Separatunt

PUBLICATION ESMATHEHATICAE

TOMUS 10.FASC.

DEBRECEN1963

FUNDAVERUNT:

A. O M T. SZELE ET O. VARGA

COMMI SSI O REDACTORUM:

J. ACZÉL, B. GY1RES, A. KERTÉSZ, A. RAPCSAK,A. RÉNYI, ET O. VARGA

SECRETARIUS COMMI SSI ON'S:

B. BARNA

L. G. Kovics

Admissible direct decompositionsof direct sums o f abelian groups o f rank one

INSTITUTUM M AT H EM AT IC U M UNIVERSITATIS DEBRECENIENSISH U N GAR IA

254

Admissible direct decompositionsof direct sums of abelian groups of rank one

By L. G. KOVAC S (Manchester)

The starting point o f the theory o f ordinary representations is MASCHKE'STheorem, which states that every representation of a finite group over a field whosecharacteristic does not divide the order o f the group is completely reducible (seee. g. VAN DER WAERDEN [6], p. 182). A partial generalization of this theorem hasrecently been given by O. GRÜN in [2], and the main step of the classical proof ofthe theorem has been generalized by M. F. NEWMAN and the author in [4]. Bothof these results arose out of a shift in the point of view: they do not refer to repre-sentations, but to direct decompositions of abelian groups, admissible with respectto a finite group o f operators (in the sense o f KUROSH [5], § 15). The aim o f thispaper is to present an extension of GRON's result, exploiting the start made in [4].The terminology follows, apart from minor deviations, that o f FUCHS'S book [1].

From [4], only a special case o f Theorem 2.2 is needed here:

Lemma. Let X be an abelian group, and G a finite group of operators on X: sup-pose that (every element of) X is divisible (in X) by the order of G, and that X hasno element (other than O) whose order is a divisor of the order of G. i f Y is an admissiblesubgroup of X which is also a direct summand of X, then Y has an admissible (direct)complement in X.

The result o f this paper is the following.Theorem. Let A be a direct sum o f abelian groups of rank one, and G a finite

group of operators on A; suppose that A is divisible by the order of G, and that A hasno element (other than 0 ) whose order is a divisor of the order of G. Then A can bewritten as a direct sum o f admissible, G-indecomposable subgroups, each o f which isa direct sum o f finitely many isomorphic groups o f rank one.

The proof splits into several steps, and occupies the rest o f the paper.(A) A is a direct sum of countable admissible subgroups each of which is a direct

sum o f groups o f rank one.PROOF. Let A = E ( C „ : ) a ) where the C'), a r e g r o u p s o f r a n k o n e , a i s a n

ordinal, and 1 runs through all the ordinals which precede cr. Denote the correspond-ing canonical projections A - -,-C2 b y y , . F o r e a c h o r d i n al p s u ch t h at p o-, o ne

makes simultaneously the following definitions. Let = s e t (p). I f i is a finiteordinal and A a countable set of ordinals preceding a, let CI = E ( C): ) . E ) , a n dlet C/ LiG be the smal lest admissible subgroup containing C. Then qt is countable;

L. G. Kovdcs: Decompositions o f direct sums o f abellan groups 2 5 5

as Cl ii G i s generated by the countably many elements cg with CE C, g E G,

is also countable. Hence Cp.' G yi, = 0 f o r a l l b u t c o u n t ab l y m a ny v a lu e s of /1 ; so t ha t

the set A Y ' defined by / 1 ' = set().: -< a, C G y ,1> - 0 ) i s c o u n t a b l e . T h i s i n d u c t i v e

definition provides an increasing chain A Z A . . . c o u n t a b l e setsof ordinals. In turn, one constructs another increasing chain by defining its generalterm Av as Av = U (AIL' v , i -< w ) , f o r e v e ry o r d in a l v w it h v 0. Th is c ha in has

the following properties:(A l) A ° is empty.(A2) I f i s a limit ordinal, Q a , then AQ = ( Apt f t , • - < C O ) =

U [ U ( A ;L: 1 1-< V , =- (Av :V <Q).

(A3) I f pt-< o-, t h e n A v ; f or p. E g Av.

(A4) I f A o-, t h e n t h e d i f fe r e nc e set — is c ou nt ab le ; for it is a subset

of U (AiA: i - < c o ) a nd each AjA is c ou nt ab le .

Correspondingly, CV =E(CA: A E A y ) d e fi n e s a n i n c r e a si n g c h ai n of p a r ti a l s um s

of E (CA: -< a ), w it h the p ro pe rt ie s:

(A l' ) CO =0;(A2') i f i s a limit ordinal, a , then CL'= U (CV: v-<o);(A3') e r — A ;(A4') i f 2, -< a, then C A-" / C i - i s a c o u n t a b le d i r e ct s um g r ou p s of r an k o ne .

