largest solution of n(G) 2 (m + p-l)!/pl(m-1)!. (7) If G is a p-group such that n(G) 2 2, Z(G) is cyclic, and if H s G then H is not isomorphic to the quaternions or the generalized quaternions then the exponent of G s pn(G)-l. The third section considers the case when n(G) = 2. Since G must either be a t-group or not a t-group we apply results about t-groups when G is a t-group and (3) when G is not a t-group to obtain the following results. (8) Let G be a nilpotent group. Then n(G) = 2 if and only if G is isomorphic to H where H is the group of (1d) or a cyclic group of order the product of two distinct primes. (9) If G is solvable, not a t-group, (A(G), DC) is transitive, and n(G) = 2 then G satisfies (3a), (3b), (3d), (3e), and (3f), and (3c) and (3g) are replaced by (c) and (g) below: (c) C(t) /G(t+1) is an elementary abelian or cyclic p-group, (8) If LO >’L1 >°.-> Ln = (1) is the lower nilpotent series of G then Li/Li+ is isomorphic to a cyclic p-group, an 1 elementary abelian p-group, the quaternions, or H where H is the Max Gilbert Gruendl group mentioned in (1d). Also only one of the factors of the lower nilpotent series is isomorphic to H. (10) Let G be a solvable, non-nilpotent t-group where G/L is the maximal nilpotent factor group of G. n(G) = 2 if and only if G has one of the following three sets of properties. (a) G/L is a cyclic group of order qm, m 2 2, |C(L)| = pq and ‘L‘ = p. (b) G/L is a cyclic group of order qr, L is a cyclic p-group, and C(L) = L. (c) G/L is a cyclic group of order qr, L is an elementary abelian p-group, and C(L) = L. rIn all cases p, q, and r are distinct primes. A PERMUTATION REPRESENTATION OF THE AUTOMORPHISMS OF A GROUP By Max Gilbert Gruendl A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1969 Z' 2~7O To Jo Ann, Robert, and Elizabeth ii ACKNOWIEDGMENTS The author is deeply indebted to his major professor, Dr. J.E. Adney for his patience and guidance in the preparation of this thesis. Our discussions about the research leading to this thesis were of immeasurable value to me. iii TABLE OF CONTENTS INTRODUCTION ............................................. CHAPTER I Basic Definitions and Some Examples ......... CHAPTER II (A(G), DG) Transitive ...................... CHAPTER III (A(G), CG) Intransitive .................... 3.1 Basic Properties of n(G) ................... 3.2 n(G) for G a p-group ..................... 3.3 n(G) = 2 .................................... BIBLIOGRAPHY ............................................. APPENDIX ................................................. iv l6 16 21 34 46 47 LIST OF FIGURES Page Figure l ........................................... 30 II. Relations: H S G E K H A(c) x 5 y mod 2 XEG Operations: {a,b,...} lsl (bk) n(G) 3(6) INDEX OF NOTATION H is a subgroup of G. H is a proper subgroup of G. A is a subset of B. A is a proper subset of B. H is a normal subgroup of G. H is a subnormal subgroup of G. H is isomorphic to K. H is A(G)-isomorphic to K (page 20). x equals y modulo 2. x is an element of G. The image of G under a 6 A(G). The image of g under a E A(G). x-lg x. The factor group of G by H. The index of H in G. The group generated by a,b,... The set containing a,b,... The order of the group G. The order of g E G. The greatest common divisor of L and k. page 16. page 20. VI III. C The number of combinations of m things taken n at a time. L(Gi,c) page 17. n(Gi,C) page 17. Groups: G is a group and H s G. A(G) The automorphism group of G. I(G) The inner automorphism group of G. N(H) {g e G|Hg= H} C(H) {g e c|hg = h for all h e H} Z(G) C(G) 6(6) The intersection of the maximal subgroups of G. Fit(G) The maximal normal nilpotent subgroup of G. Soc(G) The subgroup of G generated by the minimal normal subgroups of G. Socn(G) page 11. [a,b] a-1b-1a b. [a,x] ([h,k]‘h e H and k e K). G' [a,c]. -1 -1 c 9(1) = c' and c = [C(n ), C(n )1. CG The set of all composition series of G. DG The set of all chief series of G. (A(G), CG) page 3- (A(c>, Dc) page 3. In m1,. "’t 8 page 27. t l 1 ts m1 t1 1, page 27. page 21. VII INTRODUCTION This thesis is concerned with the problem of describing a group G given the number of orbits of the permutation representation of the automorphisms of G on the composition or chief series. The problem was first considered by A. Polimeni [4] in his thesis where he con- sidered the case when the permutation representation of the automorphisms of G on the composition series is transitive, that is, it has one orbit. From this starting point we may generalize in two directions. Since the set of chief series is contained in the set of composition series, Chapter II considers G when the permutation representation of the automorphisms on the chief series is transitive from.which results similar to Polimeni's are proven. In Chapter III, we take another -point of view and investigate G when the permutation representation of the automorphism of G on the composition series is intransitive. The basic definitions of these permutation representations and two examples are contained in Chapter I. For easy reference the Appendix lists known theorems by capitol Roman letters. Theorem A, Theorem B, etc. which are utilized in the text of the thesis. The Index of Notation should be consulted for the exact meaning of symbolism appearing throughout the thesis. The results are either called theorems or lemmas and are numbered in the same sequence by chapter, section, and order of appearance. For example, the fourth result in the first section of the third chapter is designated as Theorem 3.1.4. The definitions and examples are numbered similarly. 1 CHAPTER I BASIC DEFINITIONS AND SOME EXAMPLES In this chapter we introduce the definitions of two permutation groups associated with a group G and give some examples to acquaint the reader with the type of problem with which we will be concerned. Before proceeding further all groups in this thesis are assumed to be finite. The main idea is to study groups through investigating how their automorphism group acts as permutations on certain sets of objects associated with the group, namely, the set of chief series and composition series of the group. We let 0 .—_ = * s. G GO > C1 >...> Gn_1 > Gn (1) ( ) mean that s is a chain of subgroups of the group G which consists of the subgroups GO’G ...,Gn. It will be convenient to say that a 1, subgroup H is contained in a chain 3 given by (*) if G1 = H for some i. The first definition shows how an automorphism acts on a chain. Definition 1.1.1. Let s be a chain of a group G as in (*). Then sa is defined to be the chain of G given by a a a a a o = . . . = 1 . s . G CO > C1 > > Gn_1 > Gn ( ) Definition 1.1.2. Let s be a chain of a group G as in (*). Then 3 is a composition series of G if Gi is a maximal normal subgroup of G1 1 for i = l,...,n, and s is a chief series of G ‘ 2 if G1 is a maximal subgroup of 61-1 which is normal in G. Note that if G is a group then all composition series of C have the same length [5, page 37]. Let CG and DG denote the set of composition and chief series, respectively. Then the facts that s E C im lies that S“ E C and G p G s 6 DG implies that sa 6 DG’ a E A(G), allows us to consider the 'permutation representation of A(G) on CG and DG. Definition 1.1.3. Let (A(G), CG) be the permutation repre- sentation on CG defined as follows: if a E A(G) and s 6 CG then sq is the image of 3 under the permutation induced by a which is denoted by “d' (A(G), DC) is defined as above with DC substituted for CG and ad denotes the permutation induced by a. Let KC and KD be the kernels of the homomorphisms of A(G) G G onto (A(G), CG) and (A(G), DG) defined by 0 maps to nd and a maps to ca. Clearly {a E A(G)‘Ha 71 ll H for all quG} {a 6 A(G)‘Ha H for all Hid G} J“ which implies that KCG S KDG and I(G) S KDG. The first inclusion implies that (A(G), DC) is a homomorphic image of (A(G), CG), and the second inclusion implies that in the case of (A(G), DG) we are considering in some sense only auto- morphisms of G which are not inner automorphisms. The following two examples will be investigated as regards the transitivity of (A(G), DC) and (A(G), CG). Example 1.1.1. Q3 = (a,b) where a4 = b = 1 and a2 = b = [a,b]. C(23 = DQ3 = {81’82’83} Where s:G>(a)>(a2)>(l) l 32: c > (b) > (a2) > <1) 83: G > (ab) > (a2) > (1). (A(Qa), CQ ) is a transitive permutation group since if a and B 3 are