We continue our study of groups according to the plan outlined in the introduction of Chapter I. The chief emphasis will be on obtaining structure theorems of some depth for certain classes of abelian groups and for various classes of (possibly non
abelian) groups that share some desirable properties with abelian groups. The chapter has three main divisions which are essentially independent of one another, except that results from one may be used as examples or motivation in the others.
The interdependence of the sections is as follows.
I 3
2
�
Most of Section 8 is independent of the rest of the chapter.
1. FREE ABELIAN GROUPS
We shall investigate free objects in the category of abelian groups. As is the usual custom when dealing with abelian groups additive notation is used throughout this section. The folJowing dictionary may be helpful.
ab . . . a
+
ba . . . -a -1 e. . . . . . . . . . . 0 an . . . - . . . - . . . . na ab-1 . . . . . . . . . . . . . . . . . . .. . . . . . . . • a - b HK . . . . . . . . . . . . . . . . . . . . . . . . .
H +
KaH . . . . . . . . . . . . . . . . . . . . . . . . . . . . a
+
H70
1. F R E E A B ELIAN GROU PS
G X H . . . . . . . . . . . . . . . . . . .
G Ef:1
HH V K . . . . . . . . . . . . . . . . . . . . . . . . .H + K
IIW
Gi . . . . . . . . . . . . . . . . . . . . . . . . ._E G ..
ial i•l
weak direct product . . . direct sum
71
For any group
G
in additive notation, (m + n)a = ma + na (a e G; m,n e Z). If the group is abelian, then m(a + b) = ma + mb. If X is a nonempty subset of G, then by Theorem 1.2.8 the subgroup (X) generated by X in additive notation consists of all linear combinations n1x1 + n2x2 + · · · + nkxk (n .. e Z, Xi e X). In particular, the cyclic group (x) is f nx I n e Z } .A basis of an abelian group F is a subset X of F such that (t} F = (X); and (ii) for distinct xhx2, . . . , xk eX and ni e Z,
The reader should not be misled by the tempting analogy with bases of vector spaces (Exercise 2).
Theorem 1.1. The following conditions on an abelian group F are equivalent.
(i) F has a nonempty basis.
(ii) F is the (internal) direct sum of a family of infinite cyclic �-ubgroups.
(iii) F is (isomorphic to) a direct sum of copies ofrhe additive group Z of integers.
(iv) There exists a nonempty set X and a function L : X � F with the following property: given an abelian group G and function f : X � G, there exists a unique homo
morphism of groups f: F -+ G such that ft = f. In other words, F is a free object in the
category of abelian groups.
An abelian group F that satisfies the conditions of Theorem 1 . 1 is called a free
abelian group (on the set X). By definition the trivial group 0 is the free abelian group on the null set 0.
SKETCH OF PROOF OF 1.1. (i) ==> (ii) If X is a basis of F, then for each x € X, nx = 0 if and only if n = 0. Hence each subgroup (x) (x e X) i� infinite cyclic (and normal since F is abelian). Since F = (X), we also have F = (U (x) ). If for some z e X, (z) n (U (x)) � 0, then for some nonzero n € Z, nz = n1x1 + x.X · · · + nkxk
Xt.X
x '#z
with z,x1, . . . , xk distinct elements of X, which contradicts the fact that X is a basis.
Therefore (z) n (U (x)) = 0 and hence F =
:E
(x) by Definition 1.8.8.x�x x�)(
� ¢- z
(ii) ==> (iii) Theorems 1.3.2, 1.8.6, and 1.8. 10.