Moreover, each CV is admissible; for, Cv is generated by the elements c withc E CA, /ICA'', and, if A E Am' v , i -< c o , w h i l e g i s a n a r b i tr a r y e l em e nt of G, t he n

cg E q+ 1 Thus each CA with A a is an admissible direct summand in the

admissible subgroup ' , and so the Lemma, with X = C'-+ 1 and Y = CA, g i v e sthat CA has an admissible complement, say DA, in C4+1. I n v i e w o f D C A + 1

and (A4'), it suffices to prove that A = I (DA: A - < a ) . T h i s , i n t u r n , w i l l f o l l o w f r o m

(A3') and the general relation Cv v ) which holds for every v with v a .The validity o f this relation is proved by a simple induction: i t is valid i f v =O.because of (Al'); if it is valid for the predecessor v — I of v, then Cv = CV-1 + D _= 1 ( 41: A -< v —1)- - D,,_, E(D;,: -1,); if it is valid for every v preceding a

limit ordinal e, t h e n 0 2 — L I ( C v: Q) -= U [ I( D): v): v Q.] E(D,:

by (A2').

(B) Being a direct sum of groups of rank one, A is the direct sum of its maximalp-subgroups Ap and a torsion free subgroup A , . The Ap a r e c h a r a c t e r i s t i c a n dtherefore admissible subgroups, and so, by the Lemma with X = A, Y ' Ap ( w h e r ep runs through all primes), A, can also be chosen admissible. Moreover, both Aoand the A p are direct sums of groups of rank one. This and (A) make it possibleto assume, without loss o f generality, that A is countable and either a torsion f reeor a p-group. The torsion free case will be discussed first.

(C) I f A is torsion free and B is a subgroup o ffin i t e r a n k i n A , t h e n A h a s a d i r e c t

decomposition A = A' + A" such that A ' is o f finite rank and contains B; moreover_both A' and A" are admissible subgroups of A, and are direct sums of groups of rankone.

PROOF. In order to prove this assertion, one first notes that there is no loss o fgenerality in assuming that B is admissible and pure in A. The justification of thiscan be outlined as follows. Let B be any subgroup of finite rank and S a maximal

256 L . G. Kovdcs

independent subset of B. Consider the set SG defined by SG -=--set(sg: sE S, g E G);this is finite, f or both S and G are finite. Let 13, be the set of those elements of Awhich depend on SG; this is an admissible subgroup o f A: for, i f a, b EBG a n dg E G, t hen m a -= mis ig + + mkskgk, n b - = nts ig + + nkskgk wi th s ui ta bl e

integers m, mt, mk, n , nt, . • • , nk, m =0 =n, and e le me nt s sigt , skgk of SG;

so that mn[ (a — b)g] = .E[(m in — m n t) s igig : i m n = 0 s h ow s t h at (a — b )g is

dependent on SG and hence belongs to BG. I t is easy to see that BG c o n t a i n s B a n dis pure in A; moreover, its rank cannot be greater than t he cardinal of SG. ThusB can be replaced by BG.

Let i t be assumed therefore that B is admissible and pure in A. Consider anarbit rary decomposit ion o f A int o a direct sum of groups o f rank one:

(Cl ) A AEA ) ,with the corresponding canonical projections y,: A —CA; a n d d e fi n e a s u b s e t A( B , C l )

of the index set A by A (B, Cl ) = s e t : )E A , B y , > - 0 ) . l t i s e a s i l y s e e n t h a t t h i s

subset is finite: i f S is a maximal independent subset of B, then B consists preciselyof those elements of A which depend on S; so, if O =b EB, then nb = nisi+ + nksk

for some integrs n, nt, nk, n = 0 , a n d e l e m en t s st, sk of S; if Sy, =0, then

(nb)y „= 0 and, as C, is torsion free, n(by„) — O and n =0 imply that by, =0; so thatone has A(B, Cl ) s e t ( 2 : I EA , S y „ 0 ) w h i c h , s i n c e S i s fi n i t e , p r o ve s t he fi n i-

teness of A(B, Cl). This subset is used to define 1(B, Cl), a set of types of torsionfree groups o f rank one: p u t '.?1(1, Cl ) --- set (T(C, ): A ( B , Cl ) ) ; t his set o ftypes is clearly also finite.