defined by ad = b, ba = a, a8 = a, and bB = ab then a,B E A(Q3)’ and S“ = s and SE = s . (A(Q3), CQ ) is the symmetric group on 1 2 3 3 three symbols. Q3, of course, is the quaternions and will be shown to be the "exceptional case" to the results of Chapter III. Another important example for both Chapter II and III is the non-abelian group of order p3 of exponent p which is exhibited below. Example 1.1.2. Let G = (a,b,c) where P _ P = P _ - = ~ _ a - b c - [a,b] - [a,c] 1 and [b,c] — a. The center of G is (a), and DG consists of the p+l composition series containing (a). (A(G), DG) can be shown to be a transitive permutation group. Now (A(G), CG) has two orbits, one consisting of the elements of DC and the other consisting of the other composition series, that is, those not containing (a). Example 1.1.3. G = (a,b) where a4 = b4 = l and [a,b] = a2. Z(G) = (a2). The composition series of G are s1: G > (a2,b) > (b) > (1) ' G > (a2,b) > (azb) > (1) 82. ' G > (a2,b) > (a2) > (1) S3. ; G > (a) > (a2) > (1) : G > (a2,ab) > (a2) > (1) = c > (a2,ab> > (ab) > (1) s : G > (a2,ab) > (33b) > (1) DG = {53,34,35}, (A(G), CG) has the three orbits {sl,sz,s6,s7}, {33,85}, and {34} while (A(G), DC) has the two orbits {83,85} and {34}. CHAPTER II THE GROUP (A(G), DG) On the assumption that (A(G), DC) is a transitive permutation group we establish restrictionson the group G when G is abelian, nilpotent, supersolvable, and solvable. Theorem 2.1.1. Let G be an abelian group. Then (A(G), DC) is a transitive permutation group if and only if G is a cyclic p- group or an elementary abelian p-group for some prime p. Proof: The theorem follows from Theorem A since CG = DG when G is abelian. In order to consider the case where G is nilpotent we prove the following two lemmas. Lemma 2.1.2.. If G is a group of order p“, n 2 4, then G contains an abelian subgroup of order p3. Proof: It is sufficient to take G to have order p4 and non-abelian. Assume that G has no abelian subgroup of order p3. By considering a refinement of the upper central series we see that there is a normal abelian subgroup X of G such that ‘XI = p2. C(X) = X since X is abelian, and if C(X) >~X then for any y 6 C(X) but y d X (X,y) would be abelian of order greater than p2. Therefore G/X = N(X)/C(X) is isomorphic to a subgroup of A(X). But p2 = ‘G/X‘ does not divide |A(X)‘ since if X is elementary abelian then |A(X)‘ divides (p2 - p)(p2 - 1), if X is cyclic and p = 2 then ‘A(X)‘ = 2, and if X is cyclic and p # 2 then ‘A(X)‘ = p(p-l). (See Scott [5, pages 120 and 161]). Therefore we have reached a 6 contradiction and the lemma is proven. Lemma 2.1.3. Let G be a p-group. Then all normal subgroups of the same order are isomorphic if and only if G is isomorphic to one of the following groups: 1. cyclic p-group for some prime p, 2. elementary abelian p-group for some prime p, 3. the quaternions, that is, (a,b) such that a4 = b = l and a2 = b2 = [a,b], 4. (a,b,c) where ap = bp = cp = [a,b] = [a,c] = 1 and [b,c] = a for some prime p, p # 2. Proof: Clearly the four groups listed in the lemma have all normal subgroups of the same order isomorphic. Now let G be a p-group such that all normal subgroups of the same order are isomorphic. If G is abelian then it is clear that G is elementary abelian or cyclic, so we assume that G is not abelian. This implies that |G| 2 p3. Case 1. If ‘G‘ = p3 then since the non-abelian p-groups of order p3 are well-known [1, pages 145-146], we see that the two groups mentioned in 3 and 4 are the only groups satisfying the hypothesis. Case 2. ‘GI > p3. We now show that there are no non-abelian p-groups of order pn, n > 3, satisfying the hypothesis of the lemma by induction on n. If n = 4, the non-abelian groups of order p4 are well-known [1, pages 145-146], and it is easily checked that none of them satisfy the hypothesis. Assume n > 4 and that no non-abelian group of order pn-1 satisfies the hypothesis. Let K be a non-abelian group of order pn satisfying the hypothesis. Since K is a p-group Z(K) # (1), so there is a normal subgroup H of K such that H s ZCK) and ‘H‘ = p. Consider K/H. lK/H| = pm"1 and all normal subgroups of the same order of K/H are isomorphic. Therefore K/H is abelian by the induction hypothesis, and so K/H is cyclic or elementary abelian since K/H satisfies the hypothesis of the lemma. Since H s Z(K), K/H cannot be cyclic or K/Z(K) would be cyclic. There- fore K/H is elementary abelian and H = K'. A11 subgroups of K/H are normal so that all subgroups of K containing H are normal in K. By Lemma 2.1.2 there is an abelian subgroup A of K of order p3. Let A* = (A,H). Clearly A* is a normal abelian subgroup of K with |Af| 2 p3 since H s A? and H s Z(K). Now we proceed to * find a normal non-abelian subgroup of K with the same order as A . There are two elements k and k of K such that [k1,k2] f l 1 2 since K is not abelian. ‘(k1,k2)‘ = p3 since (k1,k2) 2 H = K' _ _ 2 * and ‘(k1,k2)/H‘ - ‘(k1H,k2H)| - p . Let B be a subgroup of K of the same order as AI which contains (k1,k2). 3* is normal non- abelian since 3* 2 (k1,k2) > H. Therefore all groups K of order pn, n > 3, must be abelian in order to satisfy the hypothesis, and so the group G of Case 2 cannot exist. Theorem 2.1.4. Let G be a nilpotent group. (A(G), DC) is a transitive permutation group if and only if G is isomorphic to one of the four groups of Lemma 2.1.3. Proof: The groups mentioned in Lemma 2.1.3 are all p-groups so that they are all nilpotent. By Theorem 2.1.1 (A(G), DC) is a transitive permutation group if G is a cyclic or elementary abelian p-group. In Example 1.1.1 we have shown that (A(G), DC) is a transitive permutation group if G is the quaternions. Now suppose that G is the group of 4, that is, G = (a,b,c) where ap = bp = cp = [a,b] = [a,c] = 1, and [b,c] = a. There are p+1 chief series of G, namely, 3 : G > (a,b) > (a) > (1) sn: c; > (a,bnc) > (a) > (1) for n = O,1,...,p-1. Let an, n = 0,1,...,p-l be the automorphism defined as follows: a-Oa-l aaan a b-oc b b“ f =1 1 O . on c or n , ,p . c-ob c-oc 0In Now 3 = 3n for n = 0,1,...,p-1, and this shows that (A(G), DC) is a transitive permutation group. To show the converse we assume that G is nilpotent, and (A(G), DC) is a transitive permutation group. G is a p-group be- cause if P and Q are a p-Sylow and a q-Sylow subgroup of G, respectively, then P and Q are characteristic subgroups of the nilpotent group C, but this contradicts the transitivity of (A(G), DG). (A(G), DG) being a transitive permutation group implies that all normal subgroups of the same order are isomorphic, and this allows us to apply Lemma 2.1.3 to finish the proof. In order to consider the situation where G is supersolvable it is easier to first look at G solvable. .The following lemmas help us to reach this end. Lemma 2.1.5. If G is a group such that (A(G), DC) is a transitive permutation group then H <1 G, K <1 G, H < K, and K/H nilpotent implies that K/H is a p-group for some prime p. Proof: Suppose that |K/H‘ is divisible by two primes p and q, 10 and let ‘P = P/H and 6-= Q/H be the p-Sylow and q-Sylow subgroups of K/H, respectively. Since K/H is nilpotent P/H and Q/H are characteristic subgroups of K/H. Since K/H < G/H, P/H and Q/H are normal in G/H and we have that P and Q are normal subgroups of G. (A(G), DG) cannot be a transitive permutation group since there are two chief series which are refinements of the normal series G > P > H > (1) and G > Q > H > (1). So we must have that K/H is a p-group for some prime p. Lemma 2.1.6. Let G be a group such that (A(G), DC) is a transitive permutation group. Then the following are true: 1. If H is a characteristic subgroup of G, then every chief series is a refinement of the normal series G > H > (1). 2. If H is a characteristic subgroup of G and K 4'G then either K S H or K 2 H. 3. If (1) = H ,H H = G are all the characteristic subgroups O 1’H2""’ n of G, then every chief series is a refinement of the normal series = ... = 1 . G Hn>Hn_1> >H1>HO () The H’s may need to be renumbered. Proof: 1. Let s be a chief series of G not containing H. Then by the transitivity of (A(G), DG) there is an a 6 A(G) such that -1 -1 sq contains H. Now 8 = sad and sad contains H since -1 H0 = H. This contradiction proves 1. 2. K.4 G and H characteristic in G implies there is a chief series containing K, and by part 1 all chief series contain H so that K S H or K 2 H. 3. Part 2 implies that 11 G = Hn > Hm1 >...