A Decomposition Theory for Finite Groups with Applications to P-Groups Citation Weichsel, Paul Morris (1960) A Decomposition Theory for Finite Groups with Applications to P-Groups. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/FCJW-5875. https://resolver.caltech.edu/CaltechETD:etd-07072006-085918 Abstract NOTE: Text or symbols not renderable in plain ASCII are indicated by [...]. Abstract is included in .pdf document. Let [...] be a set of finite groups and define [...] to be the intersection of all sets of groups which contain [...] and are closed under the operations of subgroup, factor group and direct product. The equivalence relation defined by [...] if [...] = [...] is studied and it is shown that if Qn and Dn are the generalized quaternion group of order 2n and the dihedral group of order 2n then [...] = [...]. A group G is called decomposable if [...] with [...] the set of proper subgroups and factor groups of G. It is shown that if G is decomposable then G must contain a proper subgroup or factor group whose class is the same as the class of G and one whose derived length is the same as the derived length of G. The set of indecomposable p-groups of class two are characterized and for [...] their defining relations are compiled. It is also shown that if the exponent of G is p and the class of G is greater than two then G is decomposable if G/Z(G) is a direct product. Finally the equivalence relation given above is modified and its connection with the isoclinism relation of P. Hall is investigated. It is shown that for a certain class of p-groups this relation is equivalent to isoclinism Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics) Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Dean, Richard A. Thesis Committee: Unknown, Unknown Defense Date: 1 January 1960 Record Number: CaltechETD:etd-07072006-085918 Persistent URL: https://resolver.caltech.edu/CaltechETD:etd-07072006-085918 DOI: 10.7907/FCJW-5875 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 2821 Collection: CaltechTHESIS Deposited By: Imported from ETD-db Deposited On: 31 Jul 2006 Last Modified: 19 Aug 2025 18:15 Thesis Files Preview PDF (Weichsel_pm_1960.pdf) - Final Version See Usage Policy. 2MB Repository Staff Only: item control page

A Decomposition Theory for Finite Groups with Applications to p-Groups Thesis by Paul Morris Weichsel In Partial Fulfillment of the Requirements For the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1960 ACKNOW LEDGEMENTS The author wishes to express his thanks and appreciation to Professor Richard A. Dean who acted as adviser and friend during the preparation and writing of this thesis. His many helpful suggestions and patience were of invaluable assistance. Many thanks are also due to Miss Evangeline Gibson for her expert typing of the manuscript. ABSTRACT Let {Get be a set of finite groups and define {Ge} to be the intersection of all sets of groups which contain {Ga} and are closed under the operations of subgroup, factor group and direct product. The equivalence relation defined by G,=G, if (G} = {G3} is studied and it is shown that if Q, and Dd. are the generalized qua- ternion group of order 2" and the dihedral group of order 2" then OQi= Oh. A group G is called decomposable if Ge {Aa} with tAat the set of proper subgroups and factor groups of G. It is shown that if G is decomposable then G must contain a proper subgroup or factor group whose class is the same as the class of G and one whose derived length is the same as the derived length of G. The set of inde- composable p-groups of class two are characterized and for p # 2 their defining relations are compiled. It is also shown that if the exponent of G is p andthe class of G is greater than two then G is decomposable if G/Z(G) is a direct product. Finally the equivalence relation given above is modified and its connection with the isoclinism relation of P. Hall is investigated. It is shown that for a certain class of p-groups this relation is equivalent to isoclinism. INTRODUCTION This thesis introduces a notion of decomposition for finite groups which includes direct product and subdirect product as special cases. First a closure operator on sets of finite groups is introduced which embeds a given set of groups {Gy} into its "closure", denoted by {Ge} , which is the smallest set of groups containing {Ge} and closed under the operations of direct product, subgroup and factor group. A natural question which arises in studying this operator is: What are necessary and sufficient conditions for {Gi} = {G5} ? This question seems quite hard and is not answered in this thesis. One interesting result that is obtained along these lines is that the closure of the generalized quaternion group is equal to the closure of the dihedral group of the same order. A group G is then defined to be decomposable if Ge fAu} where {Aa} is the set of all proper subgroups and factor groups of G. In classifying the various ways in which a group G may be decom- posable a mode of decomposition which we call factordirect product and which is in many ways the dual of subdirect decomposition is intro- duced. It is also shown that simple groups are indecomposable in the sense defined above. In an attempt to obtain necessary conditions for a group G to be decomposable it is shown that if Ge {Ax} then G/Z(G) « {Aae/Z(Aw)jt andif Ge fAa} then G'ec {Au}. This provides important information concerning the kinds of subgroups and factor groups that a decomposable group must have. It tells for example that a decomposable p-group of class n must have a subgroup or factor group of class n. The remaining chapters are an attempt to apply the ideas expressed above to finite p-groups. Chapter II gives several characterizations of the set of indecom- posable p-groups of class two. And in the case of p #2 all the possible defining relations for such groups are enumerated. This enumeration allows one to assert that given any finite p-group G of class two, with p #2 there exists a finite set of indecomposable groups {Ga} whose defining relations are given, such that G may be constructed in a finite number of steps from {Gal by the appli- cation of the operations of subgroup, factor group and direct product. In Chapter III an attempt is made to gain information about the decomposability of a p-group G from the knowledge that G/Z(G) is decomposable. If the class of G is greater than two and if the ex- ponent of G is p itis shownthat G is decomposable if G/Z(G) is a direct product. An example is given which shows that this decom- position is in general non-trivial. Thatis, G is neither a direct, sub- direct or factordirect product. The concluding chapter studies the connection between the closure operator and a group relation know as "isoclinism" which was introduced by P. Hallin 1940. It is easy to show that both of these ideas provide a partition of the set of finite groups into equivalence classes which share many numerical invariants. The connection between closure and isoclinism is made precise by introducing a restricted form of closure. It is then shown that for a certain class of p-groups these two concepts are equivalent. Chapter I Any algebraic decomposition theory specifies a certain set of algebras which may be called the "basis set!' and a set of operations defined on the basis set. The basis theorem then states that any algebra of the type being considered may be obtained from the basis set by some applications of the given operations. In addition if a given algebra A is to be called decomposable then the basis elements required for its construction should be contained, in some sense, in A. In the theory of finite abelian groups such a decomposition theory does indeed exist. The basis set is the set of all cyclic p-groups and the operation is that of direct product. In this case a group G may be constructed as a direct product of a certain subset of cyclic p-subgroups. There exists a decomposition theory for all finite groups in the sense that every finite group may be obtained in a finite number of steps by applying the operation of group extension to a set of simple groups. That is, for an arbitrary finite group G there exists a finite set of simple groups, {Aa}, the factors of a composition series of G, such that for some ordering of the {Aa}, a sequence {Bj may be defined as follows: B, =A,, B, is an extension of A, by Bey for i=l, ..., n and G = B,- One way of measuring the effectiveness of a given theory is its degree of 'balance''. A decomposition theory ought to divide the diffi- culty of understanding a class of algebras between the study of a fairly restricted set of them and the study of the effect of certain operations on the set. If the basis set is extremely easy to study but the operations very complex then it may be said that the theory is unbalanced. In such a case the basis theorem that results will most likely not be very informative. Exactly sucha state of affairs obtains if we restrict our attention to the set of finite p-groups for a particular prime p, and choose group extension as the operation. Here the basis set will be the set of all simple p-groups. But the only finite p-group which is simple is the group of order p. Hence all of the difficulty lies in understanding the operation of extending a p-group by a group of order p. On the other hand if we let the operation be direct product then the basis set becomes those finite p-groups which cannot be represented asa direct product. In this case little can be said concerning the basis set as such. A decomposition theory for finite groups will here be developed which lies somewhere in between the two examples given above. That is, the basis set will contain all simple groups and will be contained in the set of all groups which are not direct products. From this point on the word ''group'' will be used to mean ''finite group". Since it will be convenient to refer interchangeably to the notions of subgroup, factor group and factor group of a subgroup we define: DEFINITION 1.1 A group A is said to be an ingroup of a group G if one of the following holds: a) A is isomorphic to a subgroup of G. b) A is isomorphic to a factor group of G. c) A is isomorphic to a factor group of a subgroup of G. 1 The notation A <G will be used for the above with A<G denoting that A is an ingroup of G but A is not isomorphic with G. In this case A is said to be a proper ingroup of G. LEMMA 1.1 The relation A gG is reflexive and transitive. Proof G<G since if G=A_ then condition a) of Definition 1.1 is satisfied. Hence the relation is reflexive. To demonstrate transitivity it must be shown that ACB<C implies A<C. Denote the relation between A and B by I, the relation between B and C by II and that between A and C by III. It must now be shown that for all choices of I and Il, III is one of the relations given in the definition. The following tabulation gives the results required. I I Tir 1) a a a 2) a b c 3) a c c 4) b a c 5) b b b 6) b c c 7) Cc a c 8) c b c 9) Cc c c l. For the convenience of the reader all symbols used in this thesis are collected ina glossary on page 62. Note first that 1), 5), 6), and 7) follow from the fact that the relation "subgroup" and ''factor group" are transitive relations. 