The statement (C) wi l l be proved by induct ion on the cardinal 191(B, Cl )1 o f'.11(B, Cl). I f '.11(B, Cl ) is empty, then B = 0 and so (C) is t riv ially true. Hence onecan procede to the inductive step: let a n d let (C) be assumed to be true forevery choice of A, G, and B to which there is a decomposit ion like (Cl) which yieldsa cardinal smaller than 19I(B, C1)1. Let a be a max imal type i n C l ) , andput A , --- s e t ( : A EA , T ( C , )> - a ) , A , - = s e t () . : E A, T( C)) --- a), and A3 =

= set()L: I EA , T( C 2) a ) . T he t hr ee sets so d efi ne d are pairwise disjoint and their

union is A . Let A ' -= ( C „ : ).E Ai) ; t h e n A ' i s t h e c h a r a c t e r i st i c s u b g r ou p o f A

which is generated by the elements whose types ( in A) are greater than a; so A' isadmissible. Moreover, A ' i s a direc t s ummand o f A , f o r A =- A ' +A , wi t hAt = )E A2U A3), and B is contained in this complement At Consider the

torsion free factor group AI B ; this has a direct decomposition A I B = (A ' + B)1.13+ Ail B, with (A' +B)IB admissible. Hence the Lemma, with X =AIB and

Y = (A l + B)I B, implies t hat ( A ' + B)IB has a n admiss ible complement , sayA* IB, i n 211B. As A ' n A*( A ' + B)(-) A * = B a n d A ' fl B 0 , A * i s i n f a c t

an admissible complement o f A' i n A. Le t a be the canonical project ion o f Aonto A* corresponding to the direct decomposit ion

(C2) A = A ' H-A*.

I f a EA, and a a ' + a, wi t h a ' E A', a, EA, , t hen a l a ' a + , - - - - all , s othat A* = Acx = A ta . A s A , a v o i ds t h e k e rn e l Al of a, it is m ap pe d i s om o rp h ic a ll y

by a, so that in fact A* = A ta - = 1 ( C , 7 . : E A 2 U A 3 ) . I t i s c o n v e ni e n t n ow to c h an g e

f rom ( Cl ) t o the new decomposit ion

(C3) A = A ' + A* -= ( C ,t: At) + I ( C)a : A2 U A3) = ( Di: ), E A )

D 16

Decompos i tions o f di rec t sums o f abel ian groups 2 5 7

where I )). = CA i f ) . E Al a nd D Cc t if 2 E A2U A3. Of the c or re sp on di ng canonical

projections (53: A — DA, o ne h as to o bs er ve the f ol lo wi ng . Since BA* .

= ( D ).; ;LE A2U A3), B(52-= 0 whenever /I E A, O n the other hand, i f 2 EA 2 U A 3 ,then 5, y ;tot: f o r , t h en 0 65 = 6), by d e fin i ti o n; if a is an a rb it ra ry element of A,

then a = gay ILEA); also, ,at 0 i f y E Al a n d A y c e = AL i f y E A 2 U A 3 , s o t h a t

in this second case y ,otk= O if y =,1 and y ,to c5A yAc c i f y a n d h e n c e i t f o l l o w s

that a (5),= a w 5A = ( ay po ck : p E A) -= ay Aa. Also, if 2EA2UA3, then the kernel of cit

avoids CA a n d s o, in t hi s case , B(53 = ByAa >- O is equivalent to By, O. These observ-

ations yield the conclusion that A(B, C3) = A(B, Cl) and so %(B, C3) -=as well.

Next, consider the subgroup A2 d e fi n e d b y A2 = , ( D A: 2 E A2) . T h is s u b gr o u p

can be described as the set consisting of O and the elements of type a in the admissiblesubgroup A*; so that A2 i s c h a r a c t e r i st i c i n A * a nd h e nc e a d m is s i b le . A ls o, A2

is a direct summand in A* and so the Lemma, with X = A* and Y = A2, p r o v i d e sthat A2 h as an a dm is si bl e complement, say A3, in A*. Thus A has the admissible

direct decomposition(C4) A -= A + A2 + A3 ;let the corresponding cannonical projections A — Ai be denoted by ai, f o r i = 1 , 2 , 3 .Clearly A3 = i 4,1 3 = E( / )3 : ;,E A3) a3; as . 1.( D,1: E A3) avoids the kernel A' + A2 of

23, this subgroup is mapped isomorphically by ce3 , so that A3 = E(DAcc3: E A3).

Put EA if ;L, EAIU A2 and E3=1)3cc3 if E A3; then (C4) can be refined to the

decomposition(C5) A - -, ,E ( E), : A ) .

Like in a similar situation above, one checks that, for the canonical projectionse).: A E). corresponding to (C5), Be).-= 0 if e E A, and (x3e).=-E2=- (555)43 if 2EA3.

Put B2 -= Bo t2 and B3 =Boz3; both B2 and B3 are of finite rank, and B B2 + B3

If B3E , 0, then À.E A2 and so B3s, =Ba3EA Bkot3 shows that also B6A >- O.