> H1 > H0 = (1). (The H‘s will probably have to be renumbered.) Part 1 implies that every Hi must be contained in every chief series which gives us that every chief series is a refinement of the normal series containing all characteristic subgroups of G. Lemma 2.1.7. Let H be a characteristic subgroup of G. If (A(G), DC) is a transitive permutation group then (A(G/H), DG/H) is also. _. _* Proof: Let s and s be two chief series of G/H. We must produce _ —* * an automorphism of G/H that maps 3 to s . Let s and s be chief series of G which are refinements of the normal series con- . O l - 4 O Sisting of she preimages of the groups of s and s respectively. Since (A(G), DC) is transitive then there is an a E A(G) such a * '- that s = 8 defined by (xH)a = xaH is an automorphism of 5' -* G/H, and Sa = s which proves the theorem. The converse of Lemma 2.1.7 is not true because an automorphism of G/H cannot, in general, be extended to an automorphism of G. Definition 2.1.1. The sockel, Soc(G), of a group G is the subgroup generated by all its non-trivial minimal normal subgroups. Also, let Socn(G) be defined recursively as Soc0(G) = (l) and n+1 n n Soc (6)/Soc (G) = Soc(G/Soc (G)). From its definition the sockel of G is a characteristic sub- group of C. It can be shown that Soc(G) is a direct product of some of the non-trivial minimal normal subgroups of G [5, page 168]. This, in the case where G is a solvable group, gives us that the Soc(G) is a direct product of elementary abelian p-groups since the 12 minimal normal subgroups of a solvable group are elementary abelian [5, page 74]. Lemma 2.1.8. Socn+1(G) = (H‘H is minimal with respect to being normal and properly containing Socn(G)). Proof: The proof follows since if H is minimal with respect to being normal and properly containing Socn(G), then H/Socn(G) is a non- trivial minimal normal subgroup of G/Socn(G), and conversely. Theorem 2.1.4 assumes that G is nilpotent so in our investiga- tion of G solvable the importance of the next definition is apparent. Definition 2.1.2. Let G be a solvable group. Let L1 be defined recursively as L0 = G, and L1 is the smallest normal sub- group of Li-l such that Li-l/Li is nilpotent. The normal series = ...> = 1 G LO > L1 > Ln ( ) is called the lower nilpotent series. Theorem 2.1.9. If G is a solvable group and (A(G), D6) is a transitive permutation group then the following are true: 1. Socn+l(G)/Socn(G) is an elementary abelian p-group. 2. If K is a characteristic subgroup of G then there is an integer n such that K = Socn(G). 3. C(t)/G(t+1) is an abelian p-group, and, in particular, G/G' is elementary abelian or cyclic. 4. The Fit(G) is a p-group, and there is an integer n such that G(n) = Fit(G). 5. There are no characteristic subgroups between 6(G) and Fit(G) so that §(Fit(G)) s 9(G). 6. If Z(G) # (1) then Z(Fit(G)) 2 Z(G) 2 Soc(G). l3 7. If L0 > L1 >...> Ln = (1) is the lower nilpotent series of G, then Li/Li+1 is a p-group for some prime p, and LO/L1 is isomorphic to one of the four groups of Lemma 2.1.3. Proof: 1. Since G is a solvable group G/Socn(G) is a solvable n _ n+1 n . . group, and Soc(G/Soc (G)) - Soc (G)/Soc (G) is the direct product . n+1 n of elementary abelian groups and therefore Soc (G)/Soc (G) is . n+1 n . . abelian. By Lemma 2.1.5 Soc (G)/Soc (G) is a p-group so that it is an elementary abelian p-group. 2. Let K be a characteristic subgroup of C. By Lemma 2.1.6 there -1 is an integer n such that Socn (G) < K s Socn(G). By Lemma 2.1.6 K contains all subgroups of G that are minimal with respect to being normal in G and properly containing Socn- (G). But Lemma 2.1.8 implies that K 2 Socn(G) so that K = Socn(G). +. 3. C(t)/G(t l) is an abelian p-group by Lemma 2.1.5, and G/G' is elementary abelian or cyclic by Theorem 2.1.1 and Lemma 2.1.7. 4. Since Fit(G) is the maximal normal nilpotent subgroup of G, Lemma 2.1.5 implies that Fit(G) is a p-group. There is an integer (n) q and such that 1. G" = Fit(G), 2. S is elementary abelian or cyclic, q 3. 6(G) is the largest characteristic subgroup of G contained in G', 4. if, in fact, G" is abelian then Z(G) = (1). Proof: 1. Theorem G implies that G9 is nilpotent so that G' s Fit(G). (n) Theorem 2.1.9 says that there is an integer n such that G = Fit(G). But Fit(G) is not G since G is not nilpotent, and this gives us a that G2 =Fit(G). G” = Fit(G) must be a p-group for some prime p 15 by Theorem 2.1.9. G/G' is an elementary abelian or cyclicig-group by Theorem 2.1.9 where q # p for otherwise G would be nilpotent. p > q for by Theorem I. G has a normal Sylow subgroup for the largest prime, and if q > p then G would be a direct product of two nilpotent groups which is nilpotent. Let Sq be a q-Sylow subgroup of G. Then G = cisq and G/G9 is isomorphic to Sq and this proves 2. 3. This follows directly from part 5 of Theorem 2.1.9. 4. Ge abelian implies that all of G75 Sylow subgroups are abelian, and so G is an A—group. (An A-group is a group whose Sylow subgroups are abelian. A-groups have been extensively studied by D.R. Taunt [6]). By Theorem J in an A-group G'3 n Z(G) = (1). But this along with the facts that G‘ = Fit(G) and Z(G) s Fit(G) gives us that Z(G) = G3 n Z(G) = (1). This completes the proof. The following example shows that the conclusions of Theorem 2.1.10 are not sufficient to imply that (A(G), DC) is a transitive permutation group. Polimeni [4] shows that the converse of Theorem C is true under very limited circumstances, and we could do the same here, but no new knowledge would be gained. Example 2.1.1. Let G = (a,b,c,d) where a9 = b3 = c3 = d2 = [b,c] = 1, [a,b] = a3, [a,c] = b, [a,d] = a7, [b,d] = a3, and [c,d] = c. It may be seen that (a,b) 4 G, (a3,b,c) 4 G, and |(a,b)‘ = 1(a3,b,c)| but clearly this implies that (A(G), DC) is not a transitive permutation group. CHAPTER III (A(G), CG) INTRANSITIVE In his thesis [4], Polimeni investigated a group G where (A(G), CG) was assumed to be a transitive permutation group, and in Chapter II we have looked at G when (A(G), DC) is a transitive permutation group. Chapter III's purpose is to consider G where (A(G), CC) is not a transitive permutation group. Of course, this is a much more difficult problem and in order to attack this problem the hypothesis had to be further restricted. Chapter III is divided into three sections: .Basic Properties of n(G), n(G) for G a p-group. and n(G) = 2. Here n(G) denotes the number of orbits of (A(G), CG). 3.1. Basic Properties of n(G). Definition 3.1.1. An orbit of (A(G), CC) is {30‘OZ e A(G) and s 6 CG(s fixed)}. It is well-known that the set of orbits of a permutation group form a partition so that the set of orbits of (A(G), CG) form a partition of C . G Definition 3.1.2. Let n(G) denote the number of orbits of If n(G) = 1 then (A(G), CG) is a transitive permutation group and conversely. The following results are lower bounds on the number of orbits of (A(G), CG) depending on the structure of the lattice of subnormal l6 17 subgroups of C. To do this we need the following definition. Definition 3.1.3. If c: G = Gm > Gm >...> G >1GO = (1) is -1 l a chain of characteristic subgroups of G, then let L(Gi,c) Q1(Gi,c)) be the number of subnormal subgroups Kij of G such that Kij % Gi’ 1‘1 (Kij < Gi+l)’ and there does not exist a 6 A(G) such that ng = Kik for j # k, j,k = 1,2,...,L(Gi,c) Kij‘ = ‘Gi‘, Ki], >G Q1(Gi,c)) and i = l,2,...,m-l. In particular, if H is a proper characteristic subgroup of G and c: G > H > (1) then L(H,c) ==u,(H,c) and is called L(H) or n(H). Theorem 3.1.1. Let H be a proper characteristic subgroup of G. Then 1. n(G) 2 n(H) +'u(H)a 2. n(G) 2 n(G/H) +'L(H). Proof; 1. Let s; be a representative composition series of Ci where CE is an orbit of (A(H), CH) where i = 1,2,...,n(H). Let e 31 be a composition series of G which is a refinement of si in 6- (Si exists since H is a subnormal subgroup of C). There does not exist a E A(G) such that s? = sj where i # j, since if CY_ o i 0 ca!