4) is precisely the statement of alternative c) in Definition 1.1. Now consider 2). Here B isa factor group of C and A isa subgroup of B. Hence B=C/N and ACB. The inverse image of A in ©, callit A is a subgroup of C which contains N asa L? normal subgroup. Hence A =A,/N and A, CC. Therefore A isa factor group of a subgroup of C and hence Ill =c. Now 3) follows since A isa subgroup of a factor group of C. Hence A isa factor group of a subgroup of a subgroup of C and therefore alternative c) holds. Since it has been shown in the proof of 2) above that a subgroup of a factor group is a factor group of a subgroup 8) and 9) follow from the transitivity of the relations: subgroup and factor group. We now define the notion of closure. DEFINITION 1.2 Let {Aa} bea set of groups. A group G is said to be of rank 0 if G<A for some Ae {A,}. G is of rank 1 if G is not of rank 0 and G<AXB with A and B of rank 0. In general G is of rank n if G has not been assigned rank n-k for k>0O and G<CxXD with C and D both of rank less than n. The set of groups with assigned rank is called the closure of {As} and is denoted by Ay. ° The closure of a set of groups may also be described ina some- 2. Graham Higman, ina recent paper [1] has studied sets of groups with the property of being closed in the sense defined above. what different manner. LEMMA 1.2 If {Aa} isa set of groups then {fAs} is closed under the operations of ingroup, factor group, and direct product. Further- more, every set of groups which contains {Aa} and is closed under the operations given above contains {Ag} . Proof If Ge fAuJ then G<CXD with C, De {Ay}. Let G >H. Since the relation "<"' is transitive it follows that H <CXD. Hence He {Ay}. Suppose that E, Fe {As}. Then EXF<EXF and therefore is contained in {Ay}. Now suppose that AL isa set of groups closed under the oper- ations of ingroup and direct product and AL 2 {A} » To show that fA.} we) consider Ge fA,} of rank 0. In this case G CA«c {Aa} cL and since AJ is closed under the operation of ingroup then G c AL. Suppose that all elements of fAz} of rank less than k are contained indL. Let Ge {Au} be of rank k. Therefore G <C XD where C, D are of rank less than k in {An} - Hence C, D «AI and since a is closed under direct product and ingroup G cA. Therefore {Au} ad. The closure of a set of groups may therefore be thought of as the intersection of all sets containing the original set of groups and closed under the operations of ingroup and direct product. If closure, as defined above, is regarded as a mapping defined on the set of all subsets of finite groups then it is a closure operator in the usual sense. That is: LEMMA 1.3 If {Aa} and {Bg} are sets of groups then: a) {A} ¢ {Aa}. b) fA} = {Ky} and c) if {[A.} ¢ {B,} then {Al} c {By . Proof a) If Ge {A.} then G is of rank 0 in {Ay} and hence [A.g < {AL}. b) Clearly {A,} Cc {A,} since all elements of {Al} are of rank 0 in {A,}. Let Ge {Aa} be of rank 0. Then G <Be {Aa}. But {A.} is closed under ingroups and therefore Ge {Au} . Assume that every element of {A.} whose rank is less than some integer k is contained in fa}. Let G be of rank k in {Ac}. Then G<EXF with E, Fe {Az} of rank less than k. Therefore E, Fe fAa} and since {A,} is closed under direct products and ingroups G « {A..} . Therefore fA} c {Az} and hence {Ax} = {Aj}. c) Since {Bal is closed under the operations of ingroup and direct product, and since {Bg} > {Be} > {Au}, it follows from Lemma 1.2 that {Bg} > {Aa} A natural question which arises from the definition of closure is: Can one give necessary and sufficient conditions for (Git = {G5}, G, and G5 arbitrary groups? The answer to this question is, in general, not easy. For certain classes of groups, however, a straight- forward answer may be given. It can easily be proved that: If G, and G, are abelian groups then {Gy} = {G5} if and only if the exponent of G, is equal to the exponent of G,, written e(G,) = e(G,). One interesting fact that arises in this connection is given by: THEOREM 1.1 Let Q. be the generalized quaternion group of order 2", and let Dd. be the dihedral group of order 2” with n >3. Then ©) = G}. n-l n-2 Proof Q. = <a, b> with a’ = 1, be = aM and bab! = al, an-lo2 -l_ ol Dd. = <g,h> with g =h =1, hgh =g -. In both of these cases n>3. The theorem will be proved by showing that Do {Q}3 and € Q. € {D_}. It will then follow from Lemma 1.3 that {D3 n = {Q_}- In order to show that Dd. € {Q} it suffices to form the direct product of Q. with a cyclic group of order 4 and then consider an appropriate factor group of a subgroup of this direct product. Since Qn contains a cyclic group of order 4 asa subgroup, the group which results will be contained in {Q/}. It will then be shown that this group is isomorphic to Do A similar procedure will be adopted to show that Q, € {D} : Let H=<c> with ciel. Let G= Q XH and let N =((b°, c“]). Clearly NdG since pb” « Z(Q.). Let G, = G/N =<[a, 1] N, [b, ] N, [1,c] ND. Now consider a subgroup of G,2D =a, JN, [b,g NS and we will now show that D= D.: It is clear that (fa, l ny2n-i = ( [b,c] Nn)? = 1. Also [b,c] N fa, ] N [b,d Ne [bab , } N= fal, | N= (fa, Y n)7t, Hence the generators of D satisfy the same relations as to the gener- ators of Do If it can be shown that the order of D equals the order of Do then the isomorphism will follow. 3. Zassenhaus [2] p. 147. 10 Since (Q, x H) =4- o(D_) and o(N) = 2 it must be shown that [G,:D] = 2. We claim that G, =D+ (Lb, c“]N)D. Since any element of QQ. may be written as a wordin a and b, it can be put in the form be a. This follows from the fact that ab = bat. Hence a general element of G) has the form ay a. ; cYJN. Now it will be shown that Ib a ’ cY]N is either in D or in [b, c“JND. Since D=<[a, 1] N, [b,c] N>, [b, <]’ fa,l] N= [oe a yc] Ne D. We must now consider the four cases: a) y = 6(4), b) y = B+ 1(4), c) y= B+ 2(4), and d) ys 6 + 3(4). In case a) fp a, c’JN = [bo av, c*] Ne D. Incase b) notice that -1 1 ,c ]NeD and so [b- ,c YN [b® a” cP] = [bo ta® cP Ne D. [b But [b, c“IN[b Po ta™, cf IN = [b® a, of tly = [pe av, c JN « 2 -2n-¢ 2 2 [b;c“JND. Incase c) notice that [a, 1] N[b,c] “N =[l,c JNeD. Hence Ib” a. sch] Nf, “JN = [pe a ic Pt27y, = re a. ; c’]Ne D. Finally for case d) [b,c] N (1,c°]N = ipl, c]NeD. Hence [b,c7]N[b, cJN[bY a® 0°] N = [bP a® cP *97N = [oP a® YIN, Therefore o(D) = o(D_) and DD... Let H be defined as above and consider R = Dx H. Clearly n-2 Re {DI . Let R2S=<[g,1},[h,c]>. Let M =<[g" sc]. 2 2 gn-e oy Since [h,c]~ = [1,c"] clearly MCS. And since hg h _2n-4 n-2 =g =g it follows that M4 S. Let Q=S/M and we will now show that Q =Q,: As in the previous case the generators of Q satisfy the same n-2 relations as the corresponding generators of Q.: For (fg, UJ M)* n-2 2 ,YM= fl,c“7M = ([h, c] M)*. Also [h,c] M[g, 1] M({h, c]M)) =[g 1] =(hen'l, 1]M = (el, 1] M = ({g, J m)7). Since o(D, X H) =4 (Q.) and o(M) = 2, inorder to show that o(Q) = o0(Q,) we must show that the index of S in R is equalto 2. We claim that R=S+[1l,c]S. For anarbitrary element of R is of the form cn? o ,c¥). Clearly the element fn? oc] e S, and since f1,c7] « S it follows that for y = B(4) and y =f + 2(4) rn! eg ,c’JeS. If y=1+@(4) or ys 3 +6 (4), the 0% oY] [l,c]S. Therefore QQ. Hence we have shown that Q¢ {D} and D_ « {Q} and therefore {D,} = (Qh. The closure of a set of groups can be regarded in a somewhat different light, namely as defining a rather special group-theoretic property. Any group property is determined by some set and con- versely. The most familiar kinds of group properties have what can be called "inheritance". A property is said to be subgroup inherited if whenever a group has this property then so do all of its subgroups. In a similar way one defines factor group and direct product inheritance. It can be said that a property is considered important and interesting to study if it does have many inheritance characteristics. Commuta- tivity, nilpotence and solubility are examples. Therefore the closure of a set of groups is that set which defines the weakest property, which is subgroup, factor group and direct product inherited, that is shared by all the elements of the set. Weakest merely means that every other such property which is satis- fied by all of the elements of the set is defined by a set of groups which contains the closure of the original set. Theorem 1.1 can thus be restated. 12 Every property satisfied by Qn which is subgroup, factor group and direct product inherited is also satisfied by D, and conversely. Having now studied some properties of those operations which will be used to construct all finite groups from a given basis set we are now ready to define the notion of decomposability. DEFINITION 1.3. Let G bea group andlet {Aa} be the set of all proper ingroups of G. G is called decomposable if G« {Aa} . Ge {Ae} implies that there exist A, Be {Aa} such that G<AXB. This can happen in four ways. I. First suppose that G=AXB. Since A and B are sub- groups of G and hence are contained in {Ae} it follows that G is of rank lin {fAcl. Il. Let GC AXB. Since A and B are finite groups itcertainly follows that A and B may be chosen so that for any choice of Ay and Bi proper subgroups of A and B respectively, G EA, xB and G¢AXB,. With such a choice of A and B71, G is said to be a subdirect product of A and B. Subdirect products may be characterized in the following way: A group G is a subdirect product of groups A and B if and only if there exist non-trivial subgroups N,, N,< G such that N,NN,=1 and G/N,;% A and G/N, xB. The embedding of G into A XB is given by g —> [gN,, gN,] « G/N, XG/N,= AXB 13 for all ge G. Thus if G is asubdirect product of A and B then since A and B are factor groups of G, A and B are contained in {Aa} and hence G is of rank 1 in {Aa}. Ill. Let G=(AX B)/N. In light of the definition of subdirect product it is natural to ask: Do there exist subgroups of G which are isomorphic with A and B? THEOREM 1.2 Let G=(AXB)/N. Then there exist subgroups A 1 and B, of G suchthat 1) G#(A, xB, )/N, for some N 1 |? 