The Representation of a Lattice-Ordered Group As a Group Of
THE REPRESENTATION OF A LATTICE- ORDERED GROUP
AS A GROUP OF AUTOMORPHISMS
H. Jean Springer B. Sc. , London (university College of the West ~ndies), 1962
A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF
THE REQUIREMENTS FOR THE DEGREE OF
MASTER OF SCIENCE
in the Department
of
Mathematics
@ H. JEAN SPRINGER 1969 SIMON FRASER UNIVERSITY
JULY, 1969 EXAMINING COMMITTEE APPROVAL
,> ,> N. R. Reilly, Senior Supe6isor
A. H. Iachlan, Examining Committee
Berggren, Exam C5mmit t e"e.
T. C. Brown, Examining Committee
D. Ryebuyn, Examining Committee (ii)
The purpo e of this study is to review some of the developments in.i: the theory of lattice-ordered groups closely related to the Holland representation for lattice ordered groups. In Chapter 0, bask definitions and results required throughout this study are reviewed. Chapter 1 contains a study of regular and prime subgroups of a lattice-ordered group and concludes with the very important Holland repre- sentation theorem. In Chapter 2, the Holland representation is used to derive the very nice result: %very lattice-ordered group can be embedded in a divisible lattice-ordered group. Finally, Chapter 3 contains a study of transitive lattice ordered groups of order preserving permutations on a totally ordered set and also a discussion of a class of simple lattice ordered groups. (iii) TABLE OF CONTENTS
Page
Introduction v Notation i viii Chapter 0 Fundamental concepts 1 Chapter 1 Regular and prime subgroups of 8 an 1-group Chapter 2 An application of the Holland representat ion Chapter 3 The 1-group of o-permutations on a totally ordered set Bibliography I wish to take this opportunity to express my appreciation to Dr. N. R. Reilly for his advice, suggestions and encouragement during the preparation of my thesis. I would also like to thank
Mrs. Joyce Simpson for doing the typing.
The support received fro grant No. A-4-044is also Lattices and gro?ap< seem to provide two of the most basic tools in the study of UnLversal Algebra. Also, algebraic systems endowed with a partial or total order are important in many branches of MaKhema%fcs ,, It is therefore not surprising that there should be LncreasEng interest in the study of lattice-ordered groups, Prior to 1941, only lattice-ordered groups which are abelLan or totally ordered had been studied; the most notable contribution madcp to that time was probably due to Hahn in 1907 who developed an embedding for abelian totally ordered groups. This embeddfng was later extended to fnclude abelian lattice-ordered groups by P. F. Conrad, J. Harvey and C. Holland in 1963, In 1941, Garrett Birkhoff published a paper which appeared in the Annals of Ma%heri~atiec( 1942) and in which he fnvest'igated properties of non-abelian lattice-ordered groups. This, no doubt, formed the basis for further investfgation and, since then, many of the problems and conJeetures listed in the conclusion of that paper have been resolved. In the study which follows, some of the more recent developments in the theory of lattice-ordered groups have been reviewed, and an attempt has been made to make this presentation self-contained as far as possfble, However, a basic knowledge of group theory has been assumed. Chapter 0 contains basic results and definitions which are used throwghcxk the study with references befng given whenever proofs are omitted. These results, and a general basic theory of lattice-orderee ~TG~Scan be found in BSrh%offls Sook on "Iattice theory" and fn Fmchts book on "Pa~.tfallyordered algebraic systems " . In Chapter 1, a detailed discussion of the properties of regular and prime subgroups of an, 1-group is presented. Prime subgroups are of particular importance in obtaining representations of lattice-ordered groups. For, if M is a prime subgroup of a lattice-ordered group @, then the set of cosets of M can be endowed with a natural total order.
It follows that if M is both prime and normal, then the set of cosets of M is a totally ordered group. Thfs property was utilised by Holland ig his representation theorem which is discussed at the end of this chapter. The Holland representation of a lattice-ordered group is a representation of a lattice-ordered group as a subdirect sum of ILK where P each K is a tra,nsitfve I-subgroup of the lattice-ordered P group of all order-preserving permutations on some totally ordered set. This answered a problem originally posed by Birkhoff in the second edition of his book on latticeI theory, Though this representation throws little light on tiinternal structure of a lattice-ordered group, it is an invaluable tool in the study of the nature and occurrence of lattice ordered groups. An examp1.e of this is given in Chapter '2 when an application of the Holland representation is used to obtain an elegant embedding theorem: re very lattice- ordered group can be embedded in a divisible lattice-ordered group". Finally, Chapter 3 contains a study of the lattice- ordered group of crezr-p~cszrvf~gperm~-Z%tfons oc a tctally ordered set. This chaptw is divided into two sections.
Section I contains a study of lattice-ordered groups of order-preservhg permutations oz a totally ordered set, which are transitive on %hat set, while, in Section 2, a class of simple lattice-ordered grcra-ps is discussed. That knowledge of the propertfes of the lattice-ordered group of order-preserving permutations on a totally ordered set yields important informstfon about lattice-ordered groups in general, is clear from the Holland represen$,ation theorem. (viii)
x and y are disjoint x and y are incomparable
The positive cone of a partially ordered group G
The set of all convex 1-subgroups of an 1-group G
The set of all 1-ideals of an 1-group G
The sct of all right cosets of C The set of all o-preserving permutations on the totally ordered set S The convex 1-subgroup generated by g The convex 1-subgroup generated by M and a
The cardinal sum of 1-groups G and H
The lexicographic sm of G and H ordered from the left
G is 1-isomorphic to H
The normalizer of M in G CHATTER 0
For a general theory of lattice-ordered groups, the reader
is referred to Birkhoff ("1 and Fuchs (8). Included here are some basic definitions and resuiSs which are used throughout
this study. In general, addftfve rotatfon is used unless otherwise mentioned.
Definition 0.1
A , G, is a set C such that G is a group;
G is a partially ordered set (p.0, set) under a relation *; If a, bcG and a&, then c+asc+b and a+csb+c for every
CEG.
A p.0, group which is totally ordered is an o-group,
Definition 0,2
(a) A lattice-ordered group (1-group) is a p.0. grouF which
is also a lattice under the relation s.
(b) If G is an 1-group, then a subset H of G is an 1-subga of G if and only if H is a subgroup of @ and H is also
a sublattice of @.
In a lattice, L, if x, ycL, then denote glb{x,y] by xAy
and lub[x,y] by xVy. Deffnftion 0,3 ,
(a) Let G and H be p.0. groups. Then an o-homomorphism 8
from G into H is an fsotone group homomorphfsm. That is to say 0 is a groAp homomorphism such Chat for any
x, ycG if xsy, then x0gy0.
(b) 8 is an o-isomorphism If 0 fs a 1-1 o-homomorphism.
(c) If G and H are 1-groups, then an 1-homomorphism 8 from
G into H is an o-homomorphism such that for any x, yeG,
(XVY)~= xevye (5)
(xny) 8 = xe~ye ( ii)
(d) An 1-isomorphism 0 from an i-group G into an 1-group H is an o-isomorphism from G into H such that (i)and (ii)
of (c) hold. When G and H are 1-isomorphic 1-groups L we write G=H.
I Remark: Clearly if G is an 1-group and if H fs a subgroup of G such that for each XEH, XVOEH then it follows easily that H is an 1-subgroup of G.
Some elementary properties of 1-groups.
L(1) In any 1-group G, addition is distributive on meets and joins. That is, ifa, x, y, bcG, then,
a+(xvy) = (a+x)~(a+y),(xvy)+b = (x+b)v(y+b) (I)
a+(xAy) = (a+x)A(a+y), (xAy)+b = (x+b)A(y+b) (2)
L( 2) In an 1-group G, if a, bcG, then
aAb = -(-av-b) (3)
L( 3) As a result of ~(2)the mapping 8:G+G, defined on an I-group G such lthat xe = a-x-n-5, a, b, XEG is a
1-1 mapping such that for x, ~EGwith xly then x0ry0. Also Ff xemye then this implies xry. Such a mapping fs sometimes termed a dual isomorphism.
In any 1-group G, a-(xvy) tb = (8-x+b)~(a-y+b) (4) this follows f~omthe dud of ~(2).
L( 4) In any 1-group C, the generalisations of (I), (2) and (4) hold. That is to say
a+(Vxo)+b = ~(a+x~+b) (5)
a+(Axo)+b = A (acxD+b) (6)
a-(Vxo)+b = (a-xoib) and dually (7) I
where o ranges over some finite arbitrary index set.
Definition 0.4
I If a is an element of an 1-group, G, then a+ = avo and I
- - I a = aho; a* is called the positive part of a and a is called I the negative part of a. I
Lemma 0.5
In any 1-group G,
Proof: In any 1-group, x-(avb)+y = (x-a+y)~(x-b+y)for every a, x, y, ~EG. Therefore setting x = a and y = b, the result follows.
Corollary 0.6
In a commutative I-group, G, a-tb = avb+aAb va, beG. Corollary 0.7 , -k - In any 1-group G, Pya,@, a = a +a .
Proof: For any ~EG,by substituting o for b fn lemma 0.5, the result follows.
Definition 0.8 In an 1-group G, if ~EG,then 1 a\ - absolute value of a = av-a.
Theorem 0.9 m an 1-group C, b!aa~, (i) 1 a 120, moreover 1 a\>o unless a=o.
(ii)a'A(-a)+ = o
(iii) 1 a 1 = a+-a- = avo-a~o= avo+( -a)vo.
Proof: See Birkhoff (I).
Lemma 0.10
If G is an 1-group, let G+ = {xa~:xzo]. Then G+ is the positive cone (or partial order) on G.
(i) If u, v, wa~*then uA(v+w) SuAvCuAw. ( ii) If usG+ then for any v, waC, uv(v+w) suvv+uvw.
Proof :
(i) Since u, v, WEG', then uA(v+w) EG+. Applying (2) of
L( 1 ) , u~v+u~w= ( UAV+U)A( UAV+W)
= 2u~(u+v)~(u-t-W) A(V+W) Now u~(v+w)lu,(v+w). Hence clearly uA(vfw)l2u, u-tv, u+w, v+w. Thus u~(v*w)s(uhv)+(u+wb).