_ 3 i ‘. o si - sj where i . J then si - sj where a is a restricted to H and a” E A(H) since H is a characteristic subgroup of G. Therefore 31 and sj belong to different orbits of (A(G), CG) if i f j. Note that each si contains H. Since there are n(H) sub- normal subgroups Kj of G with the same order as H, such that there does not exist a 6 A(G) where K? = Kk’ j # k, then there are at least n(H) orbits of (A(G), CG) whose composition series do not contain H. Therefore n(G) 2 n(H) +’M(H)' 2. Let ‘;i be a representative composition series of C1 where Ci 18 is an orbit of (A(G/H), C ) with i = 1,...,n(G/H). Let 81 be G/H a composition series of G which is a refinement of the subnormal series consisting of all the preimages of the subgroups of ‘Ei in 0’ G. Again there does not exist a 6 A(G) such that si = sj for . . . . oz_ -3_- -. i f 3, Since if 31 - Sj then si - sj where a is the auto- morphism of G/H induced by a (that is, a. is defined by (gH)a = gaH, and 5.6 A(G/H) since H is a characteristic subgroup of G). Note that each 31 contains H. Since there are L(H) subnormal subgroups Ki of G with the same order as H where there does not exist a E A(G) such that K: = Kk, j r k, then there are at least L(H) orbits of (A(G), CG) whose composition series do not contain H. Therefore n(G) 2 n(G/H) +-L(H). Corollary 3.1.2. If H is a characteristic subgroup of G then 1. n(G) 2 n(H) where, in particular, there are n(H) orbits of (A(G), CG) whose composition series contain H, 2. n(G) 2 n(G/H) where, in particular, there are n(G/H) orbits of (A(G), CG) whose composition series contain H. Proof: See proof of Theorem 3.1.1. Corollary 3.1.2 is stated here to clarify the proof of the next theorem. Theorem 3.1.3. If c: G = G > G >...> G > G = (1) m m-l 1 O 1, where G1 is a characteristic subgroup of G for i - ...,m-l, then m-l . 1. [I(G) 2 n(G1)'+}.'..‘i,__1 “(Ci-3C): -1 2. n(G) 2 n(G/Gm_1) +211 Proof: 1. Consider G1. By Corollary 3.1.2 there are at least L(Gi.<:)- l9 n(Gl) Also there are at least p(G1,c) orbits of (A(G), CG) whose com- orbits of (A(G), CG) whose composition series contain G1. position series do not contain G1 but do contain G2 (Definition 3.1.3). G < G2, and G1 characteristic in G implies that there 1 are at least n(Gl) +‘p(G1,C) orbits of (A(G),.CG) whose composition u series contain G2. Now repeating this argument for Gi’ i = 2,3,...,m-l§ we obtain part 1 of the theorem. 2. Consider Gm By Corollary 3.1.2 there are at least n(G/Gm -1. _1) orbits of (A(G), CG) whose composition series contain Gm_1. Also by Definition 3.1.3 there are at least L(Gm_1,c) orbits of (A(G), CG) whose composition series contain Gm_2 but do not contain Gm-l' Gm_2 < Gm_1, and Gm“1 characteristic subgroup of G implies that there are at least n(G/Gm_l) +-L(Gm_1,c) orbits of (A(G), CG) whose composition series contain Gm-2' Repeat this argument for Gi’ i = m-2,...,2,1, and we obtain the result of the theorem. Another lower bound for n(G) is given in the next theorem by considering the number of orbits of the factor groups of a chain of characteristic subgroups of G. Theorem 3.1.4. If c: G = Gm‘> Gm >...> G1 > G0 = (1) is -l a chain of characteristic subgroups of G then n(G) 2 111:1 n is the automorphism of Gi/Gi i s = s . 1,3(1) i,k(1) induced by a (that is, -1 a. i _ a . (giGi-l) giGi-l) which, in turn, implies that Si,j(i) and Si,k(i) are in the same orbit of (A(Gi/Gi-l)"CGi/G ), so that j(i) = k(i), i-l i = 1,...,m. Therefore we have proven this theorem. Definition 3.1.4. If G is a group and s1,s2 6 CG then 31 is isomorphic to denoted by s if every subgroup of 8 § 52’ 1 32’ 1 is isomorphic to a subgroup of 32. s1 is A(G)-isomorphic to 82’ denoted by 31 !A(G)82 if there exists a 6 A(G) such that every sub- group of 82 is an image of a subgroup of 51 under a. a - a a 2 Clearly s1 A(G)82 implies s1 82. A(G) and are equivalence relations on .CG, and the equivalence classes of §A(G) are the orbits of (A(G), CG). n(G) is the number of equivalence classes of CG under and so we similarly define EKG) to be EA(G)’ the number of equivalence classes of CG under E. Trivially n(G) 2 3(6). This new equivalence relation allows us to prove two theorems similar to Corollary 3.1.2 where Hid q C or H.4 G instead of H is characteristic in G. Theorem 3.1.5. Let H 4.4 G where G is a group, then 3(H) s n(G) and, in particular, there are at least 3(H) orbits of (A(G), CG) each of which contain a composition series containing H. Proof: Let s' be a representative composition series of C' where 1 i c; is an orbit of (A(H), CH) (not necessarily all the orbits of (A(H), CH) are included) where 8; ¥ 53, if i # j, i,j = l,...,h(H). Let 81 be a composition series which is a refinement of s; in i t i . 1 § G (31 ex s s s nce Hid 4 C) Si ¥A(G)sj’ for i f j, s nce si A(G)sj implies 31 E s and this implies s; 5 3'. Therefore there are at J J 21 least 3(H) orbits of (A(G), CG) each of which contains a composition series containing H. Theorem 3.1.6. Let Hiq G where G is a group, then h(G/H) S n(G), and, in particular, there are at least 3(G/H) orbits of (A(G), CG) each of which contains a composition series containing H. Proof: Let s be a representative composition series of C; where i E] are orbits of (A(G/H), CG/H) (not necessarily all the orbtis of (A(G/H), CG/H) are included) where E] 2 E], if i # j, i,j = l,...,h(G/H). Let s be a composition series of G which is a refinement of the i subnormal series of all the preimages in G of the subgroups of s.- 1- 3' g 3' which is a contradiction. Therefore there are at least 3(G/H) 1 J orbits of (A(G), CG) each of which contains a composition series con- or i # j, for if taining H. 3.2 n(G) for G a p1group. Next we restrict our attention to p-groups, and show that in most cases n(G) 2 the class of G. Also we will find upper bounds on the exponent of G. The generalized quaternions, because of the fact that there is only one subgroup of order 2, is very crucial in this chapter so the next two lemmas investigate this important group. n-1 Definition 3.2.1. The group Qn = (a,b) where a = 1, —2 2 n - b = a2 , b 1 a b = a.1 is called the generalized quaternions for n = 4,5,... . Q3 is the quaternions. Lemma 3.2.1. n r 3. 1. Every element of Qn can be written in the form ambL where m = 0,1,...,2n”1 - 1, L = 0,1, and therefore ‘in = 2n. 2. Tamb| = 4 for all m. 22 k 2n-k-l 3. If xEQn and \x\=2,k23 thenx€(a ). n-2 4. a2 is the only element of order 2. .k S. (a2 ) is a characteristic subgroup of Qn for k = 0,1,...,n-2. k zn-k-l zn-k 6. If K s Qn and ‘K‘ = 2 then K = (a ) or K = (a ,amb) for some m. 7. If s E CQ then 3 must be of the form n t t+1 n-2 t s: Qn > (a2,amb) >...> (a2 ,amb) > (a2 ) > (a2 ) >...> (a2 ) > (1) where t = 0,1,2,...,n-2. Proof: 1. Every element can be written in the form ambL since n-1 n-1 (a) 4 Qn where m ranges from 0 to 2 - 1 since ‘a‘ = 2 n-2 and L ranges from 0 to 1 since ‘b‘ = 4 and b2 = a2 m - _ - - 2. (a b)(amb) = ama 1b am 1b since b 1a b = a 1 = ems-2b am-zb ama-mb b = b2, and so lamb] = 4 since ‘b‘ = 4. k 3. Let x E Qn and ‘x‘ = 2 , k'2 3, then x E (a) by 2. 4. By 2 any element of order 2 must also belong to (a) and so it is n-2 a2 5. (a) is characteristic in Qn since it is the only cyclic subgroup of k n 1. (a2 ) is characteristic in Qn since (a) is a cyclic order 2 2-group. 6. If K s Qn and ‘K‘ = 2k then K contains only one subgroup of order 2 and Theorem K implies that K is cyclic or K is a generalized quaternion group or quaternion group of order 2k. If k‘2 3, in the n-k-l former case, by 3, K = (a ), and, in the latter case, K must be 23 generated by an element of order 4 which, by 2, must have the form mb k-l . a for some m, and by an element of order 2 which, by 3, is 2n-k 2n-k in (a ). Therefore K = (a ,amb). If k = 2, then n-k (32 samb) = (amb) and so K is either (amb) for some m or n-k-l 2 (a ). 2t 7. Let s 6 GQ and let t be the smallest integer such that (a ) n n-2 is contained in s ~(t exists because (a ) is the only subgroup of order 2). The next largest subgroup of 3 must be of the form t (a2 ,amb) by 6. Any other subgroup K of s of order 2k contain- 2t 2n.k m' ing (a ,amb) must be of the form (a ,a b) by 6, but cut 6 K. n-k Therefore K = (a2 ,amb). Therefore s is of the form shown in 7. Definition 3.2.2. If G is a group then zn(G) is defined recurSively as ZO(G) = (l) and Zh+1(G)/Zn(G) = Z(G/Zn(G)). We will let 2 = Zi(G) when there is no chance of confusion. 1 20 s 21 s...s Zn s... is called the upper central series of G. If Zm = G for some m then the class of G, denoted by d(G) is the smallest such m. Lemma 3.2.2. If n 7‘ 3, then n(Qn) = n-l so that the exponent n02 ) n of Qn = 2 Proof: .Let s E C be of the form '5 Q 2 n 2t 2t 2n-2 s: Qn > (a ,b) >...> (a ,b) >'(a ) >...