2) A, and B, are isomorphic to factor groups of A and B respectively and 3) G=A 2B>, with A,=A,, B,* B, and A> and Bs permute elementwise. Proof For the sake of simplicity we will write AX B as AB, and ab in place of [a,b]. Let M=(ANN)U(BNAN). Since MIG we may write G = AB/N= AB/M/N/M = AM/M ° BM/M/N/M. This last equality holds because every element of AB/M is of the form abM = aMbM « AM/M: BM/M. Similarly a typical element of AM/M*BM/M is am,Mbm,M =aMbM = abM. We will now show that AM/M-BM/M is a direct product. Suppose AM/MN BM/M2M/M. Then there exists a, «A and b, ¢€ B such that a,Mz=b,M. Hence a, = b,m=b,ab and a, =a. 1 1 1 1 1 1 i But since AN Bez=l1, M is the direct product of ANN and BONN and so abe M implies that a, be N. Hence ae M and a,M = M. 14 Clearly AM/M, BM/M<AB/M and so AM/M:BM/M =AM/MXBM/M. Let A, = AM/M and B, = BM/M. Hence G = (A, XB,)/N and 1 Ay: By are isomorphic to factor groups of A and B respectively. Further A,MN, = B/N N,=1. Let A, = ALNL/N, and B, = B,N,/N,. Then G=A,B,, A,“ A,, B,&=B, and A, and B 2° 2’ i 2’ 1 2 2 permute elementwise. For Ay Nn By = 1 and hence permute element- wise. This completes the proof of the theorem. In analogy with the notion of a subdirect product this theorem leads to the definition of factordirect product. DEFINITION 1.4 A group G is called a factordirect product of A and B if for some N, G=(AXB)/N and 1#A, BCG. Thus Theorem 1.2 shows that if G=(AXB)/N then G isa factordirect product of two groups which are factor groups of A and B respectively. It will now be shown that condition 3) of Theorem 1.2 implies that G is a factordirect product. THEOREM 1.3 If G=AB and ab=ba forall ae A andall be B then G is a factordirect product of A and B. Proof Let H=AX B= {fa,b] ac A, be B}. Let N= {fcc J for all ce AN B considered as subgroups of Gh - Since A and B permute elementwise in G it follows that AN Be Z(G). Therefore NGA XB. Consider the following mapping between G and AX B/N. Goag=a,b, <-> [a,b] N. 15 a _ -l 1 To show that it is one-one let a,b, = ajb,. Then bib, =a, a. Since A and B permute elementwise and ajta, e B, aj a,b, = ~1 . . - - boa, as = b,- This implies that aja, = b5 by. Now consider the images of a,b, and a,b, in AX B/N. [a,,b,JN = f[a,,b,] N falta bi lp JN since favta poly J N=N. 1’ "1 1?) 1 "2? "1 “2 1 "2? "1 72 Therefore fa,» bJ N = fa,.bo] N. Suppose that [a,>>y] N= [a,,b5] N. Then there exists cé ANB such that [a,b] = [a5,b,] fe,c4J. Hence a, =a,c and b, =b,c . Th sta, =b) by. But this implies that aj!a,=b,b,! 1 =>2,¢ + Thus a, a, =b, bo. ut this implies that a, a, = b,b, and hence a,b, =a,b,. To show that itis a homomorphism suppose that a,b) Se [a,> by] N and a,b, Se [a,b] N. Then a,b,a,b, =a,a,b,b, and similarly fa,.b]J N [a,, bo] N = [a,2>, bj) bo] N. Hence the mapping is an isomorphism, and G2AX B/N. Since A, BCG, G isa factordirect product of A and B. It is interesting to note that subdirect and factordirect products are dualin many respects. One illustration of this duality is the fact that the relation of "subdirect product" is transitive in some sense. And a similar result may also be stated for factordirect products. First consider the subdirect case. LEMMA 1.4 Let G bea subdirect product of A and B. If HCG then either H is a subdirect product or else H is isomorphic toa 16 subgroup of A or B. Proof Clearly HCAXB. If HNA#1 and HNB#1 then H isa subdirect product, i.e. H¢(H/HNA)X (H/HNB). Suppose HNA=1. Let B, = {b, [fab] « H} and consider the mapping between H and By: H [a,b] <>» be B,. To show that this is one-one suppose that [a,b] = [a,»b5] in H. Then b, = b,. If b, = by then consider [a,,b,] and [a,b]. Clearly [a,,b,] [ayleey] = [ajay 1] « H. But HNA=1. Hence aya) = 1 and therefore [a,>b,] = [a,.b,] . Hence the mapping is one-one. Since multiplication in H is carried out componentwise the mapping is clearly a homomorphism and under the assumption that HNA=1, H>B, CB. If HNB=1 then the argument above, applied to a subgroup of A shows that H =A, CA. This completes the proof. It is natural to ask whether a factor group of a subdirect product is always either a subdirect product or isomorphic toa factor group of one of the factors. The answer to this question is in the negative as may be demonstrated by the quaternion group of order 8. For it has been shown in Theorem 1.1 that the quaternion group is a factor group of a subdirect product of the dihedral group of order 8 and the cyclic group of order 4. Clearly the quaternion group is neither a factor group of either of these nor is ita subdirect product. 1? The dual result to Lemma 1.4 for factordirect products is as follows: LEMMA 1-5 Let G bea factordirect product of A and B. If _H= G/N then either H isa factordirect product or else H is isomorphic toa factor group of A or B. with ab = ba for Proof Since G isa factordirect product G = A,B, all ace A, and be B). Let NAG. Then G/N = A,N/N-B,N/N. But A,N/N ~AL/A, AN and B,N/N* B,/B, ON, and these groups permute elementwise. If neither A,A N or B, N N=1 then H is a factordirect product. If AJM N=A, then HY B,/B Nn N, and similarly for A, ON. Analogous to the subdirect product situation a subgroup of a factordirect product need not be a subgroup of one of the factors ora factordirect product itself. In the proof of Theorem 1.1 the dihedral group of order 8 is constructed as a subgroup of a factordirect product of the quaternion group and the cyclic group of order 4. Clearly itis neither a factordirect product nor is it isomorphic to a subgroup of one of the factors. IV. The final case in the enumeration of G<AXB is G=H/N with HOAXB. A very important distinction sets this case apart from the others. For up until now it has been the case that if G<AXB then G<A,x B with Ay» By proper ingroups of G. 1’ No such conclusion will be possible here as is illustrated by Theorem 1.1. In other words, let {Hg} bea set of groups. Thenif Ge {Ha} by virtue of being a non-trivial direct product, subdirect product or 18 factordirect product then Ge {Aw} where {Aw} is the set of proper ingroups of G. In any event if G=H/N with HO AXB thena special choice of the groups A, B, H, and N may be made. LEMMA 1.6 Let G = H,/N, with H, a subdirect product of A,, By, 7 1. Then there exist factor groups A, B, H and N of A, B, H,, and N, respectively such that G*H/N, H¢&AXB and there exist R, $<@H such that: RNS=ROAN=SOAN=I1. Proof Since Ay is a subdirect product it follows that there exist S,73H subgroups R such that RA Sy = 1. Suppose that 1?’ 1 (R,AN,)U(S;AN)=M21. Let N= N,/M, H=H,/M, R=R,M/M and S= S,M/M. Clearly RMN=1 since if rmM=nm'M with re R)> m, m'te M and ne WN, then r=nm''=nr's'. This follows from the definition of M. But r'e Ry nN Ny CM and s'e« Si N, cM. Hence re N, and therefore re Ry ON ¢M, which implies that rmM=M. Ina similar manner it can be shown that SNN=1. Since Ry N Ss) = 1 itis clear that RNS =1 and since R,|M/M =R,/RN M the lemma is proved. Since the case of GCAXB already has been considered we will assume now that G ~H/N is nota subdirect product. THEOREM 1.4 Let G, H, A, B, R, S be defined as in the lemma above. Assume in addition that G is not a subdirect product and that G is not isomorphic toa factor group of A or B. Then G contains 19 a normal abelian p-group. Proof Consider the element S/)NR. Since all of these subgroups are normal RUN may be writtenas RN or NR. Assume first that NROS=1. Since the lattice of normal sub- groups of a group is modular it follows that NU(NR/AS) =NROANS=N. Now if NR #N and NS #N then NR #NS and hence H/N isa sub- direct product. If NR =N then G=H/N2#H/R/WN/R and hence G is isomorphic toa factor group of A. If NS=N then G=H/N © H/R/N/R and G is isomorphic to a factor group of B. Hence if NROMS=1 then G is either a subdirect product or G is isomorphic toa factor group of A. But itis easy to see that NRMS=1 if and only if NSMR = 1 which holds if and only if RS N= l. For if RSNN #1 then there exist r, s, and n suchthat rs=n #1, and hence s=r /n and r=ns !. Note that r #1 and s #1 for other- wise RNN#1 and SN N#1 contrary to Lemma 1.6. Hence if Y=NSOR and Z=RSMN then Y7#1 and Z# 1. The lattice diagram in Figure | below illustrates the relation- ships of the groups that have been discussed above. All of the argu- ment above and some of the argument that follows will verify that the unions and intersections given in the diagram are indeed correct. 20 X=NRO|S Y=NSNR Z=RSMN _FIGURE 1 XNY=(NRAS)N(NSOR)CROAS=1. Applying the same argu- mentto XMZ and YNZ, (i) XNY=XNZ=YNZe=1. By the use of the modular law we see: XY = (NRMS)U (NS/ R) = NSN [ (NR a S)UR] =NSMNRA RS. By symmetry the same result holds for XZ and YZ. Hence (ii) XY = XZ = YZ = XYz. Ore [3] has shown that under conditions (i) and (ii) the group XY is abelian. Consider XYUN =(NSMNR(M RS) UN=NS/ [NU(NRNS)] NS 1 [NRA (NURS) = NS/ [NRA NRS] = NSO NR. Also XYMN if [NSO NRN RS] NMN=NARS=Z. Therefore RNMSN/N XY NM N/N = XY/XYNN = XY/Z = XZ/Z= X. Hence we have exhibited a normal subgroup of H/N =G isomorphic with X, whichis abelian. Now if X had composite order then it would contain two characteristic 21 subgroups with trivial intersection. Since a characteristic subgroup of a normal subgroup is normal it would follow that G is not sub- directly indecomposable. Hence X isa p-group. This completes the proof of the theorem. Using this theorem, a statement made earlier may now be verified. Thatis, the basis set of the decomposition theory presented here contains the set of all simple groups. COROLLARY 1.4.1 If G isa simple group then G is indecomposa- ble. Proof Assume that G is decomposable. If {A«} is the set of all proper ingroups of G then Ge {Au}. Since G is simple itis clearly not a subdirect or factordirect product. Hence G=H/N and H is a subdirect product of A and B, A, Be {Aa and the ranks of A and B are less than the rank of G. Theorem 1.4 then shows that G is isomorphic toa factor group of A or B. This implies that the rank of G equals the rank of A or B, a contradiction. Hence Ge {Aa} must be false and G is indecomposable. The problem of determining whether or not a given group is inde- composable is a special case of the more general question: Givena group G anda set of groups {Ae}, what are necessary and sufficient conditions for Ge {Au} ? Three necessary conditions are given below which in the decomposability problem, that is, in the case of {Aa} equal to the set of proper ingroups of G, prove to be sufficient for some cases. 