( if) Applying ( 1 ) of L( 1 ) , uVv+uVw = (uvv+u)v(uvv+w)
= 2uv(u+v)v(u+w)v(v+w) Since uaG' and 2u, (v+w)r (uvv)+(uvw), then u,v+wg(u~v)+(u~w).Since G is an 1-group, the result foll3ws.
Lemma 0.11
In an 1-group G, ]a+b]s\al+lb/+lal for every, a, ~EG.
Proof: In any 1-group G, 1 a = a for each asG. Hence lal+Ib(+lal = / -a/+lbl+l-a1
= aVo+avo+bVo+-bvo+-avow0
2-a.vo+(a+b) vo+( -b-a) vo+avo by lemma 0.10
= -avo+I a+bl +avo
21 a+b 1 .
Similarly it follows that ( a-bl 51 a1 +I bl +I a1 .
Definition 0.12 S Two positive elements a and b in an 1-group G are called dis.joint (denoted alb) Tf and only if ahb = o.
Lemma 0.13
In any 1-group G, disjoint elements are permutable.
Proof: If a, b€G such that ahb = o, then clearly a+b = a-aAb+b = bva from lemma 0.5. But bva=avb=b-bAa+a-b+a. Thus a+b = b+a,
Lemma 0.14
Let G be an 1-group. If a, baG such that aLb, then avb = a+b.
Proof: If alb, then aAb = 0. By lemma 0.5, avb = bVa = a-a~b+b= a+b.
Definition 0.15
A p.0. group, G, is Archimedean if for a, ~EG,nasb for every
integer n implies a = o. The next theorem which is stated without proof is due to Holder. The proof can be found in Fuchs (8), P.45.
Theorem 0.16 ,(~older)
An o-group is archimedean if and only if it is o-isomorphic to a subgroup of the additive group of real numbers with the natural ordering. Thus, all totally ordered archimedean groups are commutative.
Definition 0.17
Let S be a totally ordered set. Then a 1-1 order-preserving mapping from S onto S is called an o-permutation (automsphim) . Now consider the set of all o-permutations on a totally ordered set S. This set is denoted by A(S) and is an 1-group under the following order: For ~EA(s), let f21exfzx WXES, where 1 = identity mapping on S. Verification that A(S) is an 1-group under this order is routine.
Definition 0.18
Suppose G and H are 1-groups:
The cardinal sum of G and H denoted by G0H is the direct
sum of G and H with the partial order defined by
(g,h)roag20 and h20 for gcG, hcH. To verify that WH is an 1-group is routine.
The lexicographic sum of G and H is the direct sum of G
and H with the lexicographic order defined by
(g,h)ro e either h>o or h = o and $20.
Then GxH is ordered lexicographically from the right
and the lexicographic sum is denoted by GxH.t- Similarly
if ordered lexicographically from the left we denote this ---P t- by GxH. Again the verification that GxH is an 1-graup
is routine. CHAPTER I
In this chapter, the basic results related to regular and prime subgroups of a lattice-ordered group (1-group) are stated and proved. Finally, using the fact that if C is a prime subgroup of an 1-group, G, then G/C is totally ordered, we discuss representations of 1-groups as groups of order- preserving permutations on a totally ordered set; the main result being the Holland representation. Unless. otherwise mentioned, the results are due to Conrad (5) and (6).
Definition 1 .I
A subgroup C of an 1-group G is convex if for any O Osxs-a this implies XEC. Definition 1.2 (a) A subgroup C of an 1-group G is upward directed if for every a, bcC there exists ccC such that agc and bsc. (b) A subgroup C of an 1-group G is downward directed if for every a, bcC there exists ceC such that csa, csb. (c) A subgroup C of an 1-group G is directed if it is both upward directed and downward directed. Lemma 1.3 For a subgroup C of an 1-group G, the following are equivalent: *- ( I) C is a convex 1-subg~oup; (2) C is a directed convex subgroup. of G; (3) C is convex and cVOEC fqr each ccC; (4) Let K(C) = {C+g:gG], the set of right cosets of C in G. If we define @+gsC+h to mean there exists ~EC with c+glh, then this defines a partial order on R(C) which is a lattice with (c+x)v(c+~)= ~~(xvy)for x, yeG and du.ally; (5) If caC, geG and 1g121cI then gsC. Proof: (1)*(2). Since C is a convex 1-subgroup of G, then for every a, bcC, avbcC and aAbeC. Hence C is directed and convex. (2)*(3). Since OcC, (3) is trivially implied by (2). (3)*(4). First we show that the order defined on R(C) is a partial order. Since OEC, and for each c+~ER(c), O+gsg, then it follows that C+gSC+g and I is reflexive, Consider C+gsC+h and C+hlC+g for C+g, c+~ER(c). Then there exist c,, C~ECsuch that cl+gSh and c2+hsg. Therefore c2ig-hi-c,. Since C is a convex 1-subgroup, then g-heC. Thus, C+g = C+h and so S is antisymmetric. If now C+xsC+y and C+ysC+z, with C+x, C+y, C+ZER(C), then there exist c,, c2aC such that c,+xSy and c2+ySz. Therefore, C2+C1+xsC2+y~z But c2+claC and so C+x C+xvy = (c+x)v(c+~). The dual can be shown similarly. Hence R(C) is a lattice. (4)*(5). If ceC and ga2 and (glrlcl, then, gl h 1 c I . Thus cA-c~~SCV-C. From = C+cA-cSC+g. But CVC = C. Thus C+g = C and geC. (5)*(1). If O Corollary 1.4 If A, B~@(G),where @(G) is the collection of all convex I 1-subgroups of an 1-group @ and if AGB, then the mapping A+x+B+x for XEG defines an lattice homomorphism from R(A) onto R(B). \ From the definition of the partial order on R(A) and R(B), such a mapping is well defined. Ti?o surJectiveness follows since AGB. Definition 1 .5 A convex 1-subgroup C of an 1-group G is an 1-ideal if C is also a normal subgroup of G. . Corollary 1 -6 , Let ME&(G),where &(G) fs the collection of all 1-deals of G, then the canonical mappfng of the subgroups of G containing M onto the subgroups of C/M fnduces a bijective correspondence between the convex 1-subgroups (I-ideals) of G containing M and @(G/M) (respectfvely &(G/M)) Remark: If in particular CE~(G),then in (5) of lemma 1.3, R(C) is ml-group. Notation: Consider aG' and S a sub-semi-group of G" such that OES. Then we denote the sub-semi-group of G + generated by S and a by Lemma 1 .7 (Clifford) If M is a convex 1-subgroup of an 1-group G and if ~EG+\M, then, + C(M,a) = {XEG:lxlg p for some pea ,a>] is the smallest convex 1-subgroup of G containing M and a. If' a, beGC\M, then c(M,~)Ac(M,~)= c(M,~A~). In particular, when M = 0, ~(a)= {XEG:lxlina for some positive integer n]. Proof: If x, yeC(M,a).than lxlhp and Iylsq for some p, qe<~+,a>. Applying lemma' 0. I I , we get ( X-~I~1 XI +Iy)+(XI gp+q+p~a+,a>. Therefore x-ye~(M,a)and so c(M,~)is a group. Nowif lglslcl for ~EGand csC(M,a), then there exists rca+,a> such that 1 cl Sr. Thus (g1 hr for raa+,a>. Hence ~EC(M,~)and so c(M,~)is a convex 1-subgroup conta.ining M and a and must be the smallest such. Now, consider O U, V, WEG', UA(V+W)LUAV+UAW. Hence x is less than or equal to a sum of positive elements of the form m.An miAb, aAn 1 j3 j' anb, But all such elements belong to c(M,~A~)hence c(M,a)flc(M,b)rc(M,anb) . To obtain the other inclusion, consider xe~(M,aAb). Then xgml+aAb+m2+aAb+. . .+mr-l+aAb+mr, where micM + , l Corollary 1.8 /' Let K = ~{cE@(G):c~{o]].If K {O], then G is an o-group and K is the convex 1-subgroup of G that covers zero. Proof: If K f {O] and G is not an 0-group, then there exists a, baG such that a and b are strictly positive elements and anb = 0. Since ~(a),c(~)E@(G), then ~rC(a)nC(b)= ~(a~b)= ~(0) = {O]. This contradicts the - 13- hypothesis. Therefore G mus; be an o-group. Corollary 1.9 If G does not have a proper convex 1-subgroup, then G is o-isomorphic to a subgroup of the reals, R. -Proof: By corollary 1.8, G is an o-group with no proper convex 1-subgroups. Consfder a, beG such that nasb for n = 0, -+I, -+2, ... Then b@(a). Hence ~(a)is a convex 1-subgroup of G and ~(a)# G. Thus ~(a)= {O') which implies that a = 0. Thus G fs archimedean. Hence by Holderfs theorem (~uchIs p. 45)) G is o-f somorphic to an additive subgroup of the reals. Deffnition, 1 ,I0 A convex 1-subgroup M of G is called ,regular if there exists gcG such that M is maximal with respect to not containing g, and in this case M is said to be a value of a;. &ernma 1.11 Each convex 1-subgroup ofan 1-group G is the intersection of I regular convex 1-subgroups of G. Each 0 # geG has at least one value. -proof: Let CE@(G). Let geG\C. By Zornls lemma, ~ME&(G)such that M is maximal with respect to containing C and not con- taining g. For, consider CI = CS:S~(G), SIC]. Let &a,L = {si:iel] linearly ordered by set inclusion. Then, g# U Si2C and i€I U s%E@(G) by the linear ordering. Therefore U Si is an ?LEI i~1 upper bound of ;e in a and so by Zornls lemma, 0 has a maximal element. Let M be a maximal element of a. Then M is €5 g regular and is a value of geG\\C. Now consider the set of all such M,, that is {M *DEG\C).Then, CEM for each gaG\C 0 ga= g implies that C.