> (a ) > (1). Any other composition series 8 with t being the smallest number such that t 8 contains (a2 ) is of the form 2 2t 2t 2n-2 ston > (a ,amb) >...> (a ,amb) > (a )>...> (a )> (1) by Lemma 3.2.1. .Now a e A(Qn) defined by a“ = a and 6" = nub maps st to 8. Therefore the orbits of (A(Qn), CQ ) which contains n st must contain all s such that t is the smallest integer such 24 t O 2 O I O O that 3 contains (a ). Since all composition series have such a $A(G)Sr’ number of such st's is equal to n(Qn). t ranges from 0 to n-2 so we have n(Qn) = n-l. Clearly the exponent of Qn is Zn-l so nan) t by Lemma 3.1.2, and for r $ t since st ¥ sr, the 8t that the exponent of Qn is 2 In considering the relationship of the class of a p-group G and the number of orbits of (A(G), CG), we use Theorem 3.1.3, which leads to the necessity of proving the next two results. Lemma 3.2.3. If G is a p-group and c: G = G > G >...> G > G = (1) is a chain of characteristic m m-l l 0 subgroups of G such that G/Gi is not the quaternions, the gen- eralized quaternions, or cyclic for i = 0,1,...,m-2, then L(Gi,c) 2 1. Proof: To show L(Gi,c) 2 l we need to show that there is at least one subgroup of G with the same order as G and containing G i i-l’ i = 0,1,...,m-2. G/Gi-l is not cyclic, the quaternions, or the generalized quaternions so from Theorems K and L there is at least one other subgroup of G/Gi-l of the same order as Gi/Gi-l and this subgroup's preimage in G is a subgroup with the same order as Gi and containing Gi-l' -Theorem 3.2.4. If G is a p-group and c: G = Gm‘> Gm-l >...> G1 260 = (1) is a chain of characteristic subgroups of G such that G/G1 is not the quaternions or cyclic for i = 0,1,...,m-1 then n(G) 2 m. Proof; Case 1. If G/Gi’ i = 0,1,...,m-2 is not the generalized quaternions then, by Lemma 3.2.3, L(Gi,c) 2 1 for i = l,...,m-l. Theorem 3.1.3 implies that n(G) 2 n(G/Gm_1) +'ET;1.L(G1 c) 2 l + m-l = m. 25 Case 2. If G/Gi is the generalized quaternions for i = 0,1,..., or m-Z let n be the smallest integer such that G/Gn is the generalized quaternions. Since G/Gm is not cyclic -1. |G/Gm_11 2 22 so that 2 m-l-n-l+1 = _ mél m-n+l lG/Gn| - \G/Gm_1[pfil+]ai/Gi_1| 2 2 .2 . 2 This implies that if G/Gn is isomorphic to Qk’ k 2 4, then k 2 m-n+l by Lemma 3.2.1. It now follows that if n # m-2 then n(G/Gn) = n(Qk) 2 n(Qm-n+1) = m-n-l-l-l = mm, and if n = m-2 then n(G/Gn) = n(Qk) 2 “((24) = 3 2 m-n by Lemma 3.2.2. Consider the chain of characteristic subgroups c': G.) Gn > G ...> G >1G0 = (1). A(Gi,c') 2 l for i = l,...,n n-l > 1 by the minimality of n. Applying Theorem 3.1.3 to c' we obtain n + ' -+n=. n(G) 2 (G/Gn) 21:1 L(Gi,c ) 2 m n m Theorem 3.2.5. If G is a p-group, G/me2 is not the quaternions, and m = d(G), then n(G) 2 d(G). Proof: If G is abelian then d(G) = 1, and clearly n(G) 2 d(G). Assume G is not abelian, that is, d(G) 2 2. It is well-known that 6/2 is not cyclic for i = 0,1,...,m-1. G/Z cannot be isomorphic i to the quaternions for i = 0,1,...,m-3, and m-l since d(Q3) = 2 i and d(G/Zi) = m-i. G/z cannot be isomorphic to the quaternions m-2 by hypothesis. Therefore the upper central series is a chain of characteristic subgroups of G as in Theorem 3.2.4, and this implies 26 that d(G) s n(G). The quaternions Q3, Example 1.1.l,have d(Q3) = 2 and n(Qa) = l, which points out the necessity of the hypothesis of the last theorem. n(G) can be equal to d(G) for this is the case when G is the group of Example 1.1.2. The next theorem investigates a p-group G when n(G) = d(G). Theorem 3.2.6. If G is a p-group and there does not exist a characteristic subgroup H of G such that G/H is the quaternions, then m = d(G) = n(G) implies that if K is a proper characteristic subgroup of G where G/K is not cyclic then K = Z1 for some i = 0,1,...,m-1. Proof: Let K be a characteristic subgroup of G where G/K is not cyclic and K # Zi for all i = 0,1,...,m-l. There is a largest integer n (n~< m) such that Zni< K. Now K.n Zn+' > Zn, since 1 /z = z(G/z ), x/z intersects z /z non- n n n n+1 n < Zn+1 then the series of characteristic -i...> Z 2 K11 Zn+1 > Zn >...> (1) has length n+1 m+l, but c also satisfies the conditions of Theorem 3.2.4 since G/K.n Zn is not the quaternions by the hypothesis, and G/K.n zn +1 +1 is not cyclic since if n.< m-l then G/Zm_1 is not cyclic, and if n = m-l then Kt] Zn+_ = K n G = K and. G/K is not cyclic by l . = h assumption However if K n Zn+1 2 then K 2 Zn+_ whic n+1 1 contradicts the maximality of n. This proves the theorem. This completes our discussion of the length of chains of characteristic subgroups in relation to n(G). The exponent of a p-group G is the next parameter to be considered in relation to n(G). The exponent of a group G is the smallest positive integer 27 m such that gIn = l for all g 6 G. We show that if G is'a p- group then the exponent of G is less than or equal to p with- out any difficulties, and then proceed to lower this upper bound for most p-groups G. In order to do this we must build up some notation to handle the case when G is an abelian p-group. m m Definition 3.2.3. Let t1""’ts denote an abelian p-group l s which is the direct product of the direct product of ti cyclic p- m. _ groups of order p 1, i = l,...,s. Let mi 2 mj 2 O for i s j and i,j = l,...,s. or 1 J n n Definition 3.2.4. If' (“l,..., r is such that n .> n > 0 1 i,j = l,...,r then it is said to be in for ii< j and ”i > 0 for reduced form. Lemma 3.2.7. All abelian p-groups are isomorphic to an abelian n n p-group of the form u ,...,ur) which is in reduced form. In fact, 1 r , m m if G is isomorphic to an abelian p-group of the form t ,...,t8 l s then G is isomorphic to a group of the form 1 m1 m1 mi+2 ms if m = m o ,..., , ,ooo, , t1 ti+ti+l ti+2 ts 1 1+1 m m 2 £1, . ,ts‘l if m = o , 1 8-1 m1 i-l mi+l ms _ 3. , , if t, - 0 t ’ ’t ’ t t i l i-l i+l s 1 "‘1 Using these three types of reductions t ’°"’ts can always be 1 s changed to reduced form. Proof: ‘This follows directly from Definitions 3:2:3 3nd 3.2.4. m m Definition 3.2.5. Let t1’°"’ts denote the reduced m1 m3 1 8 form of t ,.-.,t -28 Lemma 3.2.8. If G is an abelian p-group isomorphic to m l s ,..., t:1 ts m 28 m t then .‘G‘ = p' 1:; i 1. Proof: The;proof is clear because of Definition 3.2.3. Definition 3.2.6. Let h- be defined recursively as HG) = 1 and 1“1 “r _2s — n1 ni “"1 ni+1 nr 9 2 .= $ '9 _ ’ S 3 9 ul u‘r 1 1 H'1 u'i 1 1 u'i+l pr n1 n where ,..., r is in reduced form. 1 H'r Lemma 3.2.9. If G is an abelian p-group isomorphic to n1 n ,..., r , in reduced form, then 141 “Dr n n -' l r n G = n () “1, ’ur r r : o o o . =1 Proof The proof is by induction on 21:1 niui If 21:1 niui then n1 =111 = s = 1. Therefore 6 is a cyclic p-group of order p so that 3(6) = 1 and n - = 1. Now assume that all abelian p- groups of order less than p 21:1 “191 the theorem is true. Let G n n be a group isomorphic to ”l,...,ur in reduced form. There are r 1 r none of which are r subgroups of G of the order p(Ei.-.1 n5-lJ'i)"1 isomorphic to any other. Call them H , i = l,...,r, where H is i i . n1 ni 'ni-t1 nr isomorphic to ,..., 1 9 9°°°9 “'1 “i- 1 ”I. - r -— I r - nl ni ni-1 nr -nl 'nr n(G)=E.= n(H) 82.: n a'°': _1: a°°': =11 2° 9 i l i i l “1 pi 1 ur 1 “r by the induction hypothesis which proves the lemma. 29 Lemma 3.2.10. RC") = l, KG) = l for m2 1. Proof: Lemma 3.2.9 allows us to write 3(1) =11: :1 =1:.) H(“I“) =er : “‘11) =36?) - The need for the next lemma and thus all the preceding lemmas u n pl ‘HH v u H and l h 5| 6‘: n 2" will be seen in Theorem 3.2.14 where we obtain upper bounds on the exponent of a p-group G. - n 1 Lemma 3.2.11. n [(1 , 111)] Cn-l-m-l,m n 2 l, m 2 l where -- uni-loud. Oink - l 1 - 1 __ Proof: Case 1. n - l. n[(1 , in] “(m-+1) - 1 by Leanna 3.2.10, = 1. and c1-l~m-1,m = Cm,m Case ‘2. n > 1. The proof is by induction on n+m. The smallest value of n+m is 3 which means that m = 1 and n = 2. {(511-369433"ilTUi’g’fi =3(§)+H©=1+l 2’ 2+1_1,1 = 62,1 = 2. Now assume the lemma is true for n-lm < k. If n-l-m = k then - l - 1 = - n n-l l + - n l 0 n 9 n 9 n o 9 1 9 n 1 9 m_1 9 1 - n-l l — n l n( 1 ’ m) + n(1 ’ m-l) Cn-l-i-m--l,m + Cn-i-tn-1-1,m-l = + = . cn-l-m-2,m Cn-i'in-2,m-1 Cn-T-tn-l,m D-‘t-I' and C 30 In order to prove that for G a p-group the exponent of G S pn(G) we need the following. L Lemma 3.2.12. If HI = (a,b) such that ap = bp = l, and L [a,b] = ap, or H = (a,b) such that ap = bp = 1 and [a,b] = 1, 2 then 3(Hi) 2 L and n(Hi) 2L+l for i = 1,2. Proof: Consider Figure l. (a,b) 81 (a) (ahb) 82 cap) {-1 (ap ,b) SL {-1 (ep > (b) 8:,+1 (1) Figure 1. Clearly si s sj for i # j and i,j = l,...,L which proves that 'E(H1) 2 L. (ap) = 9(Hi) for i = 1,2, since 9(Hi) is the smallest normal subgroup of Hi such that its factor group is elementary p) is elementary abelian so that (ap) 2 §(Hi). P). abelian, and Hi/(a 2 Since Hi/(ap ) is not elementary abelian 9(Hi) = (a L-l not exist a 6 A(G) such that (ap ) There does L-l “ = (b) since (ap ) s 6(Hi) 31 and §(Hi) is a characteristic subgroup of Hi. This gives us that 81 aEscaped Theorem 3.2.13. If G is a p-group, Qn is not isomorphic to n(G). for i,j = l,2,...,L+l which implies that n(Hi) 2 L+l. LH_< G for‘ n 2 3, and n(G) 2 2 then the exponent of G s p Proof: Since n(G) 2 2 Theorem B implies that G is not cyclic. Assume there is an a E G such that |a‘ = pL where L > n(G) and that (a) is not contained in a cyclic subgroup of G ((a) f 6). Let b E N((a)) but b 2 (a) such that bp 6 (a). ‘(a,b)‘ = pL+1 since bp 6 (a). If b d C(a) then (a,b) is a group isomorphic to the group H1 of Lemma 3.2.12. 