Hence A (B3, C 5 ) g _ A ( B , C 3) -= A (B, Cl), so that 91 (B3, C5) g %(B, Cl); moreover,

as T(EA) = l ' (DA) = T( C2) for every 2 in A, and as A (B3, C5) A3, the type a does

not belong to 1 (B3, C 5 ) . H e n c e C , ,5 ) is a p r op e r s ub se t of 9t(B, Cl) and

therefore the induction hypothesis applies to A ' , G, B3, w i t h t h e c o n c l u s i o n t h a t

A3 has an admissible direct decomposition A3 = + W such that V is of finite

rank and contains B3, w h i l e b o t h V a n d W a re d i re c t s um s of g ro up s of rank one.

Finally, consider B2. B y t h e i n i t i a l s t ep o f t hi s p r oo f , A2 has an a dm i ss i bl e

pure subgroup U o f finite rank which contains B2. T h e s e t A ( U , C 5 ) i s a fi n i t e

subset o f A2 ; pu t U ' =E(EA: / 1 E A ( U , C 5 ) ) a n d U ",- - --I ( EA: A 2 — A (U , C 5 )) ;

then A2 U ' + U" and U U'. Now U is a pure subgroup of the direct sum U'

of finitely many groups of rank one which are all of the same type a; so that a theoremof CERNIKOV, FUCHS, KERTÉSZ, and SZELE (Theorem 46. 8 in Fucus [ l i ) impliesthat U is a direct summand of U'; hence U is a direct summand of A2 a s w e l l . A sU is admissible, the Lemma (with X = A2, Y = U ) p r o v i d e s t h a t U h a s a n a d m i s s ib l e

complement, say U*, in A2. I t f o l l o w s f r o m a t h e o re m o f B AE R ( T h eo r e m 46. 6

in Fuclls [1]) that both U and U* are direct sums o f groups o f rank one.lt remains to put these results together : A --= A ' + A2 + A3 = A ' + ( U + U * ) +

+ (V + W) = (U + V) + (Al + U* 4- W); a ll these summands are admissible sub-groups and direct sums of groups of rank one; U+ V is of finite rank; and B B2 +

258 L . G. Kw/des

B3 Ur+ V; so that A' and A" given by A' = U + V and A" = A' + W

satisfy the claims made in (C).

(D) I f A is torsion free, then A can be written as a direct sum of admissible sub-groups of finite rank such that each of the summands is a direct sum of groups of rankone.

PROOF. In view of (A), A can be assumed to be countable; moreover, only thecase when A is o f infinite rank needs investigation. Let A -= E (Ci: I ( 5 ) b e adirect decomposition of A in which all the Ci a r e g r o u p s o f r a n k o n e . A c c o r d i n g

to (C), A can be written as C' +D' in such a way that C, C ' , both C' and D'are admissible subgroups and direct sums of groups of rank one, and the rank o fC' is finite. Suppose that, fo r some positive integer n,(DI ) A = Cl + C2 + + C " + D "is an admissible direct decomposition o f A in which all the summands are directsums of groups of rank one, a ll but the last are of finite rank, and C, +

c l + + C. Let 5 be the canonical projection of A onto D", corresponding to(D1). Then C , 5 is a subgroup of finite rank in D", and so (C) provides that D"has a direct decomposition Dn = Cn + + D ' I + " s u c h t h a t O r +1 a n d Dn ' a re a d m i s si b l e

subgroups which are again direct sums of groups of rank one, Ct i +, 5,- - Cn 1 a n d

Cn + 1 is of finite rank. Thus A Cl + + Cn + Cn + 1 + Dn ', and now C, + C„ +

+ C, cl + + cn + Cn + 1, so that a decomposition like (D1) has been obtained

for n+1 in place of n. This inductive process defines a subgroup O for each positiveinteger i. I t is easily seen that the subgroup generated by the O is their direct sum,and i t contains a ll the Ci. T h e r e f o r e A = I( 0 : 1 - i -o ) ) , a nd t hi s is a d ir ec t

decomposition satisfyng the claims made in (D).(E) I f A is torsion free, then the Theorem is true.

PROOF. According to (D), it can be assumed that A is o f finite rank. In thiscase A is trivially a direct sum of G-indecomposable subgroups; it remains to provethe assertion about the structure o f its G-indecomposable summands. Let B be anarbitrary G-indecomposable summand o f A, and let B = O. First, a theorem o fBAER (Theorem 46. 7 in Focus [1]) gives that B is a direct sum of groups of rankone. Let B -= + G with all the Ci o f r a n k o n e , a n d l e t a b e a m a x i m al

element of the set of types T(C,), i=1 , . . . ,n . Put B, —Z(Ci: T ( Ci) = a ) a n d B 2 =

= E ( Ci: T ( Ci) = a) ; then B = Bi+B2. The subgroup B1 consists precisely of O

and the elements of type a in B, so that B, is characteristic in B and hence admissible.Thus the Lemma, with X= B and Y=B, , gives that B, has an admissible complementin B; as B is G-indecomposable and B, 0 , this complement can only be O. HenceB B , so that all the C, are o f the same type a.