22 THEOREM 1.5 Let {Au} bea set of groups. If Ge {A,} then G/Z(G) « fA. /Z(An)}. Proof Let G be of rank 0 in {Az}. This means that G<Ae fAu}. We now consider all the possibilities for G<A. a) Let GcA. Clearly Gu [Z(A) U Z(G)]) =GuUZ(A). Since Z(A)Z(G) = Z(G)Z(A) the modular law applies and since Z(A)NG EZ(G), GNA [Z(A) v z(G)] = ) U[Z(A)N G] = Z(G). Hence G/Z(G) =G/G N[Z(A) UZ(G)] = G U EZ(A) U Z(G)] /Z(A) U Z(G) =G UZ(A)/Z(A) U Z(G) = GU vay/ztay [aa U Z(G)/Z(A). But G U Z(A)/Z(A)C A/Z(A). Hence G/Z(G) is isomorphic to a factor group of a subgroup of A/Z(A) and Ae {Aa}. The proceeding argument may be illustrated by the lattice diagram in Figure 2. Z(A) N Z(G) = GN Z(A). FIGURE 2 b) The second case to consider is G=A/N. Let I= aZ(A) for ae A and I... =aNZ(A/N). We will show that the mapping: aN I Ly of A/Z(A) into G/Z(G) 23 is a homomorphism. Suppose that I, = 1 then aZ(A) = Z(A) and ae Z(A). Consider Ln aNZ(A/N). Since ae Z(A), aNe Z(A/N). Hence Ln=l- Now let I —~* I and ay a,N I_-~ I 3 a5 a 5N I J = a,Z(A) a,Z(A) =a,a,Z(A) = I » and I I =] ° l a, a, 2 12 aya, a,N aN a,a,N Hence the mapping is a homomorphism and therefore G/Z(G) is isomorphic toa factor group of A/Z(A). c) Finally if G2H/N and H¢cAe« {Aa} then H/Z(H) is isomorphic to a factor group of a subgroup of A/Z(A) and G/Z(G) is isomorphic to a factor group of H/Z(H). Hence G/Z(G) « {Ae /Z(Az)}. This completes the argument for G of rank 0. Notice that the above proves that if G<A then G/Z(G)<A/Z(A). Now if G is of rank n >0 then we assume that for all elements of {Ax} of rank less than n the theorem holds. But G <CxXD with C and D of rank less than n. Hence C/Z(C), D/Z(D) e {Aa /Z(Aw)}. But CXD/Z(CXxXD) ~C/Z(C) x D/Z(D) « {An /Z(Aw)}. Hence since G/Z(G) <C xX D/Z(CXD), G/Z(G)« {Aw/Z(Aw)}. The proof is then complete by induction. If the set fAw} is the set of proper ingroups of G andif G is a nilpotent group then the following holds. COROLLARY 1.5.1 Let G bea nilpotent group. If G is decomposa- ble then G contains a proper ingroup whose class (length of upper central series) is the same as the class of G. 24 Proof If the class of G is n, written c(G) =n, and Z(G) is the th r" element of the upper central series, then G/Z (G) =1 and G/Z__4(G) Dl. If Ge {Ag} andthe class of A isless than n, then A/Z,_\(A) =1 forall Ac {Ag}. Therefore {Ax /Z)_j(Au}} = 1}. But it follows from the theorem that if Ge fAx} then G/Z__,(G) e {Ae/Z__j(Aw}}. Since G/Z_,_\(G) > 1 this is a contradiction. Hence G is either not decomposable or it contains an ingroup with class n. THEOREM 1.6 Let {Aw} bea set of groups. If Ge {Aa} then G'e fA}. Proof Let G be of rank 0 in fAs}. Then G<Ac« {A} a) Suppose first that GCA. Then G'CA!' and hence G'e {Al} . b) If G=A/N then G! = (A/N)' = A'N/N=A'/A'*NN. Hence Gite fA4} . c) If G=H/N with HCA then H'CA' and G'=H'/H'ON. But H'/H'NNe {AL} and hence G'e fAl}. Now assume that the lemma is true for all groups of rank less than n. Let G be of rank n. Then G<AXB where the ranks of A and B are less than n. But (A XB)'=A'X B' and since R'<S' if R<S it follows that G'<A'XB'« {AX}. The theorem follows by induction. COROLLARY 1.6.1 Let G bea soluble group. If G is decomposa- ble then G contains a proper ingroup whose derived length equals that of G. 25 Proof Let {Aa} be the set of proper ingroups of G. If the derived length of G is d, then of) 21 put old) #1. But it follows (4-1) , from the theorem that G fald-1); » and if the derived length of every proper ingroup of G is less than d then gl4-}) € fa(d-1)} = {lf}, a contradiction. Hence either G is indecomposable or it contains a proper ingroup whose derived length equals that of G. THEOREM 1.7 Let {Aw} bea set of groups. If Ge {A} then there exists a finite subset of {Au} , say {B.}, such that the ex- ponent of G divides the lcm of the exponents of {B,}. That is e(G)| lem {e(B,)} . 1 Proof Let G be of rank 0 in A. Then G<Ae fA,}. a) If GCA then clearly e(G)| (A). b) If G hi A/N then e(G) [ (a). c) If G=H/N and HCA then e(G) | e(H)| e(A). Hence for rank 0, {B} = fal. Assume the theorem is true for all groups of rank less than n. If G is of rank n then G<AxX B, and A and B are of rank less than n. By the induction assumption there exists subsets iR;} and is, } of {Aw} such that o(A) Jem fe(R3)} and e(B)| 1em {e(S,,)}. It follows thatif G<AXB then e(G) lcm {e(R.), e(S,)}. Hence the j,k j K theorem follows by induction. From Theorems 1.5, 1.6, and 1.7 given above it appears that the decomposability question might be most accessible for p-groups. Since such groups always possess non-trivial central series and derived series. The next section is a first step in this direction. 26 Chapter II In this chapter p-groups of class two will be considered. A group G is of class twoif 1C G'C Z(G). Thatis, G is non-abelian and G/Z(G) is abelian. Since such groups are quite similar to abelian groups it is reasonable to expect that a decomposable group of class two will be decomposable in a rather simple way. This is indeed the case and a complete characterization of the indecomposable groups of class two is obtained. Incase p #2 all possible defining relations for the indecomposable p-groups of class two will be given. The following lemmas concerning nilpotent groups will be needed in this section and in those that follow. The first three of these are given without proof and may be found in Zassenhaus [2] p. 114, and P. Hall [4] on p. 34 and 50 in that order. LEMMA 2.1 If G is a nilpotent group then G'C 6(G). LEMMA 2.2 If G is nilpotentand NAG then NON Z(G)D1. LEMMA 2.3 If G is nilpotent and the class of G is n then Gin) © 2(G)- LEMMA 2.4 Let G be ap-group. G is a subdirect product if and only if Z(G) is not cyclic. Proof Assume that Z(G) is not cyclic. Then Z(G) is a non-cyclic abelian group and hence it contains subgroups Ni and N, different from 1 such that Ny Nn N, = 1. But since every subgroup of Z(G) is normal in G it follows that Ny N,<G and hence. G is a subdirect 27 product. Conversely, if G contains N N,<G such that N, Nn N,=1 1? then it follows from Lemma 2.2 that Z(G) contains subgroups M,; M, different from 1 such that M, n M, = 1. Hence Z(G) is not cyclic. LEMMA 2.5 If G is a non-abelian group and Z(G) €4(G) then G is a factordirect product. Proof Since $(G) is the intersection of all maximal subgroups of G then there exists one maximal subgroup M of G such that Z(G) €M. Since M is maximal and Z(G) is normalin G, it follows that G=M2Z(G). But these groups permute elementwise and Z(G) CG, hence G is a factordirect product. LEMMA 2.6 Let a, b, ce G. If (a,b) and (a,c) « Z(G) then (a,b) (a,c) = (a, bce) and (a,b) (c,b) = (ac,b). Furthermore (a, b)” = (a’,b) = (a,b). el -1-1 -l -1,-1 Proof Since (a,b) «¢ Z(G), (a,bc)=a Mot, abe = atc +b ba(a, b)c = atco tala, b)c = ate tac(a, b) = (a,c) (a,b) = (a,b) (a,c). Similarly -l-l,-l -l-l,-l.. -1 (ac,b)=c a b acbh=c ab abc(c,b) =c (a,b)c(c,b) = (a,b) (c, b). The formula (a,b)” = (a’,b) = (a,b’) can be verified by a simple induction argument for n>0O. In order to show that the formula holds -1 for n<0, let n= -1l andconsider (a ,b). There exists a positive -1 integer r suchthat (a,b) =(a,b). But (a’,b) =(a,b)” and (a, b)*(a, b) - (a, p)t tt - (arth -1 b) = (1,b) = 1. Hence (a,b)* = (a,b) = (a ,b), and the formula holds for all integers n. 28 LEMMA 2.7 Let G bea cyclic p-group and suppose that G= <ab> with a, be G. Thenif o(a) Do(b), G =<a>. x Proof Since ab = ba, (ab)* = ab. Since ae G there is some integer “4 such that a pM =a. Hence pb «<a>. Clearly et & since otherwise o(a’ bp”) <o(a). But since ef & , év > =C{bye <a>. Therefore G =<ap>. LEMMA 2.8 Let G be ap-group. If G has two generators and G=AXB then A and B arecyclic and hence G is abelian. Proof Consider G/G'= AX B/(AXB)'~ A/A'X B/B'. If G has two generators then G/G' has two generators. Hence G/G' is the direct product of two cyclic groups. Hence A/A' and B/B' are cyclic. Therefore A and B arecyclic and G is abelian. The main decomposition theorem for p-groups of class two may now be stated. THEOREM 2.1 Let G be ap-group of class two. Then G is decom- posable if and only if G is either a direct product, subdirect product, or factordirect product. Proof Assume that G is decomposable but is not a direct or subdirect product. This implies that Z(G) is cyclic and since c(G) = 2, G'G Z(G) so that G' is cyclic. Let Z(G) =<z> with 2P =1, G'! is generated by a product of commutators each of which is contained in G'. It follows from Lemma 2.7 that there exists a commutator, (a,b), such that G'= <(a,b)>. Let o((a,b)) = pP. Since G is decom- posable it follows from Theorem !.6 that G'« {Aly where {As is 29 the set of proper ingroups of G. Since G' is cyclic of order pe? there must be an element A, « { Au) such that e(A}) > pf » and Ay is cyclic. Ay cannot be a factor group of G, for if Ay = G/N then Ay = (G/N) = G'N/N= G'/G'N N. Lemma 2.2 states that all normal subgroups of G- have a non-trivial intersection with Z(G) and hence with G'. Hence Ay would have exponent properly smaller than e(G'). It follows from Theorem 1.7 that there exists A,, a proper subgroup of G, such that Ay =G'. For A} ©G' and since e(At) = e(G') and G' is cyclic, Aj = G'. Hence there exist u, v « Ay such that <(u, v)> = Ay: Since Ay =G! for some 6, (u, v9 = (u®, v) =(a,b). Let c= a’, d=v and (c,d) = (a,b) =m. Let A=<c,d>. It will now be shown that G must be a factor- direct product with A as one of the factors. Since A is a proper subgroup of G there exist elements {g,} such that G=<c,d,g),--- i>? and such that none of the g,'s may be omitted. Consider B = <g> see + BD: & . Let (c, g;) =m -° forall i=l, ... ,» ke It follows from Lemma -% -a -%, 2.6 that (c,d) i(c,g.)=(c,d ")(c.g.)