0n the other hand if b E C(a) then (a,b) is a group isomorphic to the group H2 of Lemma 3.2.12. In either case (a,b) is subnormal in G, and by Lemma 3.2.12 ‘h((a,b)) 2 L > n(G) which contradicts Theorem 3.1.5 and this proves the theorem. n(G). So far we have shown that exponent of 6 s p The follow- ing example gives us some hope that this upper bound can be made smaller. Example 3.2.1. Let G = H X (d) where ‘d‘ = p, and P bp = cp = [a,b] = [a,c] = 1, and [b,c] = a. H = (a,b,c) with a Z(G) = (a,d). n(G) 2 3 since the three composition series 8 : G‘> (a,b,c) > (a,b) > (a) > (1) s : 6 > (a,b,d) > (a,d) > (a) > (1) s ' G > (a,b,d) > (a,b) > (a) > (1) 3. are in different orbits of (A(G), CG) since (a,b,d) is abelian and (a,b,c) is not abelian and 2(6) = (a,d). Even though n(G) 2 3 the exponent of G is p. The fact that Z(G) in this example is 32 not cyclic leads us to the following theorem. Theorem 3.2.14. If G is a p-group such that Z(G) is not k c clic and the ex onent of G is then n G 2 C - y p p k () k+p-1.p Proof: Let g E 6 such that ‘g‘ = p . It is well-known that the number of subgroups of a particular order in a p-group is equal to l modulo p [1, page 129]. Since Z(G) is not cyclic and abelian, Theorems K and L imply that Z(G) has at least l+p subgroups of k-l order p. Since (g ) may or may not be contained in Z(G) we , 1 can construct a subgroup H of G isomorphic to c; 9 p) where H is the direct product of (g) with p subgroups of Z(G) of order p. By Theorem 3.1.5, Lemma 3.2.9, and Lemma 3.2.11 we have - — 1 n(G) 2 n(H) = n(‘l‘ , p) = ck+p_1’p. Corollary 3.2.15. If 6 is a p-group such that Z(G) is not cyclic, then the exponent of G 5 pm where m is the largest solution of n G 2 C . ( ) m+P'1:P Proof: By Theorem 3.2.14, n(G) 2 Ck+p-l p where the exponent of G k is p . If m is the largest solution of n(G) 2 C then m+p-1,p k s m and the exponent of G s pm. Corollary 3.2.16. If G is an abelian p-group such that n(G) 2 2 then the exponent of G 5 pm where m is the largest solution of n(G) 2 Cm+p-1,p' Proof: Since n(G) 2 2 and 6 = Z(G), Z(G) is not cyclic by Theorem A, and we apply Corollary 3.2.15 to prove the theorem. The next corollary gives an easy method for finding an upper bound for the exponent of C, but it does not give as sharp a bound. 33 Corollary 3.2.17. If 6 is a p-group and Z(G) is not cyclic then the exponent of G S pIn where m is the smallest integer such m that 21:1 1 2 n(G). Proof: Let the exponent of G be pk. In the proof of Theorem 3.2.14 replace H by direct product of (g) with just 2 subgroups of 2(6) of order p which can be done since p 2 2. Then we have _ ." 1 = =4k+1)! 3 14kg) = k , n(G) 2 n(H) n[(11( ,2)] Ck+2-1,2 ”(191)! 2 21:1 i. If m is the smallest integer such that 2?;1 i 2 n(G) then m k 21:1 1 2 n(G) 2 21: i. 1 Therefore m 2 k, and so the exponent of G S pm. Note that if G is a p-group with Z(G) not cyclic, this corollary implies that when n(G) 2 3 then the exponent of G S pn(G)-1. If n(G) = 2, Corollary 3.2.15 implies that the exponent of pn(G)-1 G = p = . Of course, Corollary 3.2.15 gives much better bounds for the exponent of G in most cases. When Z(G) is cyclic we can n(G)-1 which is the content of the next only give the upper bound p theorem. Theorem 3.2.18. If G is a p-group such that n(G) 2 2, 2(6) is cyclic, and if H S G then H is not isomorphic to the quaternions or the generalized quaternions, then the exponent of G S pn(G)-1. Proof: Suppose that a 6 G such that ‘a‘ = pL where L 8 n(G). Let Z(G) = (z) where ‘2‘ = pm. Then m S L because of Theorem 3.2.13. Case 1. (a) 2 Z(G) and (a) n Z(G) # (1). Suppose that L‘k m-k- l) (a) n Z(G) = (ap ) where k = l,...,m-l. Let H = (a,zp ' 34 H is abelian and m-k-l + |n|-1(‘)U(zp >1 stag... pm-k-l k Ha) n (z >| P m‘k L'k [ll'k L-k {a-k-l Since zp 5 (3p )9 2p = aSp Let b = a.SP zm-k-1 Then b é Z(G) and ‘bl = p since .. L'k’l _ _ _ L-k m-k _ L-k L'k (a sP zm k 1)p = a SP zp = a Sp asp = 1. L-l Therefore H = (a,b) where b d 2(6) and sp 6 Z(G). Case 2. (a) 2 Z(G). By hypothesis N((a)) is not the quaternions or the generalized quaternions so that there is a b E N((a)) such that ‘b‘ = p and b 6 2(6). Let H = (a,b) where b é 2(6) and L-l a? e Z(G). Case 3. (a) n Z(G) = (1). Let b e Z(G) and |b| = p. Let H = (a,b) L-l where b 6 2(6) and ap 6 2(6). In all cases we consider the L+l composition series Si’ i = l,...,L+l defined in Figure 1, page 30. Again 81 P SJ, for i r j,'i,j = l,2,...,L. Also s +_ s s for i = l,2,...,L-1. _L l i 6'1 - P P SL+1 $A(G)8L Since either a G 2(6) and b 6 2(6), or a 6 2(6) and b d Z(G) in all three cases which proves that n(G) 2 L+1 = n(G) + 1 which is a contradiction so that no such a exists and we have proven the theorem. 3.3. n(G) = 2. This section is included in this thesis as an application of the results of Chapter II to show that (A(G), DC) and (A(G), CG) are intimately related. The assumption that n(G) = 2 can be broken 35 down into two cases: 6 has a subnormal subgroup which is not normal in G, and all subnormal subgroups are normal. The former case implies that (A(G), DC) is a transitive permutation group, so we can apply the results of Chapter II. On the other hand, the latter case implies that G is a t-group, in which case we use the result of Gaschfitz [2]. We recall some definitions before we state the results. Definition 3.3.1. A group G is called a t-group if every sub- normal subgroup of G is normal in 6. Definition 3.3.2. A group G is a Dedekind group if all of its subgroups are normal. A non-abelian Dedekind group is called a Hamiltonian group. A Dedekind group is a t-group, and every nilpotent t-group is a Dedekind group. The Hamiltonian groups have been characterized, and this is the content of Theorem M of the Appendix which we will use in Theorem 3.3.2. The following lemma will make the proof of the other theorems easier. Lemma 3.3.1. If G is a p-group and n(G) = 2 then G is not abelian. Proof: Assume G is abelian. Theorem 3.2.13 says that the exponent of G is less than or equal to pz. 6 is not elementary abelian or cyclic by Theorem A, and therefore 6 contains a subgroup H 1 Case 1. G-= H. H is isomorphic to H isomorphic to (i , 1) . 2 of Lemma 3.2.12 where L = 2, and so we have n(G) 2 3 which is a contradiction. 2 2 Case 2. G>H and G is isomorphic (c,d) where cp =dp =[c,d] =1. Consider the three composition series 36 s : G > (c,dp) > (c) > (cp) > (1), 1 82: c > (c,dp) > (cp,dp) > (cp) > (1), 83: c; > (c,dp) > (cp,dp) > (dp) > (1). s1 ¥ 32 and s1 ¥ 33 since (c) ¥ (cp,dp). 32 ¥A(G)s3 then there is an a E A(G) such that (c,dp)a = (c,dp) for if 82 %A(c)33 and (cp)a = (dp). Therefore [cp]a = dpn where (p,n) = l and ca = chpk where CL,p) = 1. dpn = [cp]a = (ca]P= [chkap = ch, and it follows that (dp) = (cp) and this contradiction implies that n(G) 2 3, which eliminates this case. Case 3. G > H and there is a subgroup K of G isomorphic to 2 (1 , 5). Lemma 3.2.9 implies that n(G) 2 KKK) = 3(i , g) = C2+2_1’2 = C3,2 = 3, and we have reached another contradiction. 