(iii) ==> (i) Suppose F
"'"',E
Z and the copies of Z are indexed by a set X. For each X e X, let ()X be the element t ui } of:E
Z, where Ui = 0 for i � X, and Ux = 1 . Verifythat { Ox I x e X} is a basis of
:E
Z and use the isomorphism F "'"':E
Z to obtain a basis of F.(i) ==> (iv) Let X be a basis of F and t : X ---+ F the inclusion map. Suppose we are given a map f: X ---} G. If u e F, then u = n1x1 + · · · + nkXk (n.: e Z; Xi e X) since X
k
generates F. If u = m1x1 + · · · + mkxk, (mk e Z), then
.L:
(ni - mi)xi = 0, whencei - 1
72 CHAPTER I I T H E STR UCTU R E OF GROU PS
ni = mi for every i since X is a basis. Consequently the map J : F � G, given by J(u) = 1
(t1
n;x;)
= n, f(x,) + · · · + nk f(xk), is a well-defined functiOn such that J, = f Since G is abelian 1 is easily seen to be a homomorphism. Since X generates F, any homomorphism F � G is completely detemined by its action on X. Thus if g : F � G is a homomorphism such that g, = f, then for any x eX g(x) = g(t(x))= f(x) = f(x), whence g = J and J is unique. Therefore, by Definition I.7.7 F is a free object on the set X in the category of abelian groups.
(iv) ==> (iii). Given " : X � F, construct the direct sum
L
Z with the copies of Z indexed by X. Let Y = { 8x / x c: XJ be a basis ofL
Z as in the proof of (iii) ==? (i).The proof of (iii) ==> (i) ==> (iv) shows that
L
Z is a free object on the set Y. Since we clearly have lXI = I Y/, p ro....�L
Z by Theorem 1.7.8. •Given any set X, the proof of Theorem 1 . 1 indicates how to construct a free abelian group F with basis X. Simply let F be the direct su,m
Lz
, with the copies of Z indexed by X. As in the proof of (iii) ==> (i), { 8x I x e X I is a basis of F =Lz,
and F isfree on the set { 8x I x E XJ . Since the map t : X � F given by x � 8x is injective it follows easily that F is free on X in the sense of condition (iv) of Theorem 1 . 1 . In this situation we shall identify X with its image under " so that X C F and the cyclic sub
group (8x) = { n8x I n E Z} = ZBx is written (x) = Zx. In this notation F =
L
(8x) iswritten F =
L
Zx, and a typical element of F has the form ntX
t + · · · XEX + nkxk (nt c: Z., xi e X). xeX In particular, X = t(X) is a basis of F.Theorem 1.2. Any two bases of a free abelian group F have the same cardinality.
The cardinal number of any basis X of the free abelian group F is thus an invari
ant of F; IX/ is called the rank of F.
SKETCH OF PROOF OF 1.2. First suppose F has a basis X of finite cardinal
ity n so that F � Z
EB
· · ·EB
Z (n summands). For any subgroupG
of F verify that 2G = { 2u Iu
e G } is a subgroup of G. Verify that the restriction of the isomorphism F roo...� Z EB · · ·EB
Z to 2F is an isomorphism 2F roo...� 2ZEB
· · ·EB
2Z, whence F /2F /"'.../ Z/2ZEB
. . . EB Z/2Z /"'.../ z2 EB . . .EB
z:2 (n summands) by Corollary 1.8.1 1 . Therefore /F/2F/ = 211• If Y is another basis of F and r any integer such that / Y/ > r.then a similar argument shows that JF/2F/ > 2r, whence 2r < 2
n
and r < n. It follows that / Y/ = m < n and /F/2F/ = 2m. Therefore 2m = 211 and /X/ = n = m = / Y/ . If one basis of F is infinite, then all bases are infinite by the previous paragraph.Consequently, in order to complete the proof it suffices to show that /XI = /F/, if X is any infinite basis of F. Clearly
/X/
< /F/ . Let S = U Xn, whereXn = X X · · · X XneN*
(n factors). For each s = (xh . . . , Xn) E S let Gs be the subgroup (x�, . . .
, Xn
).