=(c.d 4g.) =1. Let 7 ; B; gi=d g,. Hence (c,gi)=1 for all i. Suppose that (d, gi) =m -B, -@ for all i. Therefore (d,c) (d, g;) =(d,c -P. gi =c “gi. Hence (d, g,') =1 for all i. Butif (c, gi) = 1 then *g!) = 1. Now let (c, g.') = (c, Fign = 1. Therefore A permutes elementwise with the group B, =<gi, sees a> Since AUB, =G, AB, =G and they permute elementwise. Since A is a proper subgroup of G it remains to verify that 30 B,C G. If By = G then c, de Z(G), acontradiction. Therefore G is a factordirect product. Hence if G is decomposable then G is either a direct product, subdirect product or factordirect product. Conversely, if G is adirect, subdirect, or factordirect product then G is decomposable. COROLLARY 2.1.1 Let G be ap-group of class two. If there exists a proper subgroup A of G such that e(A') = e(G') then G is decomposable. Proof If G is not a subdirect or direct product then G!' is cyclic and hence e(A') = e(G') implies that A'=G'. It has been shown in the proof of Theorem 2.1 that this implies that G is a factordirect product. A further characterization of these groups is given by: THEOREM 2.2 Let G be ap-group of class two. G is indecomposa- ble if and only if Z(G) is cyclic and G may be generated by two elements. Proof Let G be indecomposable. Hence Z(G) is cyclic and there- fore G*® is cyclic. It follows from Lemma 2.7 that there exists a commutator (a,b) such that G'=<(a,b)>. If <a,b> were a proper subgroup of G then G would be decomposable by Corollary 2.1.1. Hence G =€a,b>. Conversely if G=<a,b> and Z(G) is cyclic then G is certainly not a direct or subdirect product. Hence if G is decomposable it must be a factordirect product. Assume G = AB, a,b, = b a, for 31 all a; « A and b, « B. Clearly G/AN B~ (A/ANB) x (B/ANB) =AxXxB. If G is generated by two elements then G/ANMB is also generated by two elements. But G/AMB is the direct product of A and B. Therefore it follows from Lemma 2.8 that A and B are cyclic,. and since AN BCZ(G), G is abelian, a contradiction. If p #2 the defining relations of an indecomposable p-group of class two can be shown to assume one of two distinct forms. LEMMA 2.9 If G is a group of class two thenfor a, be G ulu-1) (ab)* = abba) * Proof The proof is by induction on u. Itis trivially true for u= 1. r(r-1) If (ab)’ = a™b"(b, a) * yrtl then since (b,a)« Z(G), (ab r(r-1) r(r-1) = (ab)’ab = a*b"(b,a) * ab=atb™ab(b,a) 7 But a bab = a ab’ (b’, a)b = atti yrttyyr a) = at tlyrttyy, a)". Therefore (ab)?! xf rol). r(rtl - arttyrtly a) - atti yrtly, a) 2 LEMMA 2.10 Let G be a p-group of class two, with p #2, which is generated by two elements a,, b, with o(a,) > o(b,). Then there exist a, be G such that G=<a,b> and <a>N<b> = 1. Proof Suppose ay - vf #1. Let 0 =p m and @ = pon with (p,mn)=1. Clearly <a> = <ap> and <b) >= <b>. Therefore we fo 6 may as well assume that at = bi #1 and a,6 7 0. For otherwise G would be cyclic. Since o(a,) > o(b;) it follows that « >F. Let 46 - - -p- 2 ajP . Then BP =(byajP © )P = bP alP (ajP b)) 32 a Po! «= 86 @, = 1 since a)? = b,P - Notice that a is an integer only if p #2. If <b><a > D1 then this process may be repeated and since o(b)< o(b,) it follows that there exists a choice of b such that <b> N<a)> =1l. Let a= ay and the lemma is proved. If the condition p # 2 is removed then the quaternion group of order eight is a counterexample to the lemma. THEOREM 2.3 Let G be an indecomposable p-group of class two with p# 2. Then G is generated by two elements and has defining relations either bod ie) xe} il o oO i pont ® = ul @ re) with & > 2B, or o p Y B i aP =bP =1 (a,b)P =aP , y=28-@ (a,b,a) =(a,b,b) = 1 with BSas 26. Conversely any group defined by I or Il is an indecomposable p-group of class two. Proof Since G is indecomposable it follows from Theorem 2.2 and oa 6 Lemma 2.9 that Z(G) is cyclic, G=€a,b> with aP =bP =1, a > @ and <a> N<b>= 1. a a aw Suppose that o((a,b)) =p. Then (a,b)P = (a? ,b) = (a,b? ) =1 re) which implies that a? , bP ¢« Z(G). Since Z(G) is cyclic and o(a) 60) 68) ab > 0o(b), bP e<aP>., But ~<a>N<b> = 1 and therefore bP =. ad B Hence 6 <o. But (a, b)P = (a, bP ) = 1, and hence B >o& There- B fore o(({a, b)) = pf, and aP ¢ Z(G). 33 Since G is of class two either G'e Z(G) or G'= Z(G). Case Il. Assume that G'C Z(G) and let Z(G) =<g>. Any element of G, in particular g, may be represented inthe form g = a“b™(a, b)™. Since G'¢ Z(G) it follows that <(a, bn) oe Z(G) and a“b” « Z(G). Hence by Lemma 2.7, Z(G) =Xa'b>. But (a”b’,b) = (a",b)=1, and y Vv (a“b’, a) = (b’,a) = 1. Therefore a’, b’ « Z(G). This implies that either bY’ =1 or av =i. If a” =1 then (a,b)” = (a,b) = (a%,b) = 1, Vv and a’, b’ « Z(G) which would imply that bY = 1. In any event Y=1 and Z(G) =<a">. Let u= pe m with (p,m) =1. Since Z(G) = <a"> it follows é ge 5 § that Z(G) =<aP>. Since aP e¢ Z(G), f>s. But (a,b)P = (aP ,b) b ‘4 = 1. Therefore Ss e so $= p and Z(G) = <aP> , 2 B 8 Tf ay = a” then there exists 9 such that (a,b) = (a? ) with ¢ n pte @=p on, (p,n)=1. Thenif a> = aj), (a,b) = a5 » But o(a,) = ° —, & and hence 2B+¢=@. Therefore 2pm. B Case Il. Assume that Z(G) =G!'. Since aP «¢ Z(G), and Z(G) =G! u pe Y ; _ u_ my,p* (a,b) =a’ . Let u=p'm with (m,p)=1. Hence (a,b) = (a,b) p? m m =a” . Since o(b ) = o(b) we may replace b throughout by band Yo *6 8 .* 8 - call it b. Now (a,b)? P = aP P = 1 and p- P is the order e - of aP . Hence o((a, b)) = aa = pY** A. Therefore y = 2h -a&, Finally since y >0, 2G >%. The relations (a,b,a) = (a,b,b) = 1 follow from the fact that G is of class two. It will now be shown that if a group G is defined by I or II then it is an indecomposable p-group of class two. 34 a 6 Suppose that G =<€a,b> and is defined by aP =b? =1, a-6 a-B (a,b) = aP with @ > 2B. Since (a,b) = aP it follows that - B bab = aaP . Hence ba? b=aP andtherefore aP « Z(G). C- ph Pp 6 But since a-£B >P it follows that <a* > Pka >. Hence (a,b) « Z(G) and this implies that G is of class two. For any element of G may be written as ab’(a,b)”. Let g, and g, be two u,_v U5 Vv w arbitrary elements of G. Then (g)> 8>) =(a lp ! (a,b) 4,a a 2(a,b) 2) u, Vv u, Vv uj;V5-vyuU =(a Li, a a “b a) since (a,b) « Z(G). But Z(G) 9 (a,b) beele u,v viu u, V5, v, U u, Vv, Vv u =(a,b) ) *(b,a) ) *=(a Lb *)(o 1a 4) = (a bb Lb *V(a Ib ba 2) M1 Yr 42, %2 =(a “b,a “b ) = (g1>8>)- Wy Since the generators of G have order a power of p and G is of class two, G is a p-group. . . uv w To show that Z(G) is cyclic, let the element ab (a,b) bean element of the center. Since (a,b) ¢ Z(G), then a-b”’ « Z(G). There- fore (a"b’,a) =(b’,a) =1, and hence b’ € Z(G). But then (a, b’) p~ F Vv “-8 =(a,b)” =(a ) =1, and since o(a? )=p?, bv’ =1. There- Vv Ww + “-F fore ab (a,b)” = av’ YP for every element of the center and hence Z(G) is cyclic. It follows from Theorem 2.2 that G is inde- composable. ~ P Now suppose that G =<a,b> andis defined by aP =bP =1, Y @ (a,b)? =aP (a,b,a) =(a,b,b) = 1, and y=26 -@, aS 2 f. Since (a,b) ¢ Z(G) it follows from the argument given above that G is of class two. G is clearly a p-group and it remains only to show that Z(G) 35 is cyclic. Suppose that a°b’(a,b)™ « Z(G). Arguing as above b” « Z(G) and (a,b)Y = 1. But o((a,b)) = p? and hence b’ =1. Also (a‘,b) = 1 and hence a « Z(G). Therefore since o((a,b)) = p®, it follows that p| u and ave CaP > Cc <(a,b)>. Hence ab’ (a, b) e<(a,b)> and Z(G) is cyclic. Therefore G is indecomposable. This completes the proof of the theorem. Notice that the condition p # 2 is used only in applying Lemma 2.9. If G is a 2-group which satisfies the conclusion of Lemma 2.9 then its defining relations are indeed of the form I or II above. An example of such a 2-group is the dihedral group of order eight. The main results of this section may be summarized in the following manner: Let G be an arbitrary p-group of class two with p # 2. If e(G) = p” then there exists a finite set of groups {H.} of type I and II given in Theorem 2.3 such that Ge {H.} - This follows from the fact that any group is contained in the closure of the set of its indecom- posable ingroups and that the exponent of a group cannot exceed the largest of the exponents of its indecomposable ingroups. 36 Chapter III The problem of determining when an arbitrary p-group G is decomposable can in some cases be solved by considering the decom- posability of G/Z(G). In the case of e(G) =p it will be shown in this section that if G/Z(G) is a direct product then G is decomposable. In general, this decomposition is non-trivial, i.e. G is neither a direct product, subdirect product, or factordirect product. An example will be given to illustrate this point. Two lemmas will be required before the main theorem can be proved. LEMMA 3.1 Let G bea nilpotent group of class n>2. Suppose that G/Z(G) = G= AB with ab=ba for all d¢ A and be B. That is, G is a factordirect product of A and B. Thenif A and B are the inverse images of A and B respectively, c(A) =n or c(B) =n. Proof Let A =<a ., a> and B=<6,, see; D>. Then r 1 s G =€a,; sees A, by, ees by, Z(G)>. Since G is a factordirect product of A and B it follows that (a;,b,) e Z(G) for all i, j. Hence any element g¢ G may be put inthe form g =abz with ae A, be B and ze Z(G). Since the factor z may be omitted as a multiplier of any entry in a commutator there exists a non-identity commutator of length n: (cydys se, cid.) « Z(G) with c, €<aj, -+0, a,> and d.; € <b), sees bo>- This follows from the fact that c(G) =n and Gon) CZ(G). But (c,d), --«, cd) =(c,d), +--+, ¢, doy. ¢,) (c,d), ree, CL di yp d.) since these three commutators are contained in Z(G). Since (c,,d,) e Z(G) 37 (c,d); c5d,) = (c,>cp)(d,,d,)z, for some 2, € Z(G) and hence by induction, (c,d), coe, ce -14,-1) =(c), ese, C )(d,, sees d-1)2 ) n-14y> sees dei) c.) = (cyr-es, yc )(dy,---,d, ,c,). n-l for some ze Z(G). Hence (c,d), ere, Cid ys Cc, =((cy> ero; C¢ Now since n >2 (d , ad ) = (u,v) with u, ve<b,, ce e3 b>. n-l -1 )= (uvly uv, c.) pee Hence (d,, sso, diy, ch) = ((u, v), ; cy )(u, cy, c,) = (u, c,) he, clu, cv, c,) = 1 since each of these commutators is in Z(G). Therefore (c,d); ose; cd.) = (cydyo---5¢, yd ys eMeydys-ees¢) di _ys dy) =(c,,--6,c_)(d,,-..,d_). If c(A)<n and c(B)< n it follows that 1 n 1] n (c,d), eees cd.) = 1 and therefore c(G)<n, a contradiction. Hence either c(A) =n or c(B) =n. LEMMA 3.2 Let G bea p-group and let G =<ay,> ores a> such that aj, jeeees a> CG forall i, If (a,)P=1 for ne Fie Bay all i then for all ge G, g has a unique representation of the form a a . a. b with be G'. Proof Since every element of G is a word in the a,'s it follows that every element has a representation of the form given above. eal o. Pi r Suppose that g = ay vee a, by = ap eee aL b, with x, -8, a - e bj, bs e G'. Therefore ay ceo a” Ts b, e G'. Assume that for some i p} a. - p.. Then G =<a,, ees a, pr bg: ppreer ape O.