80 G is not abelian where G is a p-group and n(G) = 2. Let H = (a,b,c) where ap = bP = cp = [a,b] = [a,c] = l and [b,c] = a. In the next result we will prove that H is one of the two nilpotent groups such that n(G) = 2. Theorem 3.3.2. Let G be a nilpotent group. Then n(G) = 2 if and only if G is isomorphic to H as defined above or a cyclic group of order the product of two distinct primes. Proof: Assume n(G) = 2 and G is nilpotent. Case 1. G has a subgroup which is subnormal but not normal in G. In this case all the chief series are in one orbit of (A(G), CG) and the composition series which are not chief series are in the other orbit. This implies that (A(G), DC) is a transitive permutation group. Applying Theorem 2.1.4 we see that G is isomorphic to H. 37 (G is not isomorphic to a cyclic p-group, an elementary abelian p- group, or the quaternions because all of the composition series of these groups are chief series.) Case 2. All subnormal subgroups of G are normal in G, that is, G is a t-group. Since G is nilpotent G is a Dedekind group. If, in fact, G is aeHamiltonian group then G is isomorphic to the direct product of Q3, A, and B where Q3 is the quaternions, A is an elementary abelian 2-group, and B is an abelian group in which every element is of odd order by Theorem M. n(Q3) = l by Theorem B, so not both A and B are trivial. Let Q3 = (a,b) where a4 = b4 = l and a2 = b2 = [a,b]. Let c E A or B with ‘c‘ = p, p a prime. Three chief series that are in different orbits of (A(G), CG) can be constructed using the facts that (a,b) is the quaternions and (a,c) is not the quaternions, and (a) is cyclic of order 4, and (a2,c) is elementary abelian of order 4 if p = 2 or it is of order 2p if p is not 2. This con- tradicts n(G) = 2 so that no Hamiltonian group has n(G) = 2. If G is an abelian group (Dedekind but not Hamiltonian), then clearly ‘G‘ is not divisible by 3 distinct primes and Lemma 3.3.1 eliminates the possibility that G is a p-group so that IG‘ is divisible by two primes. If ‘G‘ is divisible by p2 then G contains a cyclic;subgroup of order p2 or an elementary abelian subgroup of order p2. Taking another element of G of order q (the other prime dividing |G‘) three non-isomorphic chief series of G may be constructed which contradicts n(G) = 2. Therefore G is a cyclic group of order pq where p and q are distinct primes. 38 To complete the proof of the theorem we show that H and a cyclic group of order pq has two orbits. The cyclic group of order pq clearly has 2 orbits. We now consider H. In Theorem 2.1.4 we have shown that (A(H), DH) is a transitive permutation group. Therefore to show that (A(H), CH) has two orbits we show that A(H) acts transitively on the composition series of H which are not chief series. There are p2+p non-chief series of H which are exhibited below: sx: H > (a,x) > (x) > (1) where x = b,c,bc,...,b c; sx n: H > (a,x) > (axn) > (1) where n = l,...,p-l. Since all elements of H have order p and Z(H) = (a) an automorphism a of H may be constructed by letting ad, ha and c“ be such that [b“,c“] e Z(H), (b“,c“) = a, and a“ = [b“,e“]. .,bp'1 c, sb -A(G)sx by the automorphism For x = c,bc,.. ax defined by a.a [x,c] a«~ [c,b] ax b a x for x f c, ac b a c c a c c a b p-l Now again for x = b,c,bc,...,b c, sx -A(G)Sx,n by the auto- morphism ax n defined by n a” [ax :YJ n a x a ax Y ‘ Y where y is an element of H such that (X,y) = H, and [x,y] = a. 39 Therefore 3b is A(H)-isomorphic to every other non-chief series of H, and this proves that n(H) = 2. If G is assumed to be a solvable group and n(G) = 2, G is either a t-group or not a t-group. These two possibilities lead to the last two theorems of this thesis. Theorem 3.3.3. "If. G His solvable, not a t-group, (A(G), DG) m“. is transitive, and n(G) = 2, then G satisfies: ; +1 ' l. Socn (G)/Socn(G) is an elementary abelian p-group. —L« 2. If R is a characteristic subgroup of G then there is an integer n such that K = Socn(G). +1 3. G(t>/G(t ) is an elementary abelian or cyclic p-group. G 4. There is an integer n such that = Fit(G). 5. There are no characteristic subgroups between 6(6) and Fit(G) so that §(Fit(G)) S §(G). 6. If Z(G) # (1) then Z(Fit(G)) 2 Z(G) 2 Soc(G). 7. If L0 > L1 >...> Ln = (1) is the lower nilpotent series of G, then Li/L is isomorphic to a cyclic p-group, an elementary abelian i+1 p-group, the quaternions, or H where H is the group mentioned before Theorem 3.3.2. Also only one of the factors of the lower nil- potent series is isomorphic to H. Proof: Assume n(G) = 2, and G is a solvable group which is not a t-group. Since (A(G), DC) is a transitive permutation group, we apply Theorem 2.1.9 to prove I to 7. 1,2,4,5 and 6 are the same as l,2,4,5 and 6 in Theorem 2.1.9. Theorem 2.1.9 part 3 says that C(t)/G(t+l) is an abelian p-group, and Theorem 3.1.2 gives us that 2 = n(G) 2 n(G(t))'2 n(G(t)/G(t+1)). n(G(t)/G(t+1)) = 1 by Lernma (t)/G L > H n L. Case 1. HLi< G. Theorem N implies that (l) HL/L is Dedekind, (2) (IHL: L‘, ‘L‘) = 1, and (S) the inner automorphisms of HL in- duce power automorphisms on L with power n in such a way that to 41 each p dividing the lLl there is an h E H«< HL with Lb = L“ for all L 6 L where n é 1 mod p. Theorem P implies that HL is a solvable t-group and HL/L is the maximal nilpotent factor group of HL. But H14 HL implies by the minimality of G that H 2 L which is a contradiction. Case 2. G = HL. H/H n L % HL/L which is nilpotent since G/L is Dedekind. Now G/H n L = HL/H n L = (H/H n L) (L/H n L) so that Theorem Q implies that G/H n L is nilpotent. Therefore L = H n L so that H 2 L, which again is a contradiction. Lemma 3.3.5. C(L) < G. Proof: If C(L) = G then Z(G) 2 L.. Since G/L is nilpotent, then G is nilpotent which is a contradiction. Lemma 3.3.6. If G/L is abelian or the quaternions, then C(L) is abelian and so H s C(L) or H 2 L implies Hid G. Proof: L is a Hall subgroup of G by Theorem N which implies that L is also a Hall subgroup of C(L). Theorem R implies that L has Ill a complement K in C(L). C(L)/L = LK/L K/L n K = x. Since G/L is abelian or the quaternions, and C(L) < G, K g C(L)/L is abelian. Therefore since L is abelian and K s C(L), LK = C(L) is abelian. The remainder of this lemma follows because G is a t-group and C(L) 4 G. If H s C(L) then Hid C(L). C(L)‘< G since 'L‘q G. Therefore qu G. If H 2 L then H14 G since G/L is Dedekind. Lemma 3.3.7. If L is a p-group, then H 4 G implies that H s C(L) or H 2 fl“ Proof: Suppose H 2 L and Hid G, then Lemma 3.3.4 implies that for all h E H such that Lh = Ln for all L E L where n 5 1 mod p. We would like to show that H s C(L). If 1.0 H = H then 42 H s L s C(L). If I.n H1< H then 1.0 H is a Hall subgroup of H so that, by Theorem R, L n H has a complement K in H. Let k E K and L be an element of L of maximal order. Then Lk =.Ln where k|k| n|k| ‘k‘ n a 1 mod p. Therefore L = L = L implies that n 5 1 mod p. Let t be the largest integer such that n a 1 mod pt. If ‘L‘ divides pt then k E C(L). If ‘L| does not divide pt then we know that n = 1+8 pt, and n‘k| = (1 + s ptflk‘ = 1 + “(‘3 pt +LdeZkL-1182P2t + (terms in higher powers of p), also n‘k‘ = l + r‘L‘. Therefore |kls Pt = rILI - (Lkl(gkL'1) sszt +...). Since (|k|,p) = 1, and pt divides lL‘ but pt # ‘L‘, then p divides s which contradicts the maximality of t. This gives us that k E C(L) for all k E K. Clearly 1.0 K.s C(L), so H s C(L) which proves the theorem since all subgroups H such that H 2 L are normal in G. Lemma 3.3.8. If L is an elementary abelian p-group or a cyclic p-group, then G/C(L) is a cyclic group whose order divides p-l. Proof: Case 1. L is a cyclic p-group. Let ‘Ll = pr. Then A(L) is a cyclic group of order pr-1(p-1) [5, page 120]. G/C(L) is isomorphic to a subgroup of A(L) and since (‘G:L|, ‘L‘) = l, G/C(L) is a cyclic group whose order divides p-l. 43 Case 2. L is an elementary abelian p-group. Let ng be defined by .L8-= Ln8 where L E L. (This is well-defined by Theorem N). Let F; be the multiplicative group of the integers modulo p. Let ¢(g) = n8 so that m is a map from G into F2. m is a homo- ngh 8h g)nh “gun morphism since x = x = (x = x which implies ngh a ngnh mod p. The kernel of m is {g E G‘ng 5 1 mod p}. Since L is elementary abelian, the kernel of m also equals {g 6 Gng = L for all L 6 L} = C(L). Therefore G/C(L) is isomorphic to a subgroup of F:. F: is a cyclic group of order p-l, so that again we have the conclusion of the lemma. Combining the results of these Lemmas with the condition that n(G) 8 2 provides the proof of the final theorem of this thesis. Theorem 3.3.9. Let G be a solvable, non-nilpotent, t-group where G/L is the maximal nilpotent factor