In view of (B), it is possible to assume for the rest of the proof that A is a count-able p-group. I n (F) a special case will be discussed, and (G) will provide the keyto the general case.

(F) I f pA =0, then the Theorem is true.

PROOF. I f A is finite as well, this statement is trivially true. Let A be countablyinfinite; then A = I (C 1 i -< co) where all the C. are of order p Fo r each positiveinteger j, let Cl = ( C 1 i j ) , and let C1G b e t h e s u b g r o u p g e n e r a t e d b y a l l

Decompositions o f direct sums of Aa i un groups 2 5 9.

the elements of the form cg with e E C, g E G. Then all the Ci and the ClG are finite;C' C ' ' and C i C J G C i s G hold for every j; and U (Cl: 1 - < (o) = A, sothat also U ( C1G : 1 - < t o ) = A . S in ce now e ve ry s ub gr ou p of A is a direct summand

of A, and since the GIG are admissible, the Lemma (with X = 0 +1G , Y = C i G )gives that each CiG has an admissible complement, say Di,1 , i n t h e c o r r e s p o n d i n g

0 + 'G. In addition, let DI = C1G . T h e n i t i s e a s y t o s ee t h at C i G = E( D1: 1-_ _ i _ j )

holds for every j, so that A = U (GIG: - < co) = [ ' (D - < co] ==E(D E a c h Di i s a d m i s s i b l e a n d , b e i ng c o n t ai n e d in t he fi n it e C iG ,

finite. Therefore each Di i s a d i r e c t s u m o f fi n i t e ly m an y fi n it e G -i n d e co m p os a b l e

subgroups, so that the direct decomposition o f A obtained above can be refinedto one in which all the summands are finite, admissible, and G-indecomposable.This refinement satisfies the Theorem.

(G) Let T be an admissible, G-indecomposable subgroup in the socle S o f A.Then T is finite, and A has a direct summand B which is admissible, G-indecomposable,and whose socle is precisely T; moreover, B is G-indecomposable.

PROOF. I t follows from (F) that T must be finite. I f T=0, then B=0 will do;hence suppose that T>-0. Let k be one o f the ordinals 0, 1, w ; then pkA is acharacteristic and hence admissible subgroup o f A. As every subgroup o f T is adirect summand o f T, the Lemma can be applied to X = T , Y T A p k A , withthe conclusion that T npkA has an admissible complement in T. Since T is G-inde-composable, it follows that either TrIpkA= 0 or T npkA = T. I f T-, - p ' " A , l e t m =w .

I f TnpwA =0, then TrIpkA =0 for some finite ordinals k ; but not fo r all, fo rT-_-A=p°A. Hence the first of the ordinals k for which TnpkA=0, can be written

in the form in + I, and then T-- - p m A , T r I p m " A — O .

Since every subgroup of S is a direct summand of S, the Lemma can be appliedto X = S, Y-= T with the conclusion that S T + U for some admissible subgroupU. Let Uk = U np kA , for k =0, 1, w; then SnpkA T +U, for every k with

km.

Let it be agreed that w — i c o for every finite ordinalPut Bo= T . I f m 0, suppose that, for some ordinal k with k -<m, and increas-

ing chain Bo, Bk o f a d mi s s ib l e s ub gr ou ps has been defined in such a way that

T=p iBi, a nd Bi has T as its sock, for i =0, k. Let V/Bk be the

sode of p "-k-l A I Bk. A s t he s oc k T of Bk i nt er se ct s Um_k_, in 0, the subgroup W

generated by Bk and U, „_k_ , i s t h e i r d i r e c t s u m : W Bk+ Um_ k _ 1 . T he f a ct o r

group 14/IB, is an admissible subgroup in VIBk a n d , a s e v e r y s u b g r o u p o f V /Bk

is a direct summand of V/Bk, t h e L e m m a ( w i t h X = V / Bk, Y = W / Bk) p r o vi d e s t ha t

W/Bk has an admissible complement, say Bk„/Bk, in V/Bk. The subgroup Bk+1

so chosen is admissible, contains Bk, and is contained in p m-k- 1A . T h e s o d e T '

of Bk+1 contains T and is contained in S n p m-k- ' A ; h e n c e , a s S A p ' n-k- 1A =

