Abelian Groups - László Fuchs [Springer]

2 Finite and Finitely Generated Groups 83
generators, so it is free. Hence A Š U ˚ A=U (again by Theorem 1.5), where U is
a finitely generated group isomorphic to a subgroup of Q, so it is cyclic.
Stacked Basis Theorem A third proof of Theorem 2.5 is based on the following
theorem which is of considerable interest in its own right (see the more general Theorem
6.5). We say fa i g i2I is a basis of A if A D˚i2I ha i i.
Theorem 2.6. If H is a subgroup of the free group F of finite rank k, then F and H
have ‘stacked bases:’
F Dha 1 i˚˚ha k i and H Dhb 1 i˚˚hb k i
such that there are non-negative integers m 1 ;:::;m k satisfying
b i D m i a i .i D 1;:::;k/ and m i 1 jm i .i D 2;:::;k/:
Proof. We select a free basis fx 1 ;:::;x k g of F with the following extremal property:
H contains an element b 1 D n 1 x 1 CCn k x k with a minimal positive coefficient n 1 .
In other words, for another basis of F, or for another permutation of the basis
elements, or for other elements of H, the leading positive coefficient is never less
than n 1 .
The first observation is that n 1 jn i .i D 2;:::;k/. For,ifn i D q i n 1 C r i .q i ; r i 2
Z;0 r i < n 1 /; then we can write b 1 D n 1 a 1 C r 2 x 2 CCr k x k where fa 1 D
x 1 C q 2 x 2 CCq k x k ; x 2 ;:::;x k g is a new basis of F. By the special choice of
fx 1 ;:::;x k g,wemusthaver 2 D D r k D 0. The same argument shows that if
b D s 1 x 1 CCs k x k .s i 2 Z/ is any element of H, thens 1 D qn 1 for some q 2 Z.
Hence b qb 1 2hx 2 i˚˚hx k iDF 1 . We conclude that F has a decomposition
F Dha 1 i˚F 1 such that H Dhb 1 i˚H 1 ,whereb 1 D n 1 a 1 and H 1 F 1 .Using
induction hypothesis for the pair H 1 ; F 1 , we infer that F has a basis fa 1 ;:::;a k g and
H has a basis fb 1 ;:::;b k g such that b i D m i a i for some non-negative integers m i .
It remains to establish the divisibility relation m 1 jm 2 (the others will follow by
induction). Write m 2 D tm 1 C r with t; r 2 Z;0 r < m 1 .Thenfa D a 1 C
ta 2 ; a 2 ;:::;a k g is a new basis of F, intermsofwhichwehaveb 1 C b 2 D m 1 a 1 C
.tm 1 C r/a 2 D m 1 a C ra 2 2 H. The minimality of m 1 D n 1 implies r D 0. ut
With the aid of Theorem 2.6, we can reprove the implication (i) ) (ii)
in Theorem 2.5. IfA is generated by k elements, then A Š F=H, whereF is a
free group on a set of k elements. Choosing stacked bases for F and H, as described
in Theorem 2.6, we obtain
A Šha 1 i=hm 1 a 1 i˚˚ha k i=hm k a k i:
Consequently, A is the direct sum of cyclic groups: the i th summand is cyclic of
order m i if m i > 0, and infinite cyclic if m i D 0. The numbers m i are called
elementary divisors.