- 6 For clearly a; * is generated by these elements and since w= 6 (, - @..p) =l, <a, : generator. Therefore G =€aj peoo,a >= <a,>. But b3 e G' and hence is a non- i-? Aya ote a> contradicting 38 “ By the hypothesis. Hence p |@, - B. for all i and thus a, Fa, for all i. Therefore by = bs and the lemma is proved. THEOREM 3.1 Let G bea p-group of class greater than two and let e(G) =p. If G/Z(G) is a direct product then G is decomposable. Proof If Z(G) is not cyclic then G is decomposable. Assume there- fore that Z(G) is cyclic and hence has order p. Let G/Z(G) = AB, with ab=ba forall ae A and be B,. Let {ayres-s ayy be an l irredundant set of generators of A. That is, if A =a, ose a? then Capreess 8, prayers a DCA for all i. Choose {b),-2-,b.3 similarly for B: If c(G) =n >2 then according to Lemma 3.1 either c(A) =n or c(B,) =n. Let c(A) =n. Clearly An) SG(y) and from Lemma 2.3 it follows that ICA) < Z(G) and therefore An) = Z(G). Hence if Z(G) =<z> then ze A' and from Lemma 2.1 z is a non- generator of A. Therefore A =<aj; eee, a> and this is an irredundant set of generators of A. Let B =<b)> s+, b>. If ~2¢e B then B= B), but this may not necessarily be the case. In any event G= AB, and hence G = AB. It can now be shown that Ge {A,B}. Since (a,, b,) = 1 it Oe. follows that (a;,b.) =z 9. Let H= Ake XAXB. Since Z(G) = Aca) r there exists c, de A such that (c, d) =z. Let OC, &B be subsets of H defined as follows: Xe i $i; or | {[a,-¢ hese “id | 5, =0 if j fi, b= if j=iandi, j=l,...,r}. “1k ~ rk are (= {fi,a “S,...,4a * sb, | (a;.,) =z, k=l, wee, sh. 39 Let H, =U, (B>. Notice that H could have been defined as A xA,* see XA,X B with Ay =<c,d>. Thatis, H may be defined as the direct product of A, B and r copies of a group of class two. aa ~O% a, ~&,. Let N=C(z (2 ,1,...,U,f% “1,2 “,1,.0.,1, oe. x, -a . [z rr lye. lye am i for i= l,...,5>. Since the commutator Oe a4 fo lk k k (fa, teesbel... J, fea peed Sb I) = [l,....2 Sh] it follows that NC Hy and clearly N< Hy- Let G, = H,/N. It will now be shown that G, =G under the mapping induced by: 5 li 1 ri aje>[a,c s.-.,c ,1JN=g, co"4 . x , b.<>[1,4 wd “i, biIN = h, where a word on the a. and b, corresponds to the same word on 8; and he It will be shown that words are equal in G) if and only if they are equalin G. This will suffice to show isomorphism. Let m be an arbitrary element of G. It appears as a word in the a,'s and bits as follows: M =X) V¥) X2Vo°° YE with x, € A, y, « B. ery Under the mapping, the image of x) will be [x)> Cok peers - eC rly N with 6,, equal to the sum of the exponents of a, in x). Ss 9 rl . pee ey d ’ yl N with ra = a 9 The image of y, will be [1,d 1 Yh and Via equal to the sum of the exponents of b. in Yq° Therefore the image of m under the given mapping will be: 40 Higil Mtg lt otigtt oo. tt rt d Yye++ VIN. [x)X2°- “XC o . Since (a.,b.) =z J 1 J suitable powers of z to the product. That is » X, may be commuted to the left by adding b.a. = a.b. (b.,a.) = a.b. z JP 1 J J 1 1 J J 1 This operation is mirrored in G, by commuting powers of c to the fo li a . left. Thatis [l,d °,.-.,d “',b,] la, 1,-.-,1,c,1,--.,1JN - [a,.4 Moja Ftig Figg Mig “bp ]N Oc, a... . - &,, a... a. = [apd Md jtigg Jee Dea Hid "yb IN. But since eet aarti ,l,.-e,l,z rl, the element [z J coe, 1] ¢ N the above expression . a. tov 1 . may be written as [ajz Pol,...,l,c,1,...,]] fl,d seed ™,bIN. Hence every time some a, is commuted with b. in m, thereby introducing a power of z, the very same power of z is introduced aX. . in the image element in G, by commuting c with d J. Therefore a oO ; Xpoee XV ors Vy S FB xyz with X= Xp ee X or “ and Y=yyere VE then the image of m, n = [xz »>C 500058 LU v ») [1,d - 22050 r JN with i equal to the sum of the exponents of a. if m =X), e+ XY, Ss appearing in the product x and v, = » O,; 5 with wo; equal to the j=l sum of the exponents of bs in ye Hence if an arbitrary element of G is put in the form xyz defined above then the image element can be é, vd ye v put in the form [xz°,c Mg 7 see cd *,y]N. By an argument similar to the one above it can be shown that an arbitrary element of Gy may be put in the form 41 ford eo i>) Ay vd [xz ,c 6d ge eg d a ] N and its image inthe form xyz Suppose that m, me G and let n and n be their images in G)- a — — a, Let m be inthe form m= xyz as defined above, and m= xyz — —— OG -jW similarly. Thenif m=m, xyz =xyz implies that x Ie Z(G) = 71 ; -lj_ Yr ~ -1_, 2 and yy ¢ Z(G) since ANBCZ(G). Hence x “x=z °, yy #2 and @-Ge= (vy, + ¥2)(p)- It follows from Lemma 3.2 that the sum of the exponents of a; in x) x is equalto omodp, and similarly for b, in yy i. o MY) i, Now form na in G)- If n= [xz ;c¢ ad 4,eoe,c d *,ylN and n = [xz »c ad “,66.,c ad * yIN then nine ~ v0 MED -0, K&R. 9 [x te ena Re My tg 1 ed Po Py Fe Ty lyn ae Py 91-9, 2 (F,-A) Pa Fe Fp 8y Ad 1 =[x XZ 2c d Zz yo0es Cc d Zz iV vIN. Since x X=2 and yy bly ya 2 then £2 &; = op) and d.= 9, = o(p) for i=l, ..., re Hence n n ¥y-(y, tY2) Z Yo _ ,l,...,l,2 “JN#=[l,...,1]N. Therefore if m=m in G then n=n in G)- Conversely let m and m in G have images n and n in G)- Now suppose that nl n=l. This product will appear as above and hence Py - = o(p), Di -d, = o(p) for i=l, ..., r. Hence -]— -|]— &-A -l]— n =[x le ,l,...,l,y yine=fh,...,1)N. Therefore -l- oY) -l-_ Y2 _ KX X=Z and y y=2 such that -(y, + Y>) 2a -&% So(p). But -]—W @-a@ -]— -~]— @-% ~j_ -j— -]_ &_@ then x 1 ye IS = 1. Hence x ln y IS = y x lee -~] -liw— &- — —— & ~y ls lye “1, Therefore m=xyz = xyz™ =m, and the mapping is one-one. Hence it is an isomorphism and therefore G is 42 decomposable. In order to show that a group which is decomposable in the manner described above, i.e. G/Z(G) is a direct product is not in general a direct, subdirect, or factordirect product consider the following example. Let p be a prime greater than 3, andlet A= Kay; as? with li {=ah syP=zP=1, (a,,a,)=y, (ayy) = 4 (az,y) = (ap 2) = (a5,2) = 1. This group has order ps and class three. This group is discussed in Burnside [5] p. 146, and it is there shown that o(A) = r', c(A) = 3 andthat Z(A) =<z>. It is easy to see that A’ =<y,z)>. Incidentally, A is indecomposable because all proper sub- groups and factor groups have order at most p? and every group of order >? has class at most two. Consider A XADH =Cfa,, 1], [a,, a4] , Ll,y}>. First we show that [AXA:H] = 9%. Consider the set <fi, ayay > for all u, v<p. There are precisely p elements in this set and we claim that these elements form a complete and irredundant set of coset representatives for H in AXA. A general element of AXA is La} teary ; a aby v8) » Let a! =p and consider the coset [1, aw P af JH. The element fafa, a] € H and therefore [avae, aah] € [1, an F 8) H. But since [y,1], [z, 1] and [1,2] « H it follows that ay ma Py Va! ; at'ab'yV'29 is contained in the coset in question. Hence this set of coset representatives is complete. Now suppose that two of these cosets are the same. Then 43 - x [1,a},a,] = (1, ava5] h with he H. Let h= [a ab y¥a vaby 09). , eg Therefore ay * a8 y¥28 = 1 and aya, = aktP ty LT - From Lemma 3.2 it follows that = @#so0(p) andhence y= $ = o(p). Similarly i-k#j-£20(p) and hence [1, aya) y= 0, akas]. Therefore the set of coset representatives is not redundant. Hence [A XA:H] = p*. Now consider N =<[z,2 'D. Clearly N“AH. Let A, = H/N 1 = <fa,> IJ] N, [a,, ai] N, [l, y]N>. Since o(N) =p it follows that o(A,) = o(fi)/p =p”. It will now be shown that Ay satisfies the hypothesis of Theorem 3.1. a) It has already been shown that o(A,) = >” and hence A, is not isomorphic to an ingroup of A. b) A, has class three. Consider the triple commutator ({a,.1)N, [a,,a,]N, [a,-1 N) = (fy, 1] N, [a,,1] N) = Tz t, 1] N # [1,1] N. Hence c(A)) >3. But since A, is a factor group of a subdirect product of two groups of class three it follows that c(A,) <3. Hence c(A,) = 3. 1 c) Z(A,) has order p andhence A, is neither a direct = [ab yy, aP ary IN product or subdirect product. For suppose € € Z(A,)- Its commutator with [a,, JN is [ay aby 29 124) 1] N. If this ow is N then (a, ab yv2?, a1) = sate) = 1. But by direct calculation (azy » ay) = (af, a) aay. a y’)(y* ,a D = y Pn Y =1. Therefore Bay «x § 90 ; ; zo(p). Hence € =[a,2 Ly ZPqN. Its commutator with [a,,a,]N is a fg o ({a)2 jye22IN, [a,,a,] N) = [(a%e , a5) ), (y OF a DIN = f(a arya) (yas N. (1) 2 9 -0 “ a But one can easily show that (a,,a,) =y z , and (y ,a,) =z a(a-1) oe a) a 2 ~Q Hence (a,>4). ly »a,) N= [yz , zw JN= fl,1]N, & = o(p) and 9 =o(p). Therefore € = [2° 2F4N = tot =<[z, 1]N>. Thus o(Z(A,)) = p. d) Consider the subgroups R =<Ta,, 1]N, [a,,a,] N> and 9 UN and Z(A,) s=<[l,y]N, [1,2]ND. It is easy to see that A, = RS, RMS =Z(A,) and {(r,s)]r eR, se S}= Z(A,)- Hence A, /Z(A,) is a direct product. It follows from the theorem that Ay is decomposable. If Ay were a factordirect product then one of the factors would have to be a subgroup of class three. Since the smallest p-group of class three has order >, such a subgroup is of necessity maximal and hence must contain At =<Ty, 1] N, {z, 1] N>. Now the intersection of the two factors of a factordirect product must be contained in the center of the whole group, and hence the factor group A, /Z(A,) must be a direct product. Since o(A,) = >> and o(Z(A,)) =p, and A,/Z(A,) is not abelian, it must be a direct product of a group of order p and one of order >. Hence the orders of the factors in the whole group must be p* and po”. But the factor of order p- is abelian and hence in the center, contradicting the fact that o(Z(A,)) =p. Therefore Ay is not a factordirect product. The decomposability of Ay could also have been deduced from the fact that Ay « {A} and A is isomorphic to a proper subgroup of A,:. Thatis, the mapping a, ~—> la,> 1] N a, <> [a,; ay] N induces an iso- morphism. Hence A, « tA} © tAa} where tAa} is the set of proper ingroups of A. 45 It is easy to check that Ay may be defined by: cP = qP =P = (c, a)? = (c,c, a)? = (c,c,c,d) =(d,c,c,d) = (f,c,c,d) = 1 and (d,f) = (c,c,d). 