T + U, , _k_i, T ' = T+ (T' n U ,n_k_1) . On the other hand, one knows that

T ' n U ,n_k_i Bk+ 1 W = Bk, so that T' nU,n_k_l Bkn Un,_k_ , 0 ; hence

it follows that T ' = T+ 0 = T. Next, note that p Bk + 1-- B k i s a n i m m e d i a t e c o n -

sequence of the choice of Bk + 1. O n t h e o t h e r h a n d , i f b E Bk, t h en Bk

pm kA

'A) implies that b =pa for some a in p " ' ' IA ; f o r t h i s a , a + B k E V I B k =

= (Bk U „ , _k_1)1Bk+Bk+ i/ Bk, s o t h a t a u+ b ' w i th u E b ' E Bk + t, and

this shows that b = pa = pu +pb' -= pb' E p Bk + 1; h e n c e Bk--- p Bk +, . T h u s i n f a c t

260 L . G . Kov dc s

Bk=pBk+,, and therefore T=pkBk=pk+iBk+1. To sum up: Bo, Bk, Bk+, has

all the relevant properties o f Bo , B k , w i t h k + 1 i n p l a c e o f l c .

I f m is finite, this inductive process provides in a finite number of steps a sub-group B„T; i n t h is c as e, put B=B„ ,. If m =co, then the process provides a subgroup

Bk for every finite ordinal k; in this case, let B=U(Bk:13.--k-<co). In each case,

the sode of B is precisely T. in the first case, every non-zero element of T has heightin in A (for T --- p m A b u t T p m+ A = 0 ), and its h ei gh t in B is also in (for T p "11 3„ ,=

p'nB); hence [e. g. by J) on p. 78 of FUCHS [1]] B is a pure subgroup in A; moreover,B is bounded, so that a theorem of KULIKOV (Theorem 24. 5 in FUCHS [1]) impliesthat B is a direct summand of A. In the second case, every non-zero element of Tis of infinite height in B (as T = p k Bk- p k B f o r e v e r y fi n i t e k ) , s o t h a t B i s d i v i s i bl e

[see e. g. (f ) on p. 59 o f FUCHS [ I] ], a n d h e n c e , a c o r d i n g t o a t h e o r em o f B A ER

(Theorem 18. 1 in FUCHS [1]), B is a direct summand of A. By construction, B isadmissible; and the G-indecomposability of T implies that B is also G-indeco mpo-sable. This completes the proof of (G).

(H) I f A is a p-group, then A =E(C,: 2 E A) where each C), i s e i t h e r c y c l i c o r o f

the type C(p l , Let C= Z(C), : À E A , C), c y c l i c ) a n d D = E ( C ,: 2 E A , C C( r )) ;

then D is precisely the maximal divisible subgroup of A, so that D is characteristicin A and is therefore also admissible. Now the Lemma (with X = A, /7- - - D ) p r o v i d e sthat D has an admissible complement C' in A. Of necessity, C' C , so that C' isa direct sum of cyclic groups. Hence it suffices to prove the Theorem under the furtherassumption that A is either divisible or a direct sum o f cyclic groups.

(I) I f A is a divisible p-group, the Theorem is true.PROOF. In view of (F), the sode S of A can be written as a direct sum of finite,

admissible, G-indecomposable subgroups TA, with 2. running through some indexset A. According to (G), each T is the sode of some admissible, G-indecomposable,direct summand A., of A. Each BA i s o f fi n i t e r a n k , f o r i t s s o d e T)„ is fi n i t e, a nd e ac h

B, is divisible; hence each BA is a direct sum of finitely many (isomorphic) divisiblegroups of rank one (that is, of groups of the type C ( r ) ; by another theorem of BAER,Theorem 19. 1 in FUCHS [1]). The subgroup generated by the BA is their directsum, and i t is divisible; moreover, i t contains the whole sode o f A; henceA =E(B, : 2E A).

(.1) I f A is a direct sum o f cyclic p-groups, then A is a direct sum o f boundedadmissible subgroups.

PROOF. Now all the direct summands o f A which are o f rank one are cyclicgroups. I f A is of finite rank, then A itself is bounded, so there is nothing to prove.In view of (A), it can be assumed that A is countable, so that in the remaining caseA = E(C - < co) where all the C, are cyclic. Let S be the sode of A and Si t h esode o f Ci, f o r e a ch i ; t he n S =I( Si: 1 i iv). As before, VG will denote, for

each subgroup V of A, the subgroup generated by all the elements o f the form vgwith v E V, gEG; VG is always admissible; and, i f V is finite, then so is VG. SinceA is a direct sum of cyclic groups, p''A =0, and so each finite subgroup of A musthave zero intersection with pkA for some positive integer lc

Let k(1) be the first positive integer for which SIG n p k (1) A = O . S i n c e S n p k o ) A

is characteristic in A, it is also admissible. Let St = S 1G + ( S n pk ( 1 )A ) ; t h e n S l /S , G

Decompositions of direct sums o f abelian groups 2 6 r

is an admissible subgroup and a direct summand in the elementary group SI S,G,On applying the Lemma to X = SI SIG , S1 I SIG , o n e o b t a i n s a n a d m i s s ib l e

complement, say SIG , f o r S1 SiG i n S I S , G. As T, and S' g en e ra t e S. and as.