~fl[M,:geG\~]. If X&{M :gaG\C3, then xaMg for t3 €5 each ~EG\C. Thus X#G\C and so xsC. Hence, ~[M~:~EG\c)-cc and so n[~g :gsG\~] = C. Definition 1 .I2 An element a of a lattice L is called meet irreducible if a is not the greatest element in L and if a<~bb(ba~and b>a). This is more restrictive than the usual concept of finite meet irreducible (b, EL, b>a, c>a*bhc>a) . Theorem 1 .I3 Let ME@(G), then the following conditions are equivalent: M is regular; There exists M*G(G) such that MGM* and M* is contained in every convex 1-subgroup of G that properly contains M; M Is meet irreducible in @(G) which is a lattice under set inclusion. If MdG, then each of the above conditions is equivalent G/M is an o-group with a convex 1-subgroup that covers zero. Proof: , (1)*(2). Suppose M is a regular convex 1-subgroup and let M be a value of g&, Let M* = ~{C~~(G):MCG].Then M*E@(G) and M*'C for every CE{CE@(G):MCT].Since M is regular, for every CE{CEG(G):MCC],g~@. Thus ~EM*\M and so Ma*. (2)*(3). We have that M*E@(G) such that McM* where M* = n{Cc@(G):MGC]. It follows immediately from Definition 1.13 that M is meet irreducible Ln @(G). (3)*(1). By lemma 1.11, M is the fntersectlon of regular convex 1-subgroups of G. So, if M is meet irreducible, then M must be regular, (4)*(2). If MAG, then ME&(G) . Assuming (41, let K = M*/M be the convex 1-subgroup of G/M that covers zero. Then, M*E@(G) such that McM* and $ICE@(@) such that McCeM*. Thus, M* is contained in every convex 1-subgroup of G that contains M. I 1 (2)*(4), If M satisfies (2), then M*/M = ~~{c/M:cE@(G),MCC]. Since MCM*, M*/M fl M/M. By Corollary 1.8, G/M is an o-group and M*/M is the convex 1-subgroup which covers the zero. Corollary 1.14 If M is a regular convex 1-subgroup of G and a, ba2\M then ~A~EG'\M. Proof: From lemma 1.7, c(M,aAb) = c(M,a)fl@(~,b)cM.Since M is regular, ZM*~(G) such that McM* and M*W for every CE{CEG(G):McC]. Thus, C(M,aAb) = c(M,~)~Ic(M,~)~M*. If aAbcM, then M - ?(~,ar\b);l~*whkh implies that M = M*. $4 This is impossible. So a~bji~and aAbcG \M. In the following theorem, the notton of a prime 1-subgroup is introduced and a number of equivalences proved. The equivalences (49, (59, and (6) have been proved by Holland while those of ( 1 ) , (3Q,(4) and (8) have been proved by Johnson and Kist ( 12), Theorem 1.15 Let ME@(G) then the following are equivalent: If AnBcM where A, BE@(G) then AEM or BGM; If McA and McB where A, BE@(C) then McAnB; If a, ~EG+\M, then a~be~+\~; If a, bf&\M, then aAb>O; The lattice P(M)of rfght cosets of M is totally ordered; The convex 1-subgroups of G that contain M form a chain; M is the intersection of a chain of regular convex 1-subgroups. If MAG, then each of the above is equivalent to G/M is an o-group. Proof: (1)3(2). If MCA andMcBforA, B E@(G), thenMEA!lB. If M = AnB, then from (I), A = M or B = M. This contradicts the hypothesis and so McAnB, (2)*(3). If a, beG+\M, then C(M, a~b)= c(M,~)~c(M,~)=M since M~c(M,~)and MCC(M,~). Thus aAb@ and so aAbf~+\M. (3)*(4). This follows trivially. (4)*(5). Consider M+a, M+~E~(M)with a, bG\M. Then - - a = Z-taAb, b = c+aAb where aAb = s for then, - - - - aAb = (~+aAb)A(&-aAb)- aA'fj-taAb = anb. Since aAb = o, from - - (4), aEM or EM. Suppose acM. Then M-ka - M+aAbsM+b. Similarly, if EEM, then we get M4-ba43-a and so tt follows that @(M) is totally ordered, (5)*(6). Assme (5) then suppose that the convex 1-subgroups of G containing H do not form a chakn. Then, BA,BE~(G) such that McB and A\(B. Consider o (7)=3(1 ) . Asswne (7) and suppose that 3A, BE@(G) such that AnBGM, A$M and B$M. Let hi:iol] be a chain of regular (, convex l-subgroups of G such that M = fl M,. Choose afAC\M LEI I and ~EB+\M. Then 3j~Isuch that a, b{~~,that is, a, beG?'\MS. By Corollary 1.14, ~.A~EG*\M~,but A~BG(G) and so o Definition 1.16 A convex 1-subgroup of an 1-group G which satisfies any of the conditions ( I) through (7") tn the preceding theorem is called prime, The above definition is certainly equivalent to the following: An 1-subgroup M of G Ls prirne if whenever a,Ab = o where a, ~EGthen am or bcM. Tkis definition is analogous to the ring theoretis def inition of prime ideals "An ideal I of a ring R is prime if abcI*a~Ior ~EI". Remarks ( 1 ) It follows immediately from corollary 1.14 and (3) that every regular convex 1-subgroup is prime. (2) From condftfon (&), it follows that the partially ordered set of prirne subgroups of G is a root system. [A p.0. set D is a root system if for each bed, {a~o:az~]is totally ordered], (3) From condition (7),the intersection of a maximal chain of regular subgroups is a minimal prfme subgroup. Thus every prirne subgroup contains a minimal prime subgroup. (4) Every 1-automorphism IT of G induces an 1-automorphism on the lattice @(G) and so also on &(G), If M is a prirne subgroup (respectively regular), then Mn is also prime (respectively regular) . (5) The prime convex 1-subgroups of an 1-group G can be used to represent G as a group of o-permutations of a totally ordered set. The representation of 1-groups in this manner is due to @.Holland ( 10) and is discussed later. Definition 1.17 , A subgroup H of a direct sum of groups ITG is a subdirect sum A of nGA if for any x X xC-A 9cfH having x X for its component in That is to say, the prcjection IT is surjective. GA. A :H-"GX Definition 1,18 An 1-group G is representable if there exists an 1-isomorphism o of G onto a subdirect sum of nGX where each GX is an o-group. The pair (o,nGh) is called the representation of G. Definition 1.19 A group @ is an 0-group if G admits at least one total order. Example: All, free groups are 0-groups (~eumann(I 7) ) . Elementary properties- of representable groups ~(1)If a is an I-isomorphism of G into TK X ' where each K h is an o-group, then (o,nGh) is a representation of G with Gh = prAGo. [pr A Go denotes the hth projection of GO]. Proof: Since o is an 1-isomorphism and each KX is an o-group for each X, then pr X Go = GX is an 1-subgroup of Kh for each X is an 1-subgroup of ITK~ and hence is an o-group. Also, TTGh and hence (o,n~~)is a representation of Go ~(2)Every 1-subgroup of a representable 1-group is repre- sentable. Every cardinal sum of representable 1-groups is representable. , Proof: Let J e a re~res?ntable1-group. Let (0,l-r~~)be a representation of G, Lnt H be aKy l-subgroup of G. Consider J:H+G, the inclusion mapping. Then j is an 1-isomorphism of H into G. Hence, Jo:H+nDA is a? 1-isomorphism into nG A' BY P( I), since each S is an o-groq, ther-(2o,nH ) 1s a repre- X A sentation or H with HX = prA~(jo).It is easFly shown that a cardinal sum of representable I-groups is representable, for ff A and B are representabi.e 1-groups with r~presentations oaA(D~,ITB~) respectively, then (~~~o~,n~~~rr~~)is a representation of A@B, the cardL~~,lsum of A and B. (~3)A group G (not ordered) admits the structure of a t representable group if and only if it is an 0-group. roof:(+) If G is sn 0-group, G can be totally ordered and so admits the struc,ture of a representable group. () If G is representable, let (0,I-r~~)be a representation of G. Defina a well-order on A and then define a lexicographic order on nGA as f oliows : g = ( gh, ) >0 if g >O where h 1 . . . , . A \ fs the smallest index fn the well-ordering of A. This defines a total order on nGA and so on Go and since o is ar. isomor- phism, then G must be an 0-group. ~(4)G is representable if and only if it admits a class of prime normal subgroups whose intersection is lo 1. Proof: (-e=) Let [M :XcA'j be a class of prime normal subgroups A whose intersection is [o). Then (o, n G/M~) is a representatior? XeA of G where o:G+ n G/MA is defined by go = (. . . ,MX+g,. . .). XeA Indeed, from theorsm 1.15, each G/M~ is an o-group. If = the zero for G/M~, M +g = M go o where o denotes n then h h AEA for each LEA. That 1s ~EY$ for every heA. Bvt fl MA = [o] by hypothesis and so g = o. Thus afs injective, AEA ( o,rrG/~ ) is s representation of C. X () Let (0,rr~) be a representation of G. If IT is the X h proJection of ITG onto @ theri, DSIT~is an 1-homomorphism h A' of G onto G 1" Let Ker om = MA. A Then Mh-aGe If a, ~EG\M~then a(oen ) , b( aorrX)EG~. But F fs an o-group and so a( wonh) and h h b(o0nh) are comparable. Thus, MX+a and Mh+b are comparable for every a, bG, Therefore G/M is an o-group and so M is h X prtme. Also the diagram below commutes. where canonf cal, and 8 Hence n MXcKero = {o) since o is an isomorphism. Thus hcA {M ;hcA] is the requfred family of prime normal subgroups of X G. Theorem 1.20 For an 1-group the following are equivalent: ( 1) G is representable; (2) G admits a famfly of normal prime subgroups whose Lntersection is go]; a~[-xfa-t-x) = o * a = o a, xG; a, b, XEG, ahb - o * ah(-x+b+x) = o; The intersection of all conjugE&es of a prime subgroup is prime; Every minimal prima subgroup is normal; The conjugates of a prime subgroup of G are comparable; Every regular i-ideal of G is a prime subgroup. 1-ideal L of G is regular, if there exists gaG such that L is maximal among those elements of L(G) which do not contain g) '4 Remark; The equivalences of ( I), (3) and (4) are due to Lorenzen, the equivalence of ( 1 ) and ( 6) due to Sik, ( 1 ) , (6) and (7)due to Byrd and ( 1 ) , (8) due to Conrad. Before proceeding with the proof of the theorem, two lemmas on conjugate subgroups of 1-groups are stated. The proofs are routine and are not included. Lemma 1.22 X If MED(G) and if, for XEG, M denotes the conjugate subgroup of M with respect to x, then ~EG(G). Lemma 1 .22 If M is a prime 1-subgroup of G, then for any xeG, I?? is a prime 1-subgroup of G. Proof of theorem 2.20: , (l)a(2) has already beer^ dstablished (see ~(4)preceding the theorem). (I)If G is an o-group, (3) follows immediately since if an(-x+a+x) = 0, eit.her a =o or -x+a+x = o which implies that a = o. Thus, an (-x+a+x) = o =. a = ova, xeG. If G is a cardinal sum of o-groups, the result follows easily by considering components. Simflarly, the result follows if G is a subdirect sum of o-groups, Hence, if.