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Unformatted text preview: 198 3. PRODUCTS OF GROUPS where
A0 is free abelian, and
for i 1, Ai Š Zai , where
ai 2, and ai divides aj for i Ä j ;
G D B0 B1 B2 Bt ; where
B0 is free abelian, and
for i 1, Bi Š Zbi , where
bi 2, and bi divides bj for i Ä j ;
We have to show that rank.A0 / D rank.B0 /, s D t , and ai D bi for all
By Lemma 3.6.3 , we have
Gtor D A1 As D B1 B2 Bt : Hence A0 Š G=Gtor Š B0 . By uniqueness of rank, Proposition 3.5.5, we
have rank.A0 / D rank.B0 /.
It now sufﬁces to prove that the two decompositions of Gtor are the
same, so we may assume that G D Gtor for the rest of the proof.
Let a denote the period of G . By Exercise 3.6.6, as D b t D a.
We proceed by induction on the length of a, that is, the number of
primes (with multiplicity) occuring in an prime factorization of a. If this
number is one, then a is prime, and all of the bi and aj are equal to a.
In this case, we have only to show that s D t . Since aG D f0g, by
Lemma 3.6.5, G is an Za –vector space; moreover, the ﬁrst direct product
decomposition gives G Š Zs and the second gives G Š Zt as Za –vector
spaces. It follows that s D t by uniqueness of dimension.
We assume now that the length of a is greater than one and that the
uniqueness assertion holds for all ﬁnite abelian groups with a period of
Let p be prime number. Then x 7! px is a group endomorphism of G
that maps each Ai into itself. According to Lemma 3.6.4, if p divides ai
then Ai =pAi Š Zp , but if p is relatively prime to ai , then Ai =pAi D f0g.
G=pG Š .A1 A2
As /=.pA1 pA2
Š A1 =pA1 A2 =pA2 k
As =pAs Š Zp ; where k is the number of ai such that p divides ai .
Since p.G=pG/ D f0g, according to Lemma 3.6.5, all the abelian
groups in view here are actually Zp –vector spaces and the isomorphisms
are Zp –linear. It follows that the number k is the dimension of G=pG as
an Zp –vector space. Applying the same considerations to the other direct ...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
- Fall '08