SIG T1, T, and Snpko)A also generate S; moreover, T, n (S npk(1)A) =-

= S I n ( S n p ' (1 )A ) = SIG n( S n p k (1 )A ) = 0; so t ha t in fact S +

± ( S n p k( 1 )A ) .

For an inductive construction, suppose that 1 '1, Ti, k(I), k(i) have-

already been defined, in such a way that T, , Ti a r e a d m i s s i b l e s u b g r o u p s ,51 + + Si + + Ti , k (1) k (j), and S = T, + + Ti +(5n pk(f) A)

for every i with I i j . Let it denote the canonical projection of S onto S flpku)Acorresponding to the direct decomposition S + + T + (S npku)AI, andlet k( j + 1) be either k(j) or the first positive integer for which S1+ 1G i t fl p k ( i +l ) A — 0 ,

whichever is the larger. Check that Si +, G 2 1 i s a d m i s s i b l e . S i m i l a r l y t o t h e a p p l i c at i o n

in the preceding paragraph, the Lemma can be used to prove the existence o f anadmissible complement Ti + 1 o f S p k (1 1 )A i n S n p k U)A s uc h t ha t Si + 1G IE

It can easily be seen that the hypothesis carries over to I ' , , Ti ,141), I c ( j + 1).

This process defines, for each positive integer i, and admissible subgroup Trand a positive integer Ic(i). The subgroup generated by the T

i i s t h e i r d i r e c t s u m ,and it contains all the Si, s o t h a t i t i s e q u a l t o S . T h us , i f T, + + Ti is d en ot ed

by T1. one has that S= (T : co). Moreover, S = + (S pk( DA) for

every positive integerNext, let B, be a subgroup of A maximal with respect to being admissible and

having T1 f o r i t s s od e. For ano the r i nduc ti on , suppose that B1, Bi are already

defined in such a way that they form an increasing chain of admissible subgroupsand the socte o f B, is Ti whenever I i j . Then Bi i n t e r s e c t s Ti +, i n 0 , f o r i t s

socle does; so the subgroup generated by Bi a n d Ti, I s t h e i r d i r e c t s u m Bi + Ti +, ,

and its sode is Ti+ Th u s it is possible to choose Bi, , a s a s u b g r o u p w h i c h c o n t a i n s

Bi + Ti,, and is maximal with respect to being admissible and having for its

sode. This process provides, fo r each positive integer j, an admissible subgroupBi, such that these subgroups form an increasing chain, and the sode of each Bi-

is the corresponding P .Observe that, for each j. Bi a n d p k ( l ) A i n t e r s ec t i n 0 , f or t h ei r s o cl e s do so.

Therefore one can speak of the direct sum C of Bi a n d S p k ( ) A i n A . L e t U b e

the sode of AI Bi; c l e a r l y , C I Bi i s a n a d m is s i bl e s ub g ro up and a d irec t summand

in U. Hence, according to the Lemma, C/Bi h a s a n a d m i s s i b l e c o m p l e m e n t , s a y

B IB» i n U. Now B is admissible, and B n C=Bi. The sode of B contains T', and

so i t i s T i +(13 fl S n p k(D A ) ; b u t B n( S n p k( D A ) - = - B fl C T I( S A pk o A)

Bi npk(i)A =0, so that in fact the sode of B is just P. Hence, by the maximality

of Bi , i t fo ll ows that B Bi; therefore U= C/Bi. Now if E is any subgroup of A

such that Bi- < E , t h en E l Bi flU = E /Bi C/Bi >- 0, and so EFICBi: it follows

that E n p k(i )A > - O . T h us Bi is m ax im al among all the subgroups of A which intersect

p k(1)A in 0, so that a Lemma of M. ERDÉLYI (Lemma I in [3]; or, the main step

in the proof of Theorem 24. 8 in FUCHS [1]) proves that Bi i s a d i r e c t s u m m a n d

of A.Hence the union of the Bi i s p u r e i n A [ e . g . b y F ) o n p . 7 7 in F UC HS [ 1] ]; and

it contains the union o f the P , which is the whole sode o f A; so that in fact(Bi: 1 co) A [e. g. by K) on p. 78 in FUCHS [la Put B1 =131 , and apply

262 L . G. Kovdes

the Lemma (with X = Bi , / T = Bi) t o o b t a in a n a d m i ss i b l e c o m pl e m en t Bi+

for each Bi i n t h e c o r re s po n d in g Bi, ,. Then it follows readily that Bi

= ( B1 for every j, and so A = (B co) = [I CB':

co] -= ( B i : c o ) . Here a l l the Bi a r e a d m i s s i b l e s u b g r o u p s , a n d

npkii)A -- Bin p k(i)A =0 shows that they are all bounded subgroups as

well.(K) I f A is a direct sum o f cyclic p-groups, the Theorem is true.