-G 1s representable, let (a,n-G ) be a representation, and suppose a~(-x+a+x) = o h I for a, xeG. Then, aaA(-x+a+x)a = o * aa = o because aa is an element of the subdirect sum of nG X" Therefore a = o since a is an l-isomorphism. (3)3(4). Suppose anb = o. Then a, b2o. Bence, -x+b+x, x+a-xzo VxG, T3us an( -x+b+x) ~(x+a-x)nb - o . Let g = an-x+b+x, Then x+g-x = x+[an(-x+b+x) ]-x Therefore gAx+$-x = an(-x+b+x)n(x+a-x)nb = o. From (3) this implies that g = aA(-x+b+x) = o, X (4)*(5). Let M be a prime subgroup of G. Let J = flM = XEG intersection of all conjugates of Me Then JAG and JE@(G), for, if MEC(G) then, for all xaG, $E@(G) by lemma 1.21. Thus JE&(G). To show that J is prime, we argue by contradiction. Suppose J is not prime. Then 3 a, baGf\J such that aAb = o. Consider that for some b&$. Then since aAb = o, a~-g+b+g = o for all ~EG(from (4)). By lemma 1.22, IV? is prime since M is prime. Therefore (M~)~= -g+$+g is also a prime 1-subgroiip and, since bjd?, -g+b+g#~x+g Thereyore aa&fYige Th!rs af fl = J" This is a g EG contradiction. Therefore J is a prime 1-subgroup of G. (5)3(6). From (51, every mfnfmal prime subgroup must be the intersection of all its conjug3tes and hence must be normal, t M be a prime subgroup of @. Let M be a minimal prime subgroup contafned in No Then, VgeG, M - -g+M+gG-g+M+g. Since M is prlme, the convex 'r 1-subgroups contaLning M form a chain. Hence N and -g+N+g must be comparable. (7)3(8). Let MEZ(G) such that M ,is maximal among the elements of d(~)which do not contafn acG, (M is a regular 1-ideal of G.) Then 3 a value N of a such that MEN. Let T = fl 3. XEG Then MGT. But T&(G) and a#~*ajd~. Since M is maximal in &(G) with respect to not containing a, then TsM. Thus M = T. But, from (7) every conJugate of a prime subgroup of G is comparable, Therefore T is the intersection of a chain of regular subgroups and so is prime. HenceM is prime. (8)*(2). Let 8 be the family of all regular 1-Ideals of G. Then by (8), each M~misa prime subgroup. Also, fl M = [o] Me m and so (2) is satisf fed. Corollary 1.23 Every commutative 1-group is representable. Proof: This follows immediately from conditions (4) of theorem 1.20. For, if G is a commutative 1-group, then, Va, b, XEG, aAb = o=mA(b+x-x) = o*aA-x+b+x = o. Hence G is representable, Corollary 1.24 If G is representable and CGZ(G)then G/C Es representable, hat Is a homomorphfc image of a representable 1-group is representable), Proof: Using condit,ion (4) of the theorem, suppose C+a, C+~EG/C such that C+anC+b - C+aAb = C, We can write - - a = anb+x, b = aAb-6 where ahb -' o, Then, C+a~-(~+x)+(~+b)+(C9x> = (~+a~b+g)A(C-x+b+x) = ( Wanb-kz)A( ~-x+anb-+~+x) - c+~Ac-x+~+x( since anb EC&) = c+(~A-x+~+x)= C, - - This follows from (4) since aAb = o*%-x+F+x - o, G being representable. Thus G/C is representable. Corollary 1.25 Let G be a representable group, M a regular subgroup of 6, M* the convex 1-subgroup of G which covers M. Then the normaliser of M in G is also that of in partfcular, M&* and M*/M is o-isomorphic to a subgroup of the real numbers). Notation: For a group G and M or subgroup of G, denote the normaliser of M in G as NG (M). Then N G (M) = {~EG:~+M-~GM]. Proof: Since M%- covers M, then N~(M)GN,(M*). Consider LT ~EN~(M*). Then -a+M*ij. = MX, for M%N~(M*). By condition (7)of theorem 1.20, every conjugate of a prime subgroup is comparable. Hence since M is regular, M is prime and so either -aiM+aGM or MG-a+M+a. If. -aCM+ar;M, then aeNG(M). Now considering the second case, we get MG-a+M+ac-afM*+a = M*. Since MX covers M, then M = -a+MM. and so aeNG(~).Hence N~(M)= N~(M*) as required. Remark: It follows from corollary 1.25 above that if M is a maximal convex 1-subgroup of a representable group G, then M&. To see thfs, observe that by corollary 1.25, N~(M)= N~(G)= G. The Holland representation of an 1-group as a group of permutat ions Lemma I .26 Let G be an 1-group and C a prime convex 1-subgroup of G. If we define a mapping n from G into the group of all permutations of P(C) such that (~+x)gn= C+x+g where x,geG, then for each geG, ~~EA(P(c)).Also n is an 1-group homomorphism from G into A(~(c))= the 1-group of o-pres~rving permutations of P(C). Proof: By theorem 1.15, since C is a prime convex I-subgroup of G, 4(~)is totally ordered. To show that, for each gcG, ( c+x) gn=( Cd-y) gn, then C;.+xag=C+y+g. Hace C+x = C+y and gn is injective. For any C+XER(C), C+x = C+x-g+g = (c+x-g)gn. But C+X-~ER(C) and so gn is surjective. gv is order-preserving since if C+xgC+y for C+x, c+~GQ(c), then 3c~Csuch that c+xSy. Hence c+x+ggy+g and so C+xSgIC+y+g. That is (~+x)gn~(C-t-y)gl-r.Thus ~~EA(R(c)),To show that n is an l-group homomorphism, consider g, ~EG. For any c+xE?(c), (c+x)(g+h)n = @+x+g+h= (C+x+g)hn = (~+x)gnhn. Thus (g+h)n = gnhn and n is a group homomorphism. Wow, if 1 is the identity fn A(?(C) ) , we must show that for any ~EG, gnVl = (gvo)~. Take any ( C+X) EP( C) . Then ( C+X) ( gl-rv 1 ) = c+x+gvc+x = c+(x+g)vx = c+x+(gvo) = ( c+x) ( gvo)n . Thus gnVl = (g~0)l-r. The dual is shown similarly. Hence n is an l-group homomorphism. L Remark ( 1) : Gn is transitive on P(C). To see this, con- it sider any Ctx, c+~E~(c)and notice that @+y = C+x-x+y , = (c+x)(-x+y)n. Thus Gn is transitive on ?(c). (2) Ker n = fl -x+C+x, To see this, consider geKer n. XEG Then (C+x)grr = C+x VC+XE~(C). Then C+x+g = C+x and so x+g-XEC. Therefore, g~-x+C+x VXEG and so Kernc fl -x+C+x, XEG If g~ rl -x+C+x, then, xzG, 3cxeC such that g = -x+cx+x, XEG and so (c+y)gl-r = (C+y)(-y+c +y)n for any C+ycG. Therefore, Y ( c+y) gn = C+c +y = C+y. Thus gcKern and Kern = n -x+C+x. Y XEG (3) If in addition CaG, then CE&(G) and the diagram below commutes: where 6 is the canonfcal I-homomorp?.,i.sm, i the fdentity mapping, G/CWC;TT and n* is defined by (C+g)n* = grr. Clearly, if CAG, G/C is an o-group. Also, for any x, g~@, (@+x)(~+g)rr* = (~+x)g&rr*= (~+x)grrand the diagram commutes. Theorem 1.27 (~olland)The main embedding theorem. An 1-group G is 1-isomorphic to a subdirect sum of n5where each Kh is a transitive 1-subgroup of the 1-group of all o-permutations of a totally ordered set Tho Proof: Let ?ll = {ME@(@):Ma minimal prime subgroup of GI. Then, for Mep, @/M = p(M) is totally ordered. From lemma 1.26, n~ :G+A(R(M)) is an 1-homomorphism and GnM is transitive on R(M). So, define cp:+ I-T A(R(M)) such that gcp = ( . . . ,gnM,. . . ) . Then Mm cp fs an 1-homomorphfsm. cp fs also injective, for since each Men is a minimal prime subgroup, then M = n MX. Thus XEG M = Ker rr Also, for any g, ~EG,gcp = hcp-grr M = hnM, VMeR M* @(g-h)E Ker nM=M, VMER e(g-h)~n M MeR But n M = {o] = intersection of minimal prime subgroups of G. Men7 Therefore gcp = hcpog = h. Thus cp is an 1-isomorphism of G into IT A(R(M)) and the~f~refulfills the condftions of the McR theorem. Remark ( 1 ) : For the class of minimal prime subgroups of one can take any class of prime subgroup of 3 having as their intersection {o]. (2) If G is representable, then every mfnimal prime subgroup of G is normal. Hence the diagram below commutes. where tjM:e~/M is the canonical 1-homomorphism, i the identity map, cp the mapping defined in theorem 1.27, and cp* defined 5y Y 2 (. . ., M+g,. ..)cp* = (. . . ,(M+~)IT+~,.. .) where IT*~:G/M-A(R(M)) i I is defined by (M+x)((~+g)n*~)= M+x+g as in remark 3 following lemma 1.26. It -is immediate from that, remark that the diagram commutes. Theorem 1 .28 (~olland) Every 1-group is 1-isomorphfc to an 1-subgroup of A(T) where T is a totally ordered set. Proof: Let R = {ME~(G):M a minimal prime subgroup of G]. -c..l Define a well order $ on R. Let T = {M+X:MER,xeG]. Define a partial order 2 o,? T such that M+xlN-+y If M for each gG, define go as follows: (M+x)ga = Mex+g. Then as fn lemma 1.26, it is easily verif fed that o is an 1-isomorphism of G into A(T) . The following theorem, a reformulation of theorem 3 in Holland ( 10) answered the question %%at 1-groups are trans it ive groups of automorphisms of totally ordered sets?'" Theorem 1.29 For an P-group G, the following conditions are equivalent: ( 1) 3 an I-isomorphism ~:C+A(T) where T is a totally ordered set such that Go 2s transitive on T; (2) 3 a prime 1-subgroup C oQ G such that {o] is the only normal subgroup of G contained in f; (3) 3 a prime subgroup C of G such tha.",o? ks the only 1-ideal contained in C; Proof: (2)*(3). Assuming (2), if 3H#{o] such that C~HE&(~),thw H4G and this contradicts (2). (3)*(1). Assume (3). Let XG-+A(R(C)) be the l-homomorphfsm described in lemma 1.26. Then Ker IT c il cX = {o] (by (3)). i xeG Thus Ker n = {o] and IT is an 1-isomorphism of G into A(R(@)) lemma 1 26, Gn is transitive)on ?