46 Chapter IV The definition of closure given in Chapter I suggests an equiva- lence relation for finite groups. That is, G,= G, if iG} = {G,} : It is readily shown that this is indeed an equivalence relation and hence provides a partition of the set of all finite groups. This chapter will investigate the relation between this partition and a group relation introduced by P. Hall [6] known as isoclinism. DEFINITION 4.1 The groups G, and Gs are said to be isoclinic, G|~ G, if: 1) G,/Z(G)) =~G,/Z(G,), 2) G, =G3, 3) if a Z(G, )<> a ) under 1) ; Z(G ) and b, Z(G,)<b Z(G 2 2 2 2 then (a,.b,)<* (a,, by) under 2). That is, there must exist isomorphisms 1) and 2) such that 3) is satis- fied. Notice that 3) is unambiguous even though a particular choice of coset representatives has been made. For (ay, by) =(a)2), bj 2) if Z, and 2, € Z(G,). Two groups are then said to be isoclinic if the first elements of their descending central series are isomorphic and if the first factor groups of their ascending central series are isomorphic and if the latter isomorphism induces the former. Although the definition of isoclinism doesn't restrict the groups to p-groups, it becomes equivalent to isomorphism if Gy and Gs have trivial centers or if Gy = Gi} and Gs = G3. For p-groups, of course, neither of these two possibilities can occur. The discussion 47 from this point on will be primarily restricted to p-groups. It is quite easy to see that the relation of isoclinism is an equiva- lence relation and hence provides a partition of the set of groups being considered into isoclinic classes, or 'families''as P. Hall denotes them. It follows from the definition that both the elements of the descending and ascending central series of one member of a given family is common to all members and hence is a family invariant. Similarly for the derived series. All groups of a given family will therefore have the same class and the same derived length. Clearly all abelian groups belong to the family containing the element 1. And it is easy to see that if A is any abelian group then GXA-~G for any G. This follows from the fact that (GX A)' =G! and (GX A)/Z(G X A)~G/Z(G). Using these properties Hall shows the following: 1) If KOG then G~K if and only if G = KZ(G). 2) If H=G/N then G~H if and onlyif NNG'=1. The statements 1) and 2) may be restated as follows: 1') If KCG then G~K if and only if G=KA, A is abelian and A commutes with K elementwise. 2') If H=G/N then G~H if and only if there exists MIG with NN M=1 and G/M abelian. Another equivalent form is: 1") If KEG then K ~G if and only if G is a factordirect product of K with an abelian group. 2") If H=G/N then H~G if and only if G is a subdirect product of H with an abelian group. 48 Hence the isoclinism of a group with an ingroup of itself is the statement that G is decomposable in a rather special way. This suggests that there may be some relation between G,~G), and (Gq) = GH. It is not true in general that G,~G, implies {Gs = G5} . For G, ~ G, X A where A is an abelian group of arbitrarily large exponent whereas the exponent of any element of {Gj} is bounded by e(G,). Hence G,X A ¢ {G3 if e(A)> e(G,). Similarly iG} = {G5} does not necessarily imply G,~G,, sinceif G,=G), x G, itis clear that {Gi} = {G, x G}}. But if G, is non-abelian G,/Z(G,) # (G, X G,)/Z(G, X G,) =~G,/Z(G,) x G,/Z(G,) and therefore G, #G,XG,- Hence if some statement concerning the relation between isoclinism and the equivalence induced by closure is to be formulated then either the domain of groups under consideration must be restricted or the definition of {G} must be modified, or both. Before we can give the modification of closure that will be useful we define the notion of A-permissi bility. DEFINITION 4.2 If HOG and A is acyclic group then H is called an A-permissible subgroup of G if G is a factordirect product of H and A with A 1? « {A}. If H=G/N then H is called an A-per- 1 missible factor group of G if G is a subdirect product of H and Ay e {A}. Following are some of the elementary properties of this relation. LEMMA 4.1 An A-permissible factor group of an A-permissible factor group of G is an A-permissible factor group of G. 49 Proof Let K be an A-permissible subgroup of G. Then G = KA)» Ay e {A}, and K commutes elementwise with the group Aye If K = HA,, A> e {A}, and H commutes elementwise with the group A, then G= HA,A,. A,A, € fA} > itis abelian and commutes element- wise with H. LEMMA 4.2 An A-permissible factor group of an A-permissible factor group of G is an A-permissible factor group of G. Proof Let K= G/N, be an A-permissible factor group of G. Hence GCOG/N, XG/R with G/Re fA} and N,N R=1. Let H=K/M, be an A-permissible factor group of K. Hence KC K/M,X K/S, with K/S, « fA} and $;9M, =1. Let H= K/M, = G/N, /a/n, ~G/M. It must now be shown that H is an A-permissible factor group of G. Let S, =S/N,. Since G/R and K/S, ~G/S are abelian R, S 3G' and hence RMS 7# 1. Since Syn M, =l and SMNMs= Ny then MARNS=N,AR=1. Hence GOG/MXG/RNS. It only remains to show that G/RNS« {A}. Since G/Re {fA} and K/S, = G/S « {Ay and G/SNREG/RXG/S it follows that G/SNRe {A}. Hence H = K/M, “G/S is an A-permissible factor group of G. LEMMA 4.3 An A-permissible factor group of an A-permissible sub- group is an A-permissible subgroup of an A-permissible factor group. Proof Let K be an A-permissible subgroup of G. Thatis, G =KA,; with (k,a) =1 for all ke K and ae A, and A, « {A}. Let H be 1 an A-permissible factor group of K. Thatis, Ke K/N, X K/S with H= K/N)> Ny NS=1 and K/S « {A}. It must now be shown that H 50 is an A-permissible subgroup of an A-permissible factor group of G. Since Ny and S are normalin K it follows that N S 4a. |’? Hence GOG/N,X G/S with N,Q S=1. But G/S = KA,/S = K/S: A,S/S, with A,S/S =A, /A,NS« {A} and K/S« {A}. Hence G/S « {A}. Therefore G/N, is an A-permissible factor group of G. Now G/N, = KA,/N, = K/N, -° A,N)/N,- Clearly (KN,,aN,) = 1 for all kN, « K/N, and all aN, « A,N,/N, and since A,N,/N, i ~A)/A, N Ny «e {A}, He K/N, is an A-permissible subgroup of G/N,- It is not difficult to show that if A is a priori chosen to be of order equal to the exponent of G then the converse of Lemma 4.3 is valid. If now we define H to be an A-permissible ingroup of G if H is an A-permissible subgroup of an A-permissible factor group of G then the three lemmas above prove: THEOREM 4.1 The relation "H is an A-permissible ingroup of G" ig transitive. We now define the notion of A-closure. DEFINITION 4.3 Let {Ge} be a set of groups and let A be acyclic group. A group G will be said to be of rank 0 if G is an A-permissi- ble ingroup of some Ge. G is of rank 1 if it is an A-permissible ingroup of a direct product of a group of rank 0 with an element of {A}, and it is not of rank 0. In general, G is of rank n if itis not of rank k<n, and it is an A-permissible ingroup of a direct product of a group of rank less than n with an element of {A}. The set of 51 groups with an assigned rank number will be called the A-closure of {Ge}, written {Ge} AC The A-closure of a set of groups may be described in another way: LEMMA 4.4 Let {Gg} be a set of groups. Then {Gas satisfies: 1) {G3 Cc {Gat A? 2) if He {Gy} A and J is an A-permissible ingroup of H then J« {Ge J a and 3) if He {Ga}, and A, « {A} then HXA, « {Gb AY Furthermore, if Al is a set of groups which satisfies the same con-- ditions as {Ga} A’. above, then AL > {Ga} AY Proof 1) {Ge} « {Gat 9 since all elements of {Ge} are of rank 0 in {Gu} A 2) Trivially, J is an A-permissible ingroup of H XA, for any Ay e {A}, andis hence in {Gy} A 3) HX A, is an A-permissible ingroup of itself. Let AZ be a set satisfying conditions 1), 2), and 3). If G is of rank Qin {Ge} A then G is an A-permissible ingroup of some element of {Ga} and is hence in 42. Assume the lemma to be true for all elements of {Gy} a of rank less than n. Let H have rank n in {Gat a’ H is an A-permissible ingroup of RX Ay with R of rank less than n and A, « {A}. Then Re AZ and RXA, et. Hence from 2) H e Al. Thatis, {Ge} A is the intersection of all those sets of groups which contain {Ge} and which are "closed" under the operations of A-permissible ingroups, and direct product with elements of {A}. 52 If iGas y is thought of as an operator on the lattice of subsets of the set of all finite groups then it is a closure operator. That is: LEMMA 4.5 If {Gg} and {Hg} are sets of groups and A isa cyclic group then: 1) {Ga} CitGa} , and, 2) {{Ge3 sh, = {Gat , and, 3) if {Gy} < {He} then {Ga} , C {He} A Proof 1) This has been demonstrated in Lemma 4. 4. 2) From 1) it follows that tGa a _ {{G.3 A a of He {{Ge} ASA is of rank 0 then H is an A-permissible ingroup of an element of {Ga} A and therefore He {Ga} AC Assume the lemma true for He {{Ga} ALA of rank less than n. Let G be of rank n. Then G is an A-permissible ingroup of J x A, with Je {{G.} ata of rank less than n and Ay e« {A}. By the induction assumption Je {Gu} A’ Therefore LiGat ata = {Ga} AY By Lemma 4.4, J XA, «€ {Gad and Ge {Gu} ,- 3) It follows from 1) that {Gy} ¢ {Hp} c {Hg} a: Hence by Lemma 4.4 {Ge} AS {Hig} AY Now consider the special case in which {Ga} consists of a single group. LEMMA 4.6 If He {G} then {FH} , = {Gh ,- Proof It is only necessary to show that G« 4H} A for it follows from 3) and 2) above that if He {G} a then iB} {Gt ,: Assume that H is of rank n in {Gh ,- Hence H is an 53 A-permissible ingroup of J XA): J a group of rank less than n and A, € {A}. By definition H is an A-permissible subgroup of an A- permissible factor group of J X Ay > and it will first be shown that J XA, e {H} A Let S=HA, with A, « {A} and (h,a) =1 for all he H and 2 ae A,. Then JXA,C(J XA,)/NX(J XA,)/R with NOR =1, (J x A,)/R « {A} and S=(JX A,)/N. First it must be shown that JXA,« (},- Clearly SX(J XA,)/R « (S},- If ge IX A, then (IX A,)/R = {[N,gR] | all ge (J XA, If [g,N,g,R] is an arbi- trary element of (J XA,)/NX (J X A,)/R then [g,N, g5R] = lg,N, g R] IN, g) 858] = [N, gy eR] [g,N.g,R] since RD(J x A,)'. Hence (J X A,)/NX (J Xx A,)/R may be represented as a product of elementwise commuting subgroups one of whichis (J X A,)/R and the other a group isomorphic to J xX A)> Hence J XA, is an A-permissible subgroup of (J XA,)/NX(J xX A,)/R =8SX (IX A,)/R e {5} A and J XA, e {S} AS Therefore Je {5S} A’ In order to show that S « {H} consider (H X A,)/M with M= {{h,nJ | he HMA, inS}, By Theorem 1.3 S=(HXA,)/M. To show that (H X A,)/M is an A-permissible factor group of H XA, consider (H XA,)/MX(HXA,)/(HX 1). Since MN(HX1) =1 and (HXA,)/(HX1) =A « {A} it follows that HXA,C (H XA,)/M 2 a 2 X(H X A,)/(H X1) and (H Xx A,)/M is an A-permissible factor group of H xX A>: Hence Se {H KAS a = {H} A Therefore J « {H} ,- Now suppose that H is of rank 0 in £G} 4: then H is an A- permissible ingroup of G. And the argument above, with Ay = 1 and J=G shows that Ge {H} AS If the lemma is assumed to be true for 54 all groups of rank less than n in {G} A then the argument above shows that Ge {J} AE {Ht AC Hence the lemma follows by induction. The special case just treated actually represents the general situation since: LEMMA 4.7 If G¥iG}, then Gat,=Vt Ga tf. Proof Clearly v iGa} © {Gy A’ If He {Gy} a is of rank 0 then H is an A-permissible ingroup of some element of {Gy}, say G. Hence He G. Assume that for all H of rank less than n in {Ga} A? He iG}, for some G in {(G,}. If H is of rank n then H is an A-permissible ingroup of JX Ay with J of rank less than n and Ay ¢ tA}. But JX Ay e J and by the induction assumption Je G for some Ge iGy}. Therefore He G and the conclusion follows by induction. It is quite evident that the definition of A-closure was formulated so as to preserve the relation of isoclinism. That is: LEMMA 4.8 If G, « {G3}, then G, ~ G,- Proof Let G, be of rank O in { 3} A’ Hence G is an A-permissible ingroup of G>- Since isoclinism is a transitive relation it follows from 1") and 2") on page 47 that G,~G,. Suppose that the lemma is true for Ge (Got 9 of rank less than n. If G, has rank n then G, is an A-permissible ingroup of J XA, with J of rank less than n and Ay « fA}. Hence G, ~J XA, ~ J. But J~ Gs by the induction assumption. Therefore G, ~ G,. Notice that from Lemma 4. 