ThenGs roo...� Zy1 ffi · · · EB Zy, where Yt, . . . , Yt (t < n) are the distinct elements of { x., . . . , .x nJ. Therefore, I Gs/ = jZtj =
/
Z/ = � o by Introduction, Theorem 8.12.Since F = U Gs, we have /FI = IU Gsl <
/SINo
by Introduction, Exercise 8.12.seS S£8
But by Introduction, Theorems 8 . 1 1 and 8. 12,
lSI
=IXj,
whencejFI
<IXI�o
=jxj.
Therefore IF/ = /X/ by the Schroeder-Bernstein Theorem. •
1 . F R E E A B ELIAN G R OU PS 73
Pro position 1.3. Let F. be the free abelian group on the set X. and F2 the free abelian group on the 5et X2. Then F. "" F2 if and only ifFt and F2 have the same rank (that is,
IX./ = IX2/).
REMARK . Proposition 1 .3 is also true for arbitrary nonabelian free groups (as in Section 1.9); see Exercise 12.
SKETCH OF PROOF OF 1.3. If a : Ft /"'../ F2, then a{X.) is a basis of F2, whence /X./ = /a(XJ)I = IX2I by Theorem 1 .2. The converse is Theorem I. 7 .8. •
Theorem 1.4. Every abelian group G is the homomorphic image of a free abelian group of rank !X I, where X is a set of generators of G.
PROOF. Let F be the free abelian group on the set X. Then F =
L
XEX Zx and rankF = lXI . By Theorem 1 . 1 the inclusion map X --..
G
induces a homomorphism 1 : F �G
such that lx � x €G,
whence X C Im f. Since X generatesG
we must have Im 1=G.
•We now prove a theorem that will be extremely useful in analyzing the structure of finitely generated abelian groups (Section 2). We shall need
Lem ma 1.5. If{ x., . . . , Xn } is a basis of a free abelian group F and a € Z, then for all i � j { x�, . . . , Xj-t,Xj + axi,Xj+I, . . . , Xn } is also a basis ofF.
PROOF. Since x1 = -axi + (xi + axi), it follows that F = (x., . . . , Xj-t,Xi +
axi,Xi+h . . . , Xn). If ktXt + · · · + ki(xi + axi) + · · · + knXn = 0 (ki € Z), then k1x1 + · · · + (ki + k1a)xi + · · · + kixi + · · · + knxn = 0, which implies that kt = 0 for ali t. •
Theorem 1.6. If F is a free abelian group of finite rank n andG is a nonzero subgroup of F, then there exists a basis { x., . . . , Xn } of F, an integer r ( 1 < r < n) and positive integers d., . . . , dr such that d1 I d2 I · · · I dr and G is free abelian with basis
{ d1x1, . . . , drXr J .
REMARKS. Every subgroup of a free abelian group of (possibly infinite) rank a is free of rank at most a; see Theorem IV.6.1 . The notation "d1 I d2 l . . . I d," means ... d1 divides d2, d2 divides da., etc."
PROOF OF 1.6. If n = 1 , then F = (x.) /"'../ Z and
G
= (dtxt) /"'../ Z (di € N*) by Theorems 1.3.5, 1.3.1 , and 1.3.2. Proceeding inductively, assume the theorem is true for all free abelian groups of rank less than n . Let S be the set of all those integers s such that there exists a basis { Yh . . . , Yn } of F and an element inG
of the formSYt + k2Y2 + · · · + knYn (ki € Z). Note that in this case { Y2.,YhYa, . . . , Yn } is also a basis of F, whence k 2 € S; similarly k1 € S for j = 3,4, . . . , n. Since
G
� 0, we have S -:F- 0. Hence S contains a least positive integer d1 and for some basis { yt, . . . , Yn )74 C HAPTER I I T H E STR UCT U R E OF GROU PS
of F there exists v € G such that v = d1Y1 + k2Y2 +
· ·
· + knYn - By the division algorithm for each i = 2, . . . , n, ki = d1qi +ri
with 0 < ri < d�, whence v = dt(YI + l/2)'2 + . . . + Qn
Yr
) + r2Y2 + . . . +rnYn·
Let x. = Yt + Q2Y2 + . . . +lJnYn ;
then by Lemma 1 .5 W = { x� ,y2, • • • , Yn } is a basis of F. Since v E G, ri < d1 and W in any order is a basis of F, the minimality of d. in S implies that 0 =
r
2 = r3 =· · ·
= r,so that d1x1 = v e G.