(C). Thus we have ( I), ( I)Let ~:C+A(T) be an 1-isomorphism such that Go is transitive on T where T is a totally ordered set. For teT, let Ct = {ga~:t(go)= t]. Then Ct is a prfme convex 1-subgroup. To see this cotice tY~ai3.t 00 - 1 and so oeCt. Also if a, bcCt, then t(a-b)o = t(ao)(b-'a) - t. So a - bhCt. Therefore Ct is a subgroup of G. If ceCt, then t(cvo)o = t(cov1) = t(co)vt = t. Thus cVoeCt and Ct is an 1-subgroup of G. Ct is convex: Consider o therefore g+c-g jd C t and c#-g+~~+g= ctg. Hence n Ctg = lo]. gsG This completes the proof, Corollary 1 .,3O Let L be the intersection of all non-zero 1-deals of G. If L f 0, then condition (1) holds. In particular, a simple 1-group is I- isomorphic to a transitive :-subgroup of the group of all o-permutations of a totally ordered set, roof: L = n~ f {o]. Consider o aeE and let C be of x E&( G) a value of a. Then, a j! fl Cg = {oj since this is an 1-ideal ga which does not contain Lo Also, C bering regular is also prime and hence P(C) is totally ordered and a:G4A(?(C)) is an 1-isomorphism with Go being transitive on ?(C), Corollary 1.31 Let @ be an 1-subgroup of A(T) which is transitive on T. If G is representable, then G is an o-group. Proof: From theorem 1.29, 3 a prime 1-subgroup C of G such that {o is the only normal subgroup of @ contained in C. But C contains a minfmal prime subgroup M of G and since G Is I I representable, then M&. Hence M = {o] and G = G/M is an o-group. Remark: Conversely, we have that if G is an o-group, then G is 1-isomorphic to a transitive 1-subgroup of A(T) where T is a totally ordered set. This is clear for if pG denotes the group of right translations on G, then pG is an 1-subgroup 1 of A(G) . Also, pGsG and pG is transitive on G. Definition 2,1 4 integer n, there exists hcC sweh t8h3,t nh -. g B, H. Neumann ( '6j Y.es proved sh5.t every ,:;romp can be embedded fn a divfsi5i.e &rou&. As 3.n rppliea+,for, of his representation theorem for 1-grou~s,HoLand ( 10) proved the analogue to this, namely thdt 'wrery I-grou~can be embedded in a divisible 1-group"'. However, 'his proof depended on the existence, Por arbitrsrfly large ordinsls a, of q oe -sets of power K a set-theoretTe ressriction. exists a' here an qa-set of power K exactiy wi-er K is a reguiar cardinal K ce a such that 2 '~8~whene1.c-r B Definition 2,2 For a totally orderzd set i, znd for A(I), t?-e l-grou~of 0-preserving permbt2tlons of 5, iet gc~(,). Tksn for any xc~,an, If(is given by W rl m I0 (x) = [y~5:3m,n "Ltegers such that xg sysxg 1. .?2 An interval containing more than one pofnt is -a supporting and the unfon of suppol-ting fntervals is the I From this deflrLitio~,it is clear that for ~EA(&)ard any xcE, I,(x) is H eo~exsukset of L and that I,,(x)g - Sp(x). €5 6 w Also, if ycE (x), $her 1 (xp - E_(y), Her-ce these interv%ls 6, e, determine an equik21ecee relation on L 3cficed by: Let S be a totaliy ordered set in which any two non-trivial closed intervals are isornorpi.hfc., TLez A(S) is divisible. Proof: Let ~EA(S) and con side^ n an integer such that n>o, Witho~tloss 01' generaiity, we mky assume $\I and also that g has only one skpporting interval. By the hypothesis, if x, ycS and if x 3 1somorphisms Now, let cpX:(ao,an]--)(alq g] be the extension of ali the for 14irn. Then cp* is isomorphism. By assumption, the 'Pi an support of g = I (a ). Define f as follows: g 0 where if x~9(a 1, thc-r, m(x)b is tbe unique fnreger not g 0 I I Fm(x)ai necesssrily positive such that 20gm'x' Remark: If in a totally ordered set S any two non-trivial closed intervals are isomorphic, thee this 4s equivalent to saying that A(S) is doubly trar~sitfne (0-2-transitive) on S. or the definition of 0,-2-transitive see Chapter 3, Definition 3.1 .3(k)). Hence we have that if A(S) is doubly transitive on a tot,ally ordered set, S, then A(S) is divisible. &emma 2.4 If F is 5 totally ordered field, then A(F), the o-group of o-preserving permutations, fs doukly tmnsftive on F, Proof: Consider any a, by c , dcF wf th a Then ~EA(F)for if x, ycF and xa = yay this gives (x-a) (d-e) (b-a)-'+c = (p;)(d-c) (b-a)ll+c. Therefore x = y and so a is 1-1. If xsy, then x-asy-a. Hence, (x-a) (d-c) (b-a)-ls(y-a)(d-e) (b-a)-' since d-c, b-a>o. Therefore, xalya and so a is o-preserving. Also, aa = e and ba = d. Thus, 3acA(~)such that aa = c, ba = d and so To complzse the LYOG~of the main theorem, we need only show that every totar~yordered set S can be embedded in x totally ordered field F 4rL s72ch =I ~nmy th8t A(S, is I-~Esomor~hfc to an I-subgroup of A(T). Ther-, sEnc2 A(F~ fs doubly transitive by lemma '3.4, iemma 2,3 jfelds that A(F) 'is divfsibie, An application of the BoiZand embedding Lheorem then completes the proof. The rext lemma yields the required embedding, For a partially ordered set S, the subset c of f~,mA - edeal of S if and only if xec, tsx*t~c. Remark: If S is a t,otally ordered set, then the set of 213 ideals is comviete and is totaily ordered by set inclusion since for el, e2cC(s) = the set of all ideals of S, then either c14c2 or c2Sc3. Eernms 2.6 Let S be a subset of a totaiJy ordered s?t T. If every o-permutation of S can be extended to %n o-&errnutation of T, then A(S) is 1-'isomorrhic to an 1-subgroup of A(T), groof:: Let C(S) be the complete total.iy ordered set of .fdes:ls of S partially ordered by inclusion, Identify the elements h of S with the principal Ldeals of S (for any %EX, a={s~~:ssaj is the principal ideal genera-ted by a). Sf CEC(S), define Now, for any tcT, tel w5ere e = {s~s:s @ 1 [I,:~TC(S))forms I decom&osit;or. of T. Defire the relation p on C(3) as fo1;ows: for el, e2-C(s), c,pc2=Ie is isomorphic 3 to 1 . Then p deffnes an eqgfvaience relation on @(s). From each equivalence class E, we choose a representative c say. denote the identity map OK. 1 and let 8 be am Let @c,c c c,cj isomorphfsm of Tc octo Ic If cA,c0aE dfine 8 . d ? 1 I ,c2:=c1-Zc 2 fn the natural way s~chthat the up&er section of the diagram below comuCes, that fs, if Pt c I pc2 and c2pc3, foPlows naturally that be the unique automorphFsm of @(s)which extends cp such $:A(s)-----dA(@(s)) defined by cpB = cp* is an 1-isomorphism of A(S) into A(c(s)). For each CEC(S), we have cp(ccp*), for clearly Ic and I are isomorphic since cp can be exte~3ed ccp* to an o-isomorphism of T. In fact, if Y is any automorphism of T which extends cp, then Y maps Ic octo 1 To see this, ccp" " consider teIc and sirpgose that slsccpX, sg$cq*. Then - 1 - 1 - 1 - 1 slcp EC and seep (c. Therefore, slcp Now, let e be th.~i.dent1.t~ au.tomor~h~ism on S, Consider I ~EA(s). We must, skow that (q~e)~- cp"v'e ' . Take xmiI. Note r, that xcp'sxec~*s'c. Thu,s, if xcp 'sx, then, 19 I x(~~e)' ::: ~'~,c(~Ve)" = ~8~,~ - x = ~(qVc ). If x>~cp, then, I I x(cpve) ' =: ~0~,~,(~~~).++= xR e ,ccp'* =xrp =:x(cpveU). mus, I I (cpve) ' -: cp ve . Hence a, Is ar~i-isomor;@ism and the poof is complete. Notation: If G is a totally ordered grouF snd F is 3 field, tihe group algebra of C, over F i.s given 'by the set of all. formal sums Z c g where c,cF and c #o for rat most a fhi'te aeG- w k? number of geG. Define addition su.ch that X c g + C d,g = C (c +d,)g. Also, if hcF, then ~EG g€G o gcS g D A P cgg = 2 (xc,)~. Multiplicati.on is dzf iried such that g€G g€G 0 (Cc g) (2dhh) = CC d gh. Then F(G) , the group algebra of G Q g h over F is an integral domain'. To see this, suppose Bcgg, z~~~EF(G)such that Cc ,#o#C%h. Then let €5" GI = [ge~;c+o) and G2 = (~PG:~~#o).Pick gr~~Iand hfsG2 g such that for every gaG,, g'gr and for every heG2, hshf. Then, for every geGl with g gh Lemma 2.2 Any totally ordered set S may be embedded in a totally ordered field F in such a way that each automorphism of S may be extended to an o-preserving automorphism of F. -Proof: Let G(S) denote the free abelian group on. S as the set of free generators. The order on S may be extended to G(S) as follows: Znisi>o if n.