8 it follows that iG} A” iG A 55 implies G, ~ G,- The main theorem of this section will now be stated in the form of a corrected converse of Lemma 4. 8 above. THEOREM 4.2 If Gy and Gs are p-groups such that for i=l, 2 G,/Z(G,) may be generated by a set of elements of order p then G,~G, implies {G3 A= iGo a with o(A) = max te(G,)}. Proof Let G,/Z(G,) = <o, Z(G), coe, o, Z(G, )> be an irredundant set of generators of G,/Z(G,) such that o(%, Z(G, )) =p. Let C=C,X vs XC, C. =<c,> with o(c,) = o(&, ) for all i. Consider G,|xX CDH= (@ cy), wees (a, ¢,) . Since e(C)< e(G,) it follows that Ce {A}. And since G,X C = HZ(G,xX C) it follows that H is an A-permissible subgroup of Gx C. Hence H~ Gy x C ~G, and {FH = {G, xc A= iG} A: tis easy to see that the mapping a .c. Z(H) <> a&.2(G,) is the isomorphism required by the definition of isoclinism. Let N =<(a,0,)?, see, (a,c, )P. Since aP ¢ Z(G,), i N<H. Also H'! = (Gx C)'=G! andtherefore NN H'=1. For if ! 1 tiP vj,P an element of N, (%,¢)) Lee (o,¢),) , is contained in H' then rip T,P TIP T1P a} toe & k c } ese C K must be contained in G, and hence 1 k 1 k i ryP ryP cy ree CLS 1. Since all c, are independent o(c,) |r;p. But rsp o(c.) = o(e, ) and therefore o, = 1 for all i. Let M = H/N =<e,c,N, ..., &,c,N>= <A, ..., B,>. Since M is an A-permissible factor group of an A-permissible subgroup of G,XC it follows that M~G, and {M} = {G} ,- In addition o( B.) =p forall i. The isomorphism between M/Z(M) and G,/Z(G,) 56 required by the definition of isoclinism is induced by the corre- spondence B,Z(M) <= 4.Z(G)) since the mapping (B.. Bi) (@, *.) induces an isomorphism between M' and Gi. Now consider G,/Z(G =<y, Z(G, ), Z(G))> and choose 12 YE these cosets so that wileeewalep gives the isomorphism required by the definition. The argument given above for G, may be reproduced identically for G,. Hence we obtain a subgroup R= <v,, sees D> with o( 2.) =p, {R} Aa iG} A? R ~G, and such that the mapping V,Z(R) + y,Z(G,) gives the required isomorphism. lf {@, Z(G)), cee, X,Z(G,)t was chosen as an irredundant set of generators it follows that {A> oe w, ft is irredundant, and similarly {2, p 2005 is irredundant. v5 Since o( B,) = of d) =p for all i, j it follows that every element r r 1 of M has a unique representation of the form B, soe a 9, with 6, « M', and similarly for R. If g( (B., f;)) (2. 2) then & induces the isomorphism between M! and R! implied by the M/Z(M) ~R/Z(R Consider the mapping: r r r Yr 1 k ) 1 ) «K A, Lee A. at oe UL v(8,). Clearly it is well defined on all of M, and since o is uniquely defined the mapping is one-one. To show that it is an isomorphism d d e ey djte d, +e . 1 k 1 ] k ~k consider Ay 77+ By 01°F ae PS = 6," +B 0,9,8,, and 57 d ey e dite) dite, eee A 8o(8,) = 0,1... BS So, 0(0,)6(6,). Clearly 84 is formally identical to 8, with £. replaced by v.. Hence 9, = g (8) and since @ is an isomorphism TF (93)T(9 1) (85) = 7 (0,6,9,). Therefore R*=M and {G,} A= iG3 AS Using an argument quite similar to the one above it will now be shown that: THEOREM 4.3 If G, are p-groups such that c(G,) =2 and G, = <a,,b.> for i=l, 2 then G,~G, implies that {Gib . = {Gy A with o(A) = max fe(G,)} . Proof Let G, =€a,,by> and hence G,/Z(G)) =<a,Z(G,),b,2(G,). o Bp Suppose o(a, Z(G,)) =p } and o(b, Z(G,)) =p - Let C and D be two cyclic groups such that C = <c?, o(c) = o(a,) and D=<d>, 0o(d) = o(b,). Clearly C, Dé {Ay . Consider G,XCXDOH =<a,c,b,d>. Since G,XCXDe {Gy and HZ(G,X C XD) =G,xCXD then oy ‘al He {Gj} At let N =<(a,c)P ,(b,a)P > . Clearly N<H, and NNH'=1. Hence H, =G,/Ne TH}, = {G3 a and {H} A {Gi A’ ag Now H, =<(a,¢)N, (b,d4)N> = <a,,5, > and if al « Z(H,) then _ of _p* _ p* po (a,)P = 1. Forif (ay ,b,) =1, ((ayc)? ,b,4)N = (ay ,b NEN, a a But (af ,b,) « H| and NMH}=1. Therefore (aP ,b,) = 1 and a (e4 [a4 since G, =<a,,bp al « Z(G,), (a,c)? « N and (a,)P = 1, Similarly for bye Now let Gs = <a,,b5> where as and b, are chosen such that: a, Z(G,)<*a Z(G 2 ), b, Z(G) = b,Z(G,), induces 2 58 the isomorphism required by the definition of isoclinism. In an analagous manner form H, = <(azr)M, (b,s)M> =<ay,, b>, and { Ho = 1G53 A It will now be shown that every element of Ay has a unique ~ A oF, representation in the form ay by 9) with 9, € H}- For if ay by 8) a, Bp Of, - a 6,-6 a -a¢ — 2— -1l 24 1 -1 _ =a, D8, then a, =b, 2 6,8, and therefore a, 14 Z(H,). % > % m1 2 Pp, _#, But then ay = 1 and so ay ay Similarly by = by and hence 8) = 95. The identical argument shows that every element of H, has a wig’ 8 unique representation of the form as Oh with B, € HS. Let c((ay »bY)) = (a a5, 53). Then @ induces an isomorphism between Ay and Hi. Consider the mapping: By an argument identical to that given in Theorem 4.2 Ay = H, and hence iG, tat iG} A’ The detailed description which P. Hall gives of the isoclinism families involves a discussion of the so-called "stem" groups of the family. These are the family members of least order. Stem groups may also be characterized by the property: Z(G) ¢G'. In showing that there always are such groups the following construction is used: Form the direct product of a group in the family with a finitely gener - ated free abelian group and consider an appropriate factor group of a subgroup of this direct product. In this manner a stem group can 59 always be constructed. Hence if A is allowed to be the free group of one generator then there always exists a stem group G, such that for any group G~G,, (Gt = (Git A’ 60 CONCLUSION Several rather general questions have been raised in the preceding four chapters. In Chapter I the problem of determining necessary and sufficient conditions for {Gi = iG was mentioned. The interesting case to examine is that for which G) and Gs are indecomposable. Since G, is not an ingroup of G, this question is equivalent to whether or not the basis set of our decompositions theory is "independent". It was shown in Chapter I that independence does not hold, in general since both the quaternion and dihedral groups of order eight are inde- composable and yet have the same closure. Whether this is justa peculiarity of 2-groups or perhaps irregular p-groups is not clear. One certainly ought to be able to settle this question for indecomposable p-groups of class two with p # 2 since these are given explicitly in Chapter II. One rather obvious direction in which to carry the technics here developed is to study the decomposition question for finite groups of composite order. It follows from Theorem 1.4 that a decomposable group is either a subdirect product or it contains a normal abelian p- group. At this stage, however, it is not clear how useful it is to know that groups with no normal abelian p-subgroup are either indecomposa- ble or subdirect products. Certainly this question might be profitably pursued. One might begin with composite order groups which have abelian Sylow subgroups, for example. The unanswered question in Chapter II concerns the defining relations for indecomposable 2-groups of class two. Itis not very surprising that there is a greater variety of indecomposable 2-groups 61 of this type since almost any compilation of p-groups of certain orders invariably must deal with p= 2 ina special way. One strong possi- bility is that this greater variety of 2-groups is absorbed by considering the closure of a 2-group and not the group itself. For example, it is shown in Chapter II that the quaternion group of order eight doesn't fit in with either of the two general forms of defining relations given. But if D and Q are the dihedral and quaternion groups of order eight then {Q} = {D} and the defining relations for D do fit one of these given forms. Hence it seems plausible that if G is an indecomposable 2-group of class two then there exists a group in the closure of G whose defining relations fall in one of the two given categories. The main result of Chapter III provides a tool for handling a good part of the decomposability problem for p-groups of class three. A complete characterization of the indecomposable p-groups of class three seems quite accessible. The most difficult case arises when G/Z(G) is a subdirect product. The notion of A-closure which is linked up with the relation of isoclinism might be investigated directly as a classification procedure for p-groups. This assumes of course that all of the hypotheses of Theorem 4.2 are indeed necessary. It certainly would be desirable to settle this question. {Aa} A<B A<B ACB ACB {Acs <a,, [a,b] NIG “@> a 62 GLOSSARY Set of elements Aw, & contained in some index set. A is aningroup of B. A is a proper ingroup of B. A is a subgroup (subset) of B. A is a proper subgroup (subset) of B. The closure of {Au}. The group generated by the set {a,, ose) aj: Anelementin A XB with ae A, be B. N is a normal subgroup of G. A and B are isomorphic. The order of the group (element) A. The index of B in A. a is congruent to b modulo n. The intersection of A and B. The group generated by the subgroups A and B. The center of G. Z(G)/Z, _,(G) = Z(G/Z._,(G)) with Z(G) = Z,(G). The class of the nilpotent group G. The derived group of G. The n'} derived group of G with co) =G!, The integer a divides the integer b. The integer a does not divide the integer b. The Frattini subgroup of G. The nth higher commutator group of G with G2) = Gt, The commutator of a and b, i.e. avi pap. 63 A~B A is isoclinic to B. {Gy The A-closure of G. V{iGa 5 Set union of the sets Ge. oe 64 REFERENCES G. Higman, Some remarks on varieties of groups, Quart. J. of Math. (Oxford) vol. T0 (1959) pp. T65-178. H. Zassenhaus, The Theory of Groups, Chelsea Publishing Company, New York, 1949. ©. Ore, Structures and group theory I, Duke Math. J. vol. 3 (1937) pp. 149-174. P. Hall, A contribution to the theory of groups of prime-power order, Proc. London Math. Soc. vol. 36 (1933) pp. 29-95. W. Burnside, Theory of Groups of Finite Order, 2nd ed. Cambridge, 1911. P. Hall, The classification of prime-power groups, J. Reine Angew. Math. vol. 182 (1940) pp. 130-141.