Let H = (y2,}'3, . . . , Yn). Then H is a free abelian group of rank n - 1 such that F = (xt) ffi H. Furthermore we claim that G = (v)
EB
(G n H) = (dtX
t)EB
(G n H).Since { Xt,Y2, . . . , Yn l is a basis of F, (v) n ( G n H) = 0. If u = ttXl + t2Y2 + . . . +
tnYn e G (t
i
e Z), then by the division algorithm ft = dtl}t + r1 with 0 < r1 < d • . Thus G contains u - lftV = r1x1 + f2Y2 + · · · + fnYn · The minimality of d1 in S implies that rl = 0, whence 12Y2 + . . . + tnYn E G n H and u = qlv + (t2
Y
2 + . . . + lnYn)Hence G = (v) + (G n H), which proves our assertion (Definition 1.8.8).
Either G
n H
= 0, in which case G = (d1x1) and the theorem is true or G n H :;e 0. Then by the inductive assumption there is a basis { x2,X3, . . . ' Xn } of H and positive integersr
,d2,d3, . . . ' dr such that d2 I d3 I · . · I dr and G n H is free abelian with basis { d2x2, . . . , drxr } . Since F = (xt) Ef) H and G = (dtxt) Ef) (G n H),it follows easily that { x�,x2, . . . , Xn } is a basis of F and { d1x1, . . . , drxr } is a basis of G. To complete the inductive step of the proof we need only show that d1 I d2. By the division algorithm d2 = qd1 + ro with 0 <
r0
< d1. Since { x2,x1 + qx2,x3, . . • , Xn Iis a basis of F by Lemma 1 .5 and r
oX
2 + dt(X
t + qx2) = d1x1 + d2x2 e G, the minimality of dt in S implies that ro = 0, whence dt ! d2. •
Corollary 1.7. JfG is a finitely generated abelian group generated by n elements, then every subgroup H of G may be generated by m e/e1nents with m < n.
The corollary is false if the word abelian is omitted (Exercise 8).
PROOF OF 1.7. By Theorem 1 .4 there is a free abelian group F of rank n and an epimorphism 1r : F � G. 1r-1(H) is a subgroup of F, and therefore, free of rank
m < n by Theorem 1 .6. The image under 1r of any basis of 1r-1(H) is a set of at most m elements that generates 1r(1r-1(H)) = H. •
EX E R C I S E S
1 . (a) If G is an abelian group and m e Z, then mG = { mu I u € G } is a sub
group of G.
(b) If G �
L
Gi, then mG rovL
mGi and G/m G �L Gi/mGi.
i£1 iel iel
2. A subset X of an abelian group F is said to be linearly independent if n1x1 +
·
· · +n1cXJc = 0 always implies ni = 0 for all i (where n, e Z and x�, . . . , xk are distinct elements of X).
(a) X is linearly independent if and only if every nonzero element of the sub
group (X) may be written uniquely in the form n1x1 + · · · + n
k
xk (ni € Z,ni � 0,x1, . . . , xk distinct elements of X).
(b) If F is free abelian of finite rank n, it is not true that every linearly independent subset of n elements is a basis [Hint: consider F = Z].
I
1. F R E E A B ELIAN G R O U PS 75
(c) If F is free abelian, it is not true that every linearly independent subset of F may be extended to a basis of F.
(d) If F is free abelian, it is not true that every generating set of F contains a basis of F. However, if F is also finitely generated by n elements, F has rank
nt < n.
