Abelian Groups - László Fuchs [Springer]

1 Direct Sums and Direct Products 45
(E) If A D˚i B i , where each B i is a direct sum B i D˚j C ij ,thenA D˚i ˚j C ij .
This is a refinement of the given decomposition of A. Conversely,ifA D
˚i ˚j C ij ,thenA D˚i B i where B i D˚j C ij .
(F) If in the exact sequence 0 ! B ˛!A ˇ!C ! 0; Im ˛ is a summand of A,
then A Š B ˚ C. In this case, we say that the exact sequence is splitting.
Any map W C ! A satisfying ˇ D 1 C is called a splitting map; then
A D Ker ˇ ˚ Im . Of course, there is another map: ı W A ! B with ı˛ D 1 B
indicating splitting: A D Im ˛ ˚ Ker ı.
Two direct decompositions of A, A D˚i B i and A D˚j C j are called isomorphic
if there is a bijection between the two sets of components, B i and C j , such that
corresponding components are isomorphic.
We now prove a fundamental result.
Lemma 1.1. Let C Dhci be a finite cyclic group where o.c/ D m D p r 1
1 pr k
k
with
different primes p i . Then C has a decomposition into a direct sum
C Dhc 1 i˚˚hc k i
.o.c i / D p r i
i /
with uniquely determined summands.
Proof. Define m i D mp r i
i and c i D m i c .i D 1;:::;k/. Then the m i are relatively
prime, so there are s i 2 Z such that s 1 m 1 CCs k m k D 1.Thenc D s 1 m 1 c CC
s k m k c D s 1 c 1 CCs k c k shows that the c i generate C. Clearly, hc i i is of order p r i
i ,
so disjoint from hc 1 ;:::;c i 1 ; c iC1 ;:::;c k i which has order m i . Hence we conclude
that C Dhc 1 i˚˚hc k i.
The uniqueness of the summands hc i i (but not of the generators c i ) follows from
the fact that hc i i is the only subgroup of C that contains all the elements whose
orders are powers of p i .
ut
Decomposition of Torsion Groups One of the most important applications of
direct sums is the following theorem that plays a fundamental role in abelian group
theory.
Theorem 1.2. A torsion group A is the direct sum of p-groups A p belonging to
different primes p:
The A p are uniquely determined by A.
A D˚p A p :
Proof. Given A, letA p consist of all a 2 A whose orders are powers of the prime p.
Since 0 2 A p , A p is not empty. If a; b 2 A,i.e.p m a D 0 D p n b for integers m; n 0,
then p nCm .a b/ D 0, soa b 2 A p ,andA p is a subgroup of A. Ifp 1 ;:::;p k are
primes ¤ p,thenA p \ .A p1 CCA pk / D 0, since every element of A p1 CCA pk
is annihilated by a product of powers of p 1 ;:::;p k . Thus the A p generate their direct
sum in A; itmustbeallofA, as it is obvious in view of Lemma 1.1.