\o and at most finite number of nifo J Let F be the quotient field of the where s 3 = ~[s~:n~#o). group algebra Q[G(S) ] of G(S) over the field of rational numbers. Order Q[G(s)] lexicographically such that: B q,xg>o if qh>o when h = V[~EG(S):q fo). This order Q gas) " has a unique extension to an order on F (~uchsP.109 Theorem 3). Then the canonical extensions of an automorphism rp of S first to G(s), then to Q[G(s)] and finally to F are certainly o-preserving whenever cp is. Theorem 2.8 Every 1-group can be embeddeh in a divisible 1-group. -Proof: Applying the Holland representation theorem, every 1-group can be embedded in an 1-group A(T) of o-permutations of some totally ordered set T. By lemmas 2.3, 2.5 and 2.6, we may assume that T is doubly trans- itive. Then by lemma 2.2 due to Holland, A(T) is a divisible 1-group . This completes the proof. Remark: The above construction can also be used to embed any 1-group in the 1-subgroup of bounded o-permutations of a totally order field. Hence the result: 'Every 1-group can be embedded in a simple 1-group.' As mentioned in the fntroductim, the set A(S) of order preserving permuta,tfons ( c-permatatiocs or astomcr~hisms)of a totallyf~~(~), oraered set S becomes a.n 1-group ff the group operation is taken as composition and the partial order is deffned as follows: for frl if and only if xf rx WXES. Then, for f ,~&A(s) and YxgS, x(fvg) = xfvxg and x(f~g)= xf~xg, En this chapter, a theory of transitive o-permutation groups is first developed. This was due to C. Holland (11) and the theory is somewhat analogous to the general theory of permutation groups. In section 2, a class of simple 1-groups each containing an insular element (defined later) is shown to be Just the simple I-groups which can be repre- sented as o-permu~tatfons of a totally ordered set with bounded support. In conclusion, examples of such groups are given. me theory here is due to C. Holland (9). SectLon I, In thfs section, unless otherwise stated, G fs an 1-sitbgrou~ of A(s) , the 1-group of o-permutations on a totally order set S. Multiplicatfve notation is used 51: discussions of 1-permutat ion groups. Definftion 3.1.1 Let G be an 1-subgroup of A(s). A convex congruence on S (with respect to G is an equival&nce relation -. on S such that (i) x-y*xg-yg Wge@ (if) if xrysz and x-z, then x-yo A convex congruence is non-trivial if one of the equivalence classes c~ntainsmore ,than one element, and fs not all of se Lemma 3.1.2 ~f - is a convex congruence on S, t;hm s/- is total~yordered by letting (x-)s(y-) if xsy or xwy, mere is a natural l-homomorphism of G into A(s/-) such that for gcG, ~+~'EA(s/-) with (x-)gi = xg-. Proof: s as defined above is reflexive for clearly x-x and so (L)s(L)~ NOW, if (x-is(y-) and (y-)g(x-), then, either xsy and ysx or xmy or y-X. In any of these cases, it follows that (x-) = (y-) and so s is antisgmmetric. If (x-)a(y--) and (y-)s(z-), then consider the following cases: w(1,:x Case:x Case (3)x--y and y Case (4): (x-y) and y-z which implies xlz and so (x@)~(z-) Hence 5 is transitive and therefore a partial order. Since S is .totally ordered, for any (x-), (y-)~s/-, either x (xw) = (p).Thus s/- fs totally ordered by S as defined in the lemma. NOW, eonstder ~~:C+A(S/-) defined such that g0 - &where (x--)gl = (xg)-. 0 is a homomorplzism, for, consfder any g,h~Gthen (xw)g8h8 - ((xg)--)he = (xgh)-=(xm)(gh)ee Therefore (@I)@ = g0h0. Consfder the identfty permutation 1 EG and take any gcG. Then, (x-) (gv~)e = X(~VI)-- = (x~vx)--= xx= (x-lgyx-) I =(~-)~evl' I Therefore (gvl)8 = g8Vl ' = a~'vl'where 1 is the identity in I I A(s/--). Similarly ($~1)6=gA1 avid so 8 is an 1-homomorphfsm. Definition 3.1.3 (a) G is transitive on S if for each x,ycS there exists geG such that xg = y, , (b) G is o-2.-transitive on S if for each x,y,z,w~S, if x and yg = w. Example: (a) If G is a totally ordered group, the group pG of right translations of G is an 1-subgroup of A(G) and is transitive on G. Notice that for any x,y~G,~p-~+~=x-x+y=y. Notice also that G is not o-2-transitive (b) An o-2-transitive 1-group: If C is a totally ordered field, then A(F) is o-2-transitive emma ma 2,4). Definition 3.1.4 G is o-primitive on S if there exist no non-trivial convex congruences on S. Remark: It 5s shown later 'that any o-2-transitive I-group is o-primit ive. Definition 3.1.5 G is weakly o-primitive on S If, whenever -- is a non-trivial convex congruence on S, 9 Ifgc~sach Shat x-xg VXEX (i,e. the natural 1-homomorphism 8 of lemma 3.1.2 fails to be injective). In this case S is said to be minimal for 6. Remark: If G is not weakly o-primitive. Then for some convex congruence 0 is ingective and we say S can be reduced Definition 3.1.6 @ is a representing subgroup of G if C is a convex prime 1-subgroup of G which contains no 1-ideal of @ other than f3b At this point recall Theorem 1.29 which can be restated as follows: an 1-group G has a representing subgroup if and only if 3 an 1-isomorphism a of G into A(T) where T is a totally ordered set and such that Ga acts transitively on IT. Lemma 3.1 .7 There exists a one-to-one 1-homomorphism is:A ( S)-A(??) where denotes the completion of S by Dedekind cuts (without end-points) . Froof: For ~EA(S)and s&!, ,define i such t,hat S a(gis) = v(xg:x&, xsa]. 'Then clearly gisaA(5j. Also, any two elements of A(?$ which agree on S must agree on 5 also (from the dqfinitfon above) and so mast be equal. is is injective for consider g,~Keri Then gi = IF' s •‹ s For every aeS, agis = a - V(xg:xG, xca] = ag since g preserves order. Hence g = 1 and so f is one-to-one. To show that is is S S an 1-homomorphism, take any XES and g,heA(~). Then, x(gis) (hi,) = x gh = x( (gh)is). Therefore (gis) (his) and (gh)is agree on S and hence must be equal. Also, x( (gvh)is) = x(gvh) = xgvxh = x(gis)vx(his) = x(gisvhis). Again (gvh)is and (gisvhis) agree on S and therefore must be equal. Similarly, (ghh)is - gisAhis and so i is a S 1 - 1 1-homomorphism. Notation: For xfz, gaA(S) instead of ~(6%~)one usually writes xg and it is assumed that A(s)EA(~). Remark: It is clear that for every ~EA(S)3 EEA(~) such thatat'\ s = a. However, the converse of the lemma does not hold. That is to say there does not exist a 1-1 function mapping A(~)+A(s). To 'see this, consider the totally ordered set of all rational numbers Q and the dedekfnd completion of - the rationals to the reals, Q. Then ~~EA(Q)defined such that However, a canno3 t; restrikted to Q since all positive rational numbers which sre not perfect squares are mapped to irrational numbers. Lemma 3.1.8 Let G be transitive on S, and E >e a convex subset of S, If aeE is such -that if gcG, ag~Ethen Eg - E, then E determines a convex congruence -- on S defined by x-y if for some g4, x,y~Eg. Proof: For each gcG, Eg is convex since E is convex, Since EG such that eg = x. But G is transitive on S. Hence 3heG 1 - 1 such that ah = e. Then, ahgf-' = egf - = xf sE since xeEf. Thus, ~h~f-'= E which gives Ehg = Ef. But since ah = eE', then Eh = E. Therefore Ehg = Eg = Ef. So - is an equivalence relation on S. FEnally, if x,y~Eg,then for any fcG, xf, yfzEgf. Since for every geG9 Eg is convex, then -- is a convex congruence on S. Notation: Consider G an I-subgroup of A(S) for some totally ordered set S. Then for any acS, Ga = [gcA(~):a5 = a]. Lemma 3.1.9 If N is a convex congruence on S, a totally ordered set, if aeS, and if C = [g~~:ag-a1, then C is EL convex prime 1-subgroup of G and GarC. Conversely, if G is transitive on S and C fs a convex 1-subgroup of G containing ua ' then the relation x--y if for some g~@,xg, ygwC is a convex congruence on S, Proof: Clearly CsG. For any g,hsC2 then qg-aeh, Since P - - l - 1 is a convex congruence, then agh A and so gh EC. Therefore @ is a subgroup of G, For any gcC, a(gv1) - agva-a and so gvl E@. Similarly, gAl EC and so C is an 1-subgroup of G, C is convex, for consider l ~GCimplies ag4, then a-h. Therefore hcC. C is a prime I-subgroup for, suppose f,g~Gsuch that fAg = 1, then afAag = a and since S is totally ordered, either af = a or ag = a. Thus either fcC or ~GCand @ is prime, Slnce Ga = {ge~:ag= a] then clearly G aW. Conversely, let E = a@. Then, for f,g~C,af, ag~aC. Therefore, if for some XES, afgxsag, then since G is trans- itive, 3hcG such that x = ah. Thus a((f~h)~g)- a((f~g)v(g~hY)-x. Also, $Af~(g~f)\$(gPh)sg. Sfnce C fs convex, then (fVh.)hg~C and so x~aC. Therefore E is a convex subset of S. To show that -- is a convex congruence, we apply lemma 3.3.8. Hence it must be shown that if for some gG, ag~Ethen Eg = E, Suppose for some gaG, agczaC = E. Then ~ECsuch that - 1 - 1 ag = ah. Therefore agh = a and so gh cGacC. Since haC, then geX. Hence Eg = aCg = aC = E, Therefore E determines a convex congruence -- on S defined as stated fn the lemma, Also, C = [ gcG:ag,a]. This is easily verified. Remark: It follows easfly 'from the fact that the convex 1-subgroups 01' G co~taininga prime I-subgroup C form a chain heor or em I. l5), that ff G is transitive on S, then the convex congruences on S form a tower, Thfs Is proved In detail later. Corollary 3. I .10 For each XE~,Gx is a convex prime 1-subgroup of G. Proof: Replace S by 3 and assume as earlfer mentioned that A(S)CA(~). Then define ,on such that ~y if and only if - x = y. Clearly -Is a convex congruence on S. But G, = [ gaG:xg = x 1. Hence by lemma 3.1.9, Gx is a convex prime 1-subgroup of G. Lemma 3.1.11 Let aaS, and let K be a convex subgroup of C containing Ga and such that ?or any XES ~EKwith xsaf. Then K = G, Proof: We have KEG. Let l Thus gvf = ((gVf)f-l)f = hfeK and also IsgSgVfeK. Since K 'Is convex, then ~EK. Hence K=G, In the following definitions, S is a totally ordered set and T is a subset of S. $ For x,y~S,x and y are T-connected if for every gcX either xg, yg~Tor xg, ygp!~. Definition 3.1.13 T is bounded if Ba,bcS such that %st&for all ~,ET. Definition 3.1.14 T is dense in S if whenever a Theorem 3.1.15 If G is transitive on S, the following are equivalent: ( 1 ) G is o-primitive on S; (2) For each a~z,Ga is a maximal convex 1-subgroup of G; (3) For each ass, G, is a maximal convex 1-subgroup of C; (4) If xez, xG is dense in g; 5 If T is a convex bounded subset of S, then no two different elements of S are T-connected. Proof: ( I)). Let aez. By Corollary 3.1.10, Ga is a convex prime 1-subgroup of G. Let C be a convex 1-subgroup of @ such that Gac C. Let l Thus, by lemma 3,1.11, C = G. Hence G, is msxfmal. )()This is immediate. 