3 . Let X = { ai I i e I J be a set. Then the free abelian group on
X
is (isomorphic to) the group defined by the generators X and the relations (in multiplicative notation) { aia;ai-tai-1 = e I i,j e /} .
4. A free abelian group is a free group (Section 1.9) if and only if it is cyclic.
5. The direct sum of a family of free abelian groups is a free abelian group. (A direct product of free abelian groups need not be free abelian; see L. Fuchs
[ 1 3, p. 1 68].)
6. If F =
L
Zx is a free abelian group, and G is the subgroup with basisxeX
X' = X - { xo } for some x0 e X, then F/G .:::: Zxo. Generalize this result to ar- bitrary subsets
X'
of X.7. A nonzero free abelian group has a subgroup of index n for every positive integer n.
8. Let G be the multiplicative group generated by the real matrices a =
(� �)
and b =
G �}
If H is the set of all matrices in G whose (main) diagonal entries are 1 , then H is a subgroup that is not finitely generated.9. Let G be a finitely generated abelian group in which no element (except 0) has finite order. Then G is a free abelian group. [Hint: Theorem 1 .6.]
10. (a) Show that the additive group of rationals Q is not finitely generated.
(b) Show that Q is not free.
(c) Conclude that Exercise 9 is false if the hypothesis "finitely generated" is omitted.
1 1 . (a) Let G be the additive group of all polynomials in x with integer coefficients.
Show that G is isomorphic to the group Q* of all positive rationals (under multiplication). [Hint: Use the Fundamental Theorem of Arithmetic to con
struct an isomorphism.]
(b) The group Q* is free abelian with basis {pjp is prime in Z}.
12. Let F be the free (not necessarily abelian) group on a set X (as in Section 1.9) and G the free group on a set Y. Let F' be the subgroup of F generated by
{ aba-tb-1 I a,b e F} and similarly for G'.
(a) F' <1 F, G' <1 G and FjF', GI G' are abelian [see Theorem 7.8 below].
(b) F/F' [resp. GIG'] is a free abelian group of rank IX/ [resp. I YJ]. [Hint:
{ xF' I x e X I is a basis of F/ F'.]
(c) F "-.; G if and only if lXI = I Yl . [Hint: if <p : F "-.; G, then <P induces an isomorphism F 1 F' "-.; GI G'. Apply Proposition 1 .3 and (b). The converse . is Theorem 1 .7.8.]
76 CHAPTE R I I T H E STR U CTU R E OF GROU PS
2. FI N ITELY GENERATED ABELIAN GRO U PS
We begin by proving two different structure theorems for finitely generated abelian groups. A uniqueness theorem {2.6) then shows that each structure theorem provides a set of numerical invariants for a given group {that is, two groups have the same invariants if and only if they are isomorphic). Thus each structure theorem leads to a complete classification {up to isomorphism) of all finitely generated abelian groups. As in Section 1 , all groups are written additively. Many of the results (though not the proofs) in this section may be extended to certain abelian groups that are not finitely generated ; see L. Fuchs [1 3] or I. Kaplansky [1 7] .
All of the structure theorems to be proved here are special cases of corresponding theorems for finitely generated modules over a principal ideal domain (Section IV.6).
Some readers may prefer the method of proof used in Section IV.6 to the one used here, which depends heavily on Theorem 1 .6.
;
Theorem 2.1. Every finitely generated abelian group G is (isomorphic to) a finite direct sum of cyclic groups in which the finite cyclic summands {if any) are of orders m1, . . . , fit, where m1 > 1 and m1 I m2 l · · · lmt.