PROOF. In view of (J), it may be assumed that A is bounded; say, p"A =O. LetThe sode of A be S. For i = l , n , let Ti b e a n a d m i s s i b l e c o m p l e m e n t o f S n p i A

in S n p i- 1A ; s u ch c o mp l em en t s ex is t, for each SnpiA and Snpi-lA is charac-

teristic in A and is therefore admissible, and each subgroup of S is a direct summandin every subgroup o f S which contains it, so that the Lemma can be applied toX — S npi A, Y S n p1 - 1A . T h e n S = T1 + + T, a nd S + + Ti+

+ n piA ) f or e ve ry i. As in the proof of (J), one constructs admissible subgroups

B', B " such that

(1(1) A = B1 + . . . + B " ,the socle o f _8' + + B is precisely T, + + Ti, a n d(K2) B i npi A =- 0,whenever I

One checks that the sode of Bi is precisely Ti, a n d t h a t a s f o l l o w s .

The assertion is trivial for i= 1 ; in fact, = Tt. L e t 1 a n d t e Ti. T h e n

t E p1 - 1. 4 ; say , t =p i-l a, a A. Write a as bi + + b„, according to (K1). By (K2),

_ pi -l b i 1_0, so that

(K3) t pi - la , + pi - 1 bn .

On the other hand, (K3) is a decomposition of t corresponding to (K1), and t E B1 ++ Bi, so that one must have t = p i

-l - bi. T h i s p r o v e s t h a t S i n ce

now the sode o f Bi cmtains Ti a n d i s c o n t a i n e d i n T , + + Ti, i t i s i n f a ct

Ti+[Bi n (T1 + + Ti _1)]; but Bi (T1 + + Ti_ 1) Bi n(B1 + + = 0,

and so the socle of Bi is precisely Ti. A l s o , ( K 2 ) i m p l i e s t h a t p ( p -t 1 B i ʻ

) 0 , s o t h a t

the converse inclusion has already been seen, so thatIt follows that every non-zero element in the sode o f Bi is o f height i— 1 in

so that Bi is a direct sum of isomorphic cyclic groups of order pi. This and (K1)imply that it can be assumed without loss of generality that A is a direct sum ofisomorphic cyclic groups; each of order pm, say. The proof will be completed underthis additional hypothesis.

, I n view of (F), the sode S of A is a direct sum o f finite, admissible, G-inde-composable subgroups T,. According to (G), each T is the sode of some admissible,G-indecomposable direct summand B, o f A. The B, are then also direct sums ofcyclic groups of order pm, as is every direct summand of A; the subgroup generatedby the B, is their direct sum, and it contains the whole sode S of A: it is also a directsum of cyclic groups of order pm, so that it must be the whole of A. Finally, eachB2. is finite, for its sode T, is finite.

Decompositions o f direct sums o f abelian groups 2 6 3

The steps (B), (E), (H), (I ), ( K ) together prove the Theorem.

Remark A f t e r the preparat ion o f the paper had been completed, PROFESSORREINHOLD BAER k indly called the attent ion of the author to the fact that a combi-nation of results of KULIKOV and KAPLANSKY implies that every direct summandof A is a direct s um of groups o f rank one; this would a l low some minor cutsin the present proof .

References

[1] L. FUCHS, Abel ian groups, Budapest, 1958.[2] a GRÜN, Einige Si.tze über Automorphismen abelscher p-Gruppen, Abb. M ath. Sent. UlliV-

Ham5urg 24 (1960), 54- 58.' t3] L. Kov,kcs, On subgroups of the basic subgroup, Pub!. M ath. Debrecen 5 (1958), 261- 264.[4] L . G. Kovikcs and M . F . NEWMAN, Direct complementation i n groups wi th opeiatos, Ar rch•

Math. 13 (1962), 427- 433.[5] A . G . KuRos0. Theory o f groups, Volume 1., New York, 1955.

, [6] 13. L . VAN DER WAERDEN, Algebra 11, Dr i tte Auflage, B e r l i n —G ö t t i n g e n— H e i d e l b e r g , 1 9 5 5 .(Received March 15, 1963.)