group of G. n(G) = 2 if and only if G has one of the following three sets of properties: 1. G/L is a cyclic group of order qm, m 2 2, ‘C(L)‘ = pq and |L| = p. 2. G/L is a cyclic group of order qr, L is a cyclic p-group, and C(L) = L. 3. G/L is a cyclic group of order qr, L is an elementary abelian p-group, and C(L) = L. In all three cases p, q, and r are distinct primes. Proof: Assume n(G) = 2. Theorem N says that G/L is Dedekind and L is abelian. By Theorem 3.1.2, 2 = n(G) 2 n(G/L) so that n(G/L) = 1 or 2. Theorem B, Theorem 3.3.2, and the fact that G/L is Dedekind implies that G/L is isomorphic to a cyclic q-group, an elementary abelian q-group, the quaternions, or a cyclic group of 44 order the product of two primes q and r. (G/L ¥ H since H is not Dedekind). Again Theorem 3.1.2 implies that 2 = n(G) 2 n(L). L abelian together with Theorem 3.3.1 gives that n(L) = 1 so that L is isomorphic to a cyclic p-group or an elementary abelian p-group by Theorem A. Because of Lemma 3.3.8 G/C(L) is cyclic so that in the two cases where G/L is an elementary abelian q-group or the quaternions, \G/C(L)| = q or 2. Under these same conditions Lemma 3.3.6 implies that C(L) is abelian. By TheoreuiN there is also a non-abelian normal subgroup of G containing L of the same order as C(L). This accounts for at least two orbits. Since ‘C(L)| is divisible by two primes, and C(L) is abelian, clearly at least one more orbit exists which is a contradiction. Therefore G/L cannot be an elementary abelian q-group or the quaternions. In the remaining possibilities G/L is cyclic so that C(L) may be L. In the case where G/L is a cyclic q-group and L is a cyclic p-group, if C(L) = L, Lemma 3.3.7 implies that G has only one chief series which contradicts n(G) = 2. If G/L is a cyclic q-group, L is an elemen- tary abelian p-group, and C(L) = L, then Lemma 3.3.7 implies that all chief series contain L, so that if s 6 CG then they can be 1’82 written in the form 81: G = (g,L) > (gq,L) >...> (L) > (a1,...,am_1) >...> (a1) > (1) )>...> (b1) > (1) s2: G = (g,L) > (gq,L) >...> (L) > (b1,...,bm1 = = = g = n where (gL) G/L and L (a1,...,aup (b1,...,bu), and a1 a1 and bi = bn by Theorem N. (Note that n does not depend on which 1 element of L is conjugated by g). Let a be defined by a: = b1 and ga = g. a E A(G) since it preserves the relations of G. -Since 45 8‘1 a cyclic q-group and C(L) > L, then C(L) is divisible by both p and q. If C(L) is divisible by p2 or q2 then C(L) contains = s2, n(G) = l and this eliminates this possibility. If G/L is at least three non-isomorphic chief series, and since G is a t-group and C(L)‘4 G ‘theycan be extended to non:isomorphic chief series of G which implies that |C(L)| = pq. This group where G/L is a cyclic q-group, ‘C(L)‘ = pq, and ‘L‘ = p is the group mentioned in 1. In the case of a group as in 1, C(L) is cyclic so by Lemma 3.3.7 G has exactly two non-isomorphic chief series: one containing C(L) and L, and the other containing C(L) and the q-Sylow subgroup of C(L). This proves that such a group does have n(G) = 2. The only possibilities left are where G/L is a cyclic group of order qr where q and r are primes. If C(L) > L, then there are at least three non-isomorphic chief series in G in the same way as in the case when G/L was elementary abelian q-group or the quaternions, so that C(L) = L. . With C(L) = L1 and G/L cyclic of order qr we have the two groups of 2 and 3. If L is a cyclic p—group as in 2, then clearly G has n(G) = 2 when ‘G/L‘ = qr, L is an elementary abelian p-group, and C(L) = L. There are, by Lemma 3.3.7, two kinds of chief series: those containing the normal subgroup of index r and those contain- ing the normal subgroup of index q. In either case, all chief series of that kind can be shown to be in the same orbit in the same way as we proved that if G/L is a cyclic q-group, L is an elementary abelian p-group, and C(L) = L, then n(G) = 1. This completes the proof of the theorem. BIBLIOGRAPHY BIBLIOGRAPHY Burnside, W., Theory_of Groups of Finite Order (New York: Dover Publications, Inc., 1955). Gaschfitz, W., "Gruppen, in denen das Normalteilersein transitiv ist," Journal ffir die Reine und Angewandte Mathematik, 198 (1957), 87-92. Hall, M., The Theory of Groups (New York: Macmillian, 1959). Polimeni, A., A Study of the Automorphisms and Chains of a Finite Group, Thesis for the Degree of Ph.D., Michigan State University, 1965. Scott, W.R., Group Theory (Englewood Cliffs, New Jersey: Prentice Hall, 1964). Taunt, D.R., "0n A-groups," Proceedings of the Cambridge Philo- sophical Society, 45 (1949), 24-42. 46 APPENDIX APPENDIX Theorem A [4, page 11]. Let G be an abelian group. Then -(A(G), CG): is transitive if and only if G is an elementary abelian p-group or a cyclic p-group, for some prime p. Theorem B [4, page 14]. Let G be a nilpotent group. Then [(A(G), CG) is transitive if and only if G is one of the following: 1. G is an elementary abelian p-group for some prime p, 2. G is a cyclic p-group for some prime p, 3. G is isomorphic to the quaternion group of order 8. Theorem C [4, page 17]. If G is a supersolvable, non-n11- potent group and A(G), CG) is transitive then G = Sp-Sq, p and q primes, where SD is a p-Sylow subgroup of G and Sq is a q-Sylow subgroup of G. Furthermore if q < p it follows that 1. S1) = G', 2. Every subgroup of Sp is normal in G, 3. Sp is either cyclic or elementary abelian, 4. Sq is elementary abelian or cyclic, 5. Z(G) = (1). Theorem D [4, page 26]. If (A(G), CG) is transitive and 2 does not divide ‘G‘ then the following are true: 1. The lower nilpotent series of G coincides with the derived series of G, (k) of the derived series for G is complemented (k-l) 2. Each term G in G by the relative system normalizer of G in G, 47 48 3. There exists r subgroups H1,H2,...,Hr, r the length of the derived series for G, such that a. Each H1 is either a cyclic pi-group or an elementary abelian pi-group for some prime pi, b. G(k) .H for k = 0,1,...,r-l, 1" k+1 c. HiHj ' HjHi for all 1,] and H1 0 H - HrHr_ j = (1) if i # j, 4- Z(G) ‘ (1). 5. 9(G) ' (1) if r > 2, 6. G(k)/G(k+1) is an elementary abelian group for k = 2,3,...,r-l. Theorem E [5, page 167]. If G is a finite group then Fit(G) 2 §(G). Theorem F [5, page 170]. If G is a finite solvable group then Fit(G)/6(a) = Soc(G/§(G))- Theorem G [5, page 157]. If G is a supersolvable group then G' is nilpotent. Theorem H [3, page 159]. A supersolvable group G has a normal series G = C > C >...> Ck = (l) in which every C 0 l infinite cyclic or cyclic of prime order, and if C 1_1/Ci is either i__1/C1 and Ci/C1+1 are of prime orders pi and p1+1 we have p1 s p1+1. Theorem I. A finite supersolvable group G has a normal p- Sylow subgroup if p is the largest prime divisor of ‘G‘. Proof: .The proof follows from Theorem H. Theorem J [6]. If G is an A-group then G' n Z(G) 3 (1). Theorem K [1, page 132]. If a group G, of order ~2m, has a single subgroup of order 28 (s > 1), it must be cyclic; if it has a single subgroup of order 2, it is cyclic or isomorphic to Quf 49 Theorem L [1, page 131]. If G, of order pm, where p is an odd prime, contains only one subgroup of order ps, 3 = l,...,m-l, then G is cyclic. Theorem M [3, page 190]. A Hamiltonian group is the direct product of a quaternion group with an abelian group in which every element is of finite odd order and an elementary abelian 2-group. Theorem N [2, page 88]. Let G be a solvable t-group, G/L be the maximal nilpotent factor group of G. Then 1. G/L is Dedekind, 2. |L\ is odd, 3. L is abelian, b (\G/Ll, |L|) = 1, 5. The inner automorphisms of G induce power automorphisms on L with power n, (n, ‘L‘) = l, in such a way that to each prime divisor p of ‘L‘ there exists an n such that n i 1 mod p. Theorem P [2, page 89]. If G is a group, L a normal sub- group of G, and 1 and 4 of Theorem N hold along with (5*) the inner automorphisms of G induce power automorphisms on L, then G is a solvable t-group. If (5) of Theorem M replaces (5*), then G/L is the maximal nilpotent factor group of G. TheoremQ [5, page 166]. If A and B are normal nilpotent subgroups of a group G, then AB ’is also a normal, nilpotent sub- group of G. Theorem R [5, page 224]. If H is a normal Hall subgroup of a finite group G, then H has a complement. IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII \IHlllllllHlIIIWIWIN"010!le\Ifl‘mlfllfllflflH 3 1293 03