PROOF. If
G =;e
0 andG
is generated by n elements, then there is a free abelian group F of rank n and an epimorphism 1r : F �G
by Theorem 1 .4. If 1r is an isomorphism, then
G
� F � Z EB · · · EB Z (n summands). If not, then by Theorem 1 .6 there is a basis { Xt, . . . , Xn I of F and positive integers dt, . . . , dr such that I < r < n,n
d1 I d2 l · · · I dr and f dtXI, . . . , drXr } is a basis of K = Ker 1r. Now F =
L (xi)
andr i - 1
K = i = l
L (dixi),
where(xi)
'"'-./ Z and under the same isomorphism(
di
xi) � diZn
= { diu I u e Z } . For i = r + I , r + 2, . . . , n let di = 0 so that K =
L
i = l(
dix
i).Then by Corollaries 1.5.7, 1.5.8, and 1.8. 1 1
If di = I , then Z/diZ = Z/Z = 0; if di > I , then Z/diZ r-v Zdi; if di = 0, then Z/diZ = Z/0 � Z. Let m t , . . . , n1t be those di {in order) such that di � 0, I and let s be the number of dt such that
di
= 0. ThenG
'"'-./ Zml EB · · · EB Zm t EB {Z EB · · · EB Z),where m1 > I , 1n1 I n12 ., I · · · I m, and (Z EB · · · EB Z) has rank s. •
Theorem 2.2. Every finitely generated abelian group G is (isomorphic to) a finite direct sun1 of cyclic groups, each of which is eirher infinite or oforder a power ofaprime.
SKETCH OF PROOF. The theorem is an immediate consequence of Theorem
2 . 1 and the following lemma. Another proof is sketched in Exercise 4. •
T
j
'2. F I N I TElY GEN ERATED ABEliAN G ROU PS 77
Lem ma 2.3. If m is a positive integer and m = Ptn1P2n2 . · · Ptnt (ph . . . , Pt distinct primes and each ni > 0), then Zm f"V Zp1nl EB ZP2n� ffi · · · EB ZPtnt.
SKETCH OF PROOF. Use induction on the number t of primes in the prime decomposition of m and the fact that
Zrn
f"V
Zr EB Zn whenever (r,n) = 1 'which we now prove. The element n = n1 e Zrn has order r (Theorem 1.3.4 (vii)), whence Zr f"V (nl ) < Zrn and the map 1/;1: Zr -}> Zrn given by k � nk is a monomor
phism. Similarly the map 1/;2: Zn � Zrn given by k � rk is a monomorphism. By the proof of Theorem 1.8.5 the map 1/; : Zr EB Zn � z rn given by (x ,y) )---4 1/;I(x) + 1/;2(y) =
nx + ry is a well-defined homomorphism. Since (r,n) = I , ra + nb = 1 for some a,b € Z (Introduction, Theorem 6.5). Hence k = rak + nbk = 1/;(bk,ak) for all k e Zrn and 1/; is an epimorphism. Since IZr EB Zn l = rn = IZTn l, 1/; must also be a monomoi]Jhism. •
Corollary 2.4. JfG is a finite abelian group of order n, then G has a subgroup of order m for every positive integer m that divides n.
k
SKETCH OF PROOF. Use Theorem 2.2 and observe that
G f"V L Gi
impliesi - 1
that
I GI
=! Gtl1 G2I ·
·· I Gkl
and for i < r, pr-2Prf"V
ZPi by Lemma 2.5 (v) below. •REMARK. Corollary 2.4 may be false if
G
is not abelian (Exercise 1.6.8).In Theorem 2.6 below we shall show that the orders of the cyclic summands in the decompositions of Theorems 2.1 and 2.2 are in fact uniquely determined by the group
G.
First we collect a number of miscellaneous facts about abelian groups that will be used in the proof.Lemma 2.5. Let G be an abelian group, m an integer and p a prime integer. Then each of the following is a subgroup ofG:
(i) mG = { m u I u € G} ;
(ii) G[mJ = t u € G
I
mu = 0} ;(iii) G(p) = { u € G I
l
ui = pn for some n > 0 } ; (iv) Gt = { u € G l f ul is finite} .In particular there are isomorphisms
(v) Z,n[P] ""' Zp (n > 1 ) and pmzpn f"V Zpn-m (m < n).