3)(4). Let XE~. Def ine ,on S by zdy if $~EG such that xg lies between z and y. Then - is a convex congruence on S, for it is easily verified that .- is an equivalence relation and, if Z-y, suppose for some ~EGzhfih, then ggeG such that xg lies between zh and yh. Me may assume without loss of generality that zh Now, let a&. By lemma 3.1.9, C = {gcG:ag-a] is a convex 1-subgroup of G containing G,. If G = C, then since G is transitive on S, aC = S and all elements of S are equivalent. But xez so this is impossible. Hence G& and so Ga = C. If acbeS, then since G is transitive, H~EG such that ag - b and so EC= G,. Hence a = b. Therefore xG is dense in E. Let T be a convex bounded subset of S. Let (4)*(5). - x = glbT~S. If a, beS and the open interval (a,b)f#, then, since xG is dense in 5, 3g~Gsuch that a Sfnce x = glb T, then ag-'fj!~ - but bg If = teT and so a and b are ,& T-connected. If now a(b and gxs~strictly between a and b, then, since G is transitive on S 3ceS such that a (5)*( 1 ) . Suppose G is not o-primitive on S. Then 3 a non- trivial congruence on S, and there are at least two congruence classes say T and Q where q T is a bounded convex subset of S, Since 3 some congruence classes with more than one element, by transitivity 3 a,beT with a$'b. But, for every ~EG,either af, bfcT or af, bfk~, Hence a and b are T-connected and this contradicts (5). Therefore G must be o-primitive. This completes the proof. Corollary 3.1 .16 If G is o-2-transitive on S, then G is o-primitive on S. Proof: Suppose G is not o-primitive on S. Then from con- dition (5) of theorem 3.1 .15, if T is a convex bounded subset of S such that for x,yeS\T, x but bgh#~, This is a contradiction. Now, if for every geG. age bg#T', consider the following: Case (2) agag agh = t and b& = y (since G is o-2-transitive). Thus aghfT, bgh#~and again this is a contradiction. Case , (3) ag o-2-transitive, 3h~Hsuch that agh = t and yh = bg. Then ag Remark : The converse of this corollary is false as the example following the next corollary shows. Corollary 3.1.17 If S = 3 and G is transitive on S, then G is o-primitive on - -Proof: G is transitive on 5 and so for XE~,xG = S. Hence, xG is dense on 5. It follows immediately from theorem 3. I.15 that G is o-primitive on S. Remark: G is o-primitive on S G is transitive on S. To - - see this, consider the following example: Consider the totally -53- ordered group of real numbers R under addition. Let G = { pq~~(~)xpq = x+q, XER and q a rational number] . Then G is o-primitive. However, G is not transitive on R for given x,y~Rwith x rational and y irrational, there exists no P EG such that xP = y, Clearly since G is not q q transitive, then G is not doubly transitive. Hence this example also shows that G is o-primitive d~ is doubly transitive. Corollary 3.1.18 If G is transitive and o-primitive on S, then for every XG:, G is a representing subgroup of G. X -Proof : By corollary 3.1 .I0 for each ~€5,Gx is a convex prime 1-subgroup of G. Therefore we need only show that $ 1-ideal of G, except {I 1. From Theorem 3.1 .15, for each xog, xG is dense in 5. Now, if I &EG~, then ZaeS such that afag. Since xG is dense in 3, for some fcG, xf lies between a and ag, say asxfsag. Then asxfsagsxfg and so xf+xfg or xfxfgf-I . Hence fgf-I $G,. Therefore Gx contains no normal subgroup of G except {I ) and Gx is a representing subgroup of G. Lemma 3.1.19 Let -be a convex congruence on S and let H = [g~~:x-xgVXES 3 . Then H is an 1-ideal of G, Proof: Clearly N ts a subgro'up of G. For any hcH, x-xh \dx~S. In particular, since xgsS V~EG,then xguxgh. Thus x-ghg-' 1 and so ghg- EH for every gcG and h€H. Hence HAG. Also for all XES, x(hV1) = xhVx-x since ~EH. Therefore hV1cH. Similarly, hAI€H and so H is a swklat-bice of G. H is convex, for if l 1-ideal of G. Theorem 3.1.20 @ be trans it ive on S. Then the following are equivalent : G is weakly o-primitive; For each aeS, Ga is a maximal representing subgroup of f G; If XEEand xG is not dense in 'S, then 3l+gsG such that Y=G, Yg - y; If T is a convex bounded subset of S, and if Ba, bcS, a#b, with a and b T-connected, then 3 ?+~EG3 XES, x and xg are T-connected. Proof: (1)*(2). Since G is transitive on S, and G~(s),then by Theorem 1.29, G contains a representing subgroup. Also, from the proof of that theorem, Ga = {g:ag = a] is a repre- senting subgroup of G. It remains only to show that G, is maximal. Let C be a convex 1-subgroup of G and Ga GC, and let- be the convex congruence on S determined by C as in lemma 3.1.9. If"- is tr'ivial, then as in the proof of theorem either C = G or C! = G. If --is non-trivial, then 3.1.15, a 31fgG such that x~xgfor every xcS. Therefore {I ]+H = {gcG:x-xg ~(xEs]. But by lemma 3.1.19, H is an 1-ideal of G. Moreover HcC. Thus C is not a representing subgroup of G. Therefore Ga is a maxinal representing subgroup ((3). If XE~and xG fs not dense in E, let a& and define ,on S such that z-y if $BEG such that xg lies between z and y. Then, as in part (3) of the proof of 3.1.15, ,is a convex congruence on S and G a c @ = Ig~C+:ag-3 where C is a convex prime 1-subgroup of G. Since G, is a maximal repre- senting subgroup, C contains an I-ideal H#{I] of G. Consider l Then, akm'gk = bgk>xfk>bk = a. Hence K-' gk C. *is contra- dicts the existence of ~#{l]. Therefore xfg = xf \d xffx~. (3)44) . Let x = glb~~3where T is a convex bounded subset of S. Suppose a, bcS, a0and a and b are T-connected. Since G is transitive, 3feG such that af = b. Then b and bf are T-connected and so a and bf are T-connected. Also, a0= af0f. By Theorem 3.1.15, &xg~xGsuch that a 3lfg~Gsuch tha.t for all y~xG,yg = y. Without loss of generality, g>l . Let zeS, Clearly ,d ~EGsuch that z (&)a(1 ) . Let N be a non-trivial convex congruence on S. As in the proof of part (5) theorem 3.1.15, there are at least two congruence classes say Q and T where q Definition 3.1.21 The support of an element he~(*s),denoted ~(h),is given by o(h) = { XES:*+X 3. The lemma which follows gives a necessary condition for the 1-group of all o-permutations on a totally ordered set S if transitive on S, not to be totally ordered. Lemma 3.1 .22 If A(S) is transitive on S, a totally ordered set, and not totally ordered, then A(S) contains an element g#l of bounded support. -Proof: Suppose ~EA(S)and a, b, ces with aO xg = xf if assand xg = x otherwise. Then I+~EA(S) and g has bounded support. If no such f exists, then since A(S) is not totally ordered, 3f ,fgaA(s) such that f pf2 = 1 and f1#lff2. Then, 3ac3 such that a(fIvf2) = a. In fact, the fixed points of flvfe form a closed bounded interval with respect to the interval topology say [y,$']rg. Suppose CI (f1)~ [xsA:x (1) "1f S is a Dedekfnd complete totally ordered set, if S is not discrete, and if A(S) is transitive on S, then A(S) is o-2-transitfve on S". This result is due to Treybig. (2) "1f S is a totally ordered set and if A(S) is transitive on S and if A(s)~is a maximal convex 1-subgroup of A(S) , then A(S) is totally ordered or A(S) is doubly transitive on s". This result is due to Lloyd (15). Theorem 3.1.23 If G is an 1-subgroup of A(s), if G is transitive and o-primitive on S, and Tf G contains an element l support is bounded below (or above) then G is o-2-transitive Proof: Consider l x G is transitive on S, 31 = ~''AP.Then cf = cd*/icf" = d and also feGx Hence, for x, c, d€S and x To prove the assertion that !"i~ X, c, dcS and x Suppose YE~and x Therefore, ZrcG such that xaf But b#o(f-lgf) implies that bf - 1 gf = b. mereyore - 1 - 1 aff-I = a Now, let U = cGx If U has an upper bound, let y = lubcGx. As before, ~EG, such that y Definition 3.1 .24 G is locally o-primitive on S when there is a unique minimal non-0 element of @, the lattice of all convex congruences on S with respect to G. Definition 3.1.25 The congruence classes of a unique minimal non-0 cor congruence are called the primitive segments of S. Lemma 3.1.26 If G is transitive on S, the convex congruences on S form a tower. Proof: If afS, by Corollary 3.1.10, Ga is a convex prime 1-subgroup of G. Therefore by Theorem 1.15, the set of convex 1-subgroups of G containing Ga form a tower under inclusion. Hence it is necessary only to show that the correspondence established in lemma 3.1.9 between the convex congruences on S and the codvex 1-subgroups of G containing Ga is 1-1 and order preserving. Let 0 and p be any two convex congruences on S with respect to G. Let Co and Cp be convex 1-subgroups of G each containing Ga and determined by Ga. Ifo#p, suppose C, = Cp. Then =,yes such that xoy let xdy. But then 3gaG such that xg, ygaaCa= aCp. Hence xpy. This yields a t contradiction. Therefore ~af~p. On the other hand, if C and @' are convex 1-subgroups I of G containing G, and such that C and F determine the same congruence, then, for each gaC, 3fe~' such that af = ag. Therefore gf-I eGac C/ and so gad J . Thus CEC' . Similarly it follows that C'G C and so C = d' . That t)ne correspondence preserves order is clear. For, if o,pa eand if' C, GCp, then asp and conversely. Definition 3.1 .27 A convex 1-subgroup K of G is said to -cover the convex 1-subgroup H of G if K properly contains H and there is no other convex 1-subgroup of G between K and H. Theorem 3.1.28 Let G be transitive on S. The following are equivalent: ( 1) G is locally o-primitive on S; (2) For each aeS, G, is covered by a convex 1-subgroup ~(a)of G; (3) 3 aaS such that G, is covered by a convex 1-subgroup ~(a) of G. I In (2) and (3))the ~libg~01Q~(a) is unique. Proof 1...