Let H and Gi (i E I) be abelian groups.
(vi) lfg : G �
L
Gi is an isomorphism, then the restrictions ofg to mG andG[m]respectively are isomorphisms icl mG f"V
L
U.I mGi and G[m) :=:::.E Gi[m).
(vii) Iff : G � H is an isomorphism, then the restrictions off to Gt and icl O(p) re- spectively are isomorphisms Gt l"'oV Ht and G(p) f"V H(p).
78 CHAPT E R I I T H E STR U CTU R E OF G ROU PS
SKETCH OF PROOF. (i)-(iv) are exercises; the hypothesis that G is abelian is essential (Sa provides counterexamples for (i)-(iii) and Exercise 1 .3.5 for (iv)).
(v) pn
-
1 € Zpn has order p by Theorem 1.3.4 (vii), whence (pn-
l)"--' zp
and (pn
-1)< Zpn[p]. If u e ZPn[p], then pu = 0 in Zpn SO that pu = 0 (mod pn) in Z. But pn I pu impliespn-1 I u. Therefore, inZPn' u E (p11-1) and ZPn[P] < (pn
-
1)
. For the second statement note that pm E Zpn has order pn-m by Theorem 1.3.4 (vii). Therefore pmZ
Pn
= (pm) ""Zpn-m· (vi) is an exercise. (vii) If f :G -+ His a homomorphism and x € G(p) has order pn, then pnf(x) = f(pn
x
)=
f(O) = 0. Therefore f(x) E H(p). Hence f : G(p)�
H(p ). If f is an isomorphism then the same argument shows that f-1 : H(p)�
G(p). Sinceff-
1 = l 11cp> andf-
1f = 1a<r>h G(p) "" H(p). The other conclusion of (vii) is proved similarly. •
If G is an abelian group, then the subgroup G, defined in Lemma 2.5 is called the
torsion subgroup of G. If G = G, then G is said to be a torsion group. If G, = 0, then G is said to be torsion-free. For a complete classification of all denumerable torsion groups, see I. KapJansky [17].
Theorem 2.6. Let G be a finitely generated abelian group.
(i) There is a unique nonnegative integer s such that the number of infinite cyclic summands in any decomposition ofG as a direct sum ofcyclic groups is precisely s;
(ii) either G is free abelian or there is a unique list of (not necessarily distinct) positive integers mh
.
. . , mt such that m1 > 1 , m.I
m2 l · · · I mt andwith F free abelian;
(iii) either G is free abelian or there is a list of positive integerj Pt81, • • •
,
Pk8k, which is unique except for the order of its· members, such that p1, .. .
, Pk are (not necessarily distinct) primes, s1,. . .
, Sk are (not necessarily distinct) positive integers andwith F free abelian .
PROOF. (i) Any decomposition of G as a direct sum of cyclic groups (and there is at least one by Theorem 2.1) yields an isomorphism G "" H
EB F,
where H is a direct sum of finite cyclic groups (possibly 0) andF
is a free abelian group whose rank is precisely the number s of infinite cyclic summands in the decomposition. IfL
:
H�
HEB F
is the canonical injection (h � (h,O)), then clearly t(H) is the torsion subgroup of HEB F.
By Lemma 2.5, Gt "" t(H) under the isomorphism G ""H EB
F.Consequently by Corollary 1.5.8, G/ Gt
1"'..1 (F EB
H)/ t(H)1"'..1
F. Therefore, any decomposition of G leads to the conclusion that G/ Gt is a free abelian group whose rank is the number s of infinite cyclic summands in the decomposition. Since G I Gt does not depend on the particular decomposition and the rank of GIG, is an invariant by Theorem 1 .2, s is uniquely determined.(iii) Suppose G has two decompositions, say
T
G ""
L zni EB F
andi= 1
G =