* ( I ) (2). By lemma 3.1.26, the corwex congruences on S form a tower. From ( 1 ), 3: minimal non-0 element of the lattice of all convex congruences on S with respect to G. From the proof of 3.1.26, it follows that for aeS, Ga is covered by a convex 1-subgroup ~(a)of G. (2)) This is immediate. Also ~(a)must be unique as the proof of 3.1.26 indicates. 3) I ) If 3acS such that Ga is covered by a unique convex 1-subgroup ~(a)of G, -then it is immediate from the proof of 3.1.26 that 3! non-0 minimal convex congruence on S corresponding to ~(3).Thus G is locally o-pr imi Live. - Remark: If G is an 1-group, let l#gc~. and let C he a regular 1-subgroup of' G not containing g. Then C is prime. If K = n{S: SE@(G) such that c ~[g}a), then K covers C. Theorem 1.27 can now be restated as follows: 11An 1-group G is 1-isomorphic to a subdirect sum of .riK where each K is a S;ymsitiv~local2.v o-~rimi.tive P B 1-subgroup of A(S ) where each S - K$c@, C being a 8 8- B regular 1-subgroup of K If. P In conclusfon, we present a diagram of implfcations together with some examples, o-2-transitive trans it ive and 31<~cGsuch that transitive o(g7 is bounded below Example s (1) An o-primitive 1-group which is not o-2-transitive: The example which follows Corollary 3.1.17 is sufficient. Also the o-group of right translations of a totally ordered abelian group which is "full" in the sense of Cohn (4) is also o-primitive but not o-2-transitive. (2) A weakly and locally o-primitive 1-group which is not o-primitive: Let S be the totally ordered set of reals with the integers removed. Then A(S) has the desired properties . (3) An 1-group which is locally o-primitive but not weakly o-primitive: Let G be a non-archimedean totally ordered 1-group without 1-ideals. That such 1-groups exists is clear from the example given by A. H. Clifford (3). Let I#~EG, C the regular 1-subgroup not containing g, and K the minimal convex subgroup containing g. Then C and K are both representing subgroups of G, and K covers C. Hence by theorem 3.1.28, G is locally'o-primitive on the set of right cosets G/C, but by theorem 3.1.20, G is not weakly o-primitive. (4) An 1-group which is weakly o-primitive but not locally o-primitive: Let G be a totally ordered abelian group which has no smallest proper convex subgroup. Then I is a representing subgroup, and by Theorem 3.1.20, G is weakly o-primitive on G/[I] , but by Theorem 3.1.28, G is not locally o-primit ive. (5) An 1-group which is neither weakly nor locally o-primi- tive: Let G be the 1-group of example 3. Every convex subgroup of G not equal to G is a representing subgroup, and 3 no smallest proper convex subgroup. Hence by 3.1.20 and 3.1.28, G is netther weakly nor locally o-primitive on Section 11. An 1-group is simple if it has no non-trivial 1-ideals. Definition 3.2.1 Let G be an 1-group. For Igf, gcG, f is right of & if for all Definition 3.2.2 An element gcG is insular if for some conjugate g* of g, D* is right of g. Definition 3.2.3 I If G is an 1-group of o-permutations of a totally ordered set S, then for ~EG,f is kounded if its support, a(f) lies in a closed interval of S. Lemma 3.2.4 Let G be a transitive 1-group of o-permutations on a totally ordered set S. An element KgsG is insular og is bounded. Proof: () Suppose for l so xK-'h-lg = xk-'h-'. Thus %--'g*~ = &-'h-'ghlt = xe Hence a(k-'lg*$-) lies completely outside [a ,b] ror every ISKEG. Therefore $AX-'~*K= 1 for all IrlfaG. Hence g is insular. () Conversely, suppose l 3l z = zh-'g*h = wm2wm = z. ,Hence wgKh = wkh and so w = wg. - 1 Therefore o(g) is bounded below by zK- , . Remark: Clearly, if $:wA(s) fs an 1-isomorphism of an 1-group G onto a transftfve 1-subgroup of A(s), then l Lemma 3.2.5 If G is an 1-group of o-permutations of a totally ordered set S, then the set H = {g~G:gis bounded) is an 1-ideal of G. Proof: Clearly, if 1 = identity of G, then a( I ) = 4 and so _I__ IEH. If g, heH, ghfl, then suppose o(g) and a(h) are contained in the closed intervals [a,b] and [e,d] of S respectively. Then sinc,e o(gh)co(g)kp d, h) , it follows that /' dm) z[a,b] ub,d] ck,bvd] . Therefore gh~H. Clearly if gh. = 1, then ghcH. Thus -H is a subgroup of G. For any hcH, o(hv1) - [ x~~:x(h~l)#x1 = ( x~~:xh>x]~~(h) Since h is bounded, then hv1 is bounded and so hV1cH. Similarly o(h~l)m(h)and so hAleH. Thereyore -of G. Suppose for I~EH3geG such that l G is a simple 1-group cont,ainfng an insular element H (j. is a transitive o-primitive 1-group of bounded o-permutations of a totally ordered set, S, -Proof: Let g be an insular element of the sfmple 1-group G. By corollary 1.30, G fs 1-Isomorphic to a transitive 1-group of o-permutations of a totally ordered set, By lemma 3,2.4, since gcG is insular, then g is bounded. By lemma 3.2.5, if H = { g~G:g is bounded], then H is an 1-ideal of G. Since lfge~and since G is simple, then. H = G. That fs, every element of g is bounded and so G is a transftive 1-group of bounded o-permutatfons of S. Now, G may not be o-primitive, so let -be any convex congruence on S, Then from lemma 3.1.19, HI= (gG: x-xg Vxe~] is an 1-ideal of G. Since G is simple, the H = {I] or H = G, For I+~EG, let cr(f)Ile in the closed interval [a,bl of S. Then if for any non- trivial congruence -, a-b , then f EH and so H - @. But this is impossible since G fs transitive on S. It follows that the union of any tower of proper convex congruences on S is a proper convex congruence. Therefore, by ZornOslemma, there is a maximal proper convex congruence p on S, Also, ~/pis totally ordered as in lemma 3.1.2. Let p :G A(s/~ ) be the natural 1-homomorphism of lemma 3.1.2. Then, for ~EGand xcS, (xg)p = (xp)gp. Also, since H = { I], $l#fe~such that V xeS, xpxf and so $ is 1-1. It follows that G is 0-primitive on s/~and so is a transitive o-primi- tive 1-group of bounded o-pkrmutations on s/~. Conversely, let G be a transitive o-primitive 1-group of bounded o-permutations on a totally ordered set S. Let (I )#N be a 1-ideal of G. Define - on S such that for x,yeS, ~-~.ye~l To see this, notice that if xq, let l Now, if x,y, zeS, xqsz and X-Z, then Sl N#{I], then 3 at least one congruence class E containing more than one element. But G is o-primitive, therefore E = S and must be the only congruence class. Let l interval b,b] of S. Since a-b, then 3l geN. Therefore G = N, and G is simple. By lemma 3.2.4, every positive element of g is insular. This completes the proof. Corollary 3..2,. 7 If G is a simple 1-group with an insular element, then every positive element of G is insular, -Proof: This is immediate from the theorem. Corollary 3e2.8 If G is a simple 1-group with an insular element, then for every l Proof: By Corollary 3.2.7, every 1 I ~xamplek ( 1) If F is a totally ordered field, then A(F) is doubly transitive. Hence A(F) is o-primitive. Therefore if G is the 1-group of bounded o-permutations of F, then by theorem 3.2.6, G is simple. In particular this is true if the field is the field of real numbers, a result obtained originally by Holland ( 10). (2) A non-totally ordered simple 1-group which does contain an insular element: Let G be the 1-subgroup of A(R), where R is the totally ordered field of real numbers, such that G = [aeA(~):xcr+I = (x+l ) u VXER]. Then G is transitive on R for given any x, yeR with x G. Therefore, for any IR G which contains g must also contain any such f. Hence G is simple. Using the results of section I, C. Holland also showed in (11) that under suitable but rather general conditions, if an 1-group G is 1-isomorphic to a transitive 1-subgroup of A(S) and to a transitive 1-subgroup of A(T) , where S and T are totally ordered sets, then TCFwhere 3 is the completion of S by Dedekind cuts (without end-points) and the action of G on T is obtained by first extending the action of G on S to an action on and then cutting back to T. -70- BIBLIOGRAPHY Birkhoff, Garrett, : 'lattice Theory ', Third (New) Edition, Amer%can Mathematical Society Colloquim Publications, Vsl,25, New York 'i9670 Bernau, S. J, : Tree abeltan lattice groups 1, Math, Ann. 180(1969) 48-59 (to appear), Clifford, A. H,: IAn non-comutative ordinally simple linearly ordered groupP, Proc. Am, Math. Soc. 2 ( 1952) 902-9030 Cohn, Po M.: 9Growps of order automorphisms of ordered sets', Mathematfka 4 (1957) 41-50. Conrad, P. 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