Abelian Groups - László Fuchs [Springer]

80 3 Direct Sums of Cyclic Groups
(2) Prove the following converse of Theorem 1.5: a group F is free if it has the
property that whenever B < A and A=B Š F, thenB is a summand of A.
(3) Give a presentation of Z.p 1 /, and one of Q.
(4) Let A be presented by a set of generators and defining relations, and assume
that the set of generators is the union of two disjoint subsets, fb i g i2I and fc j g j2J ,
such that each of the defining relations contains only generators from the same
subset. Then A D B ˚ C,whereB is generated by the b i ,andC by the c j .
(5) Let A be presented by a set of generators and defining relations, and B by
a subset of these generators and defining relations. Show that letting the
generators of B correspond to themselves qua generators of A induces a
homomorphism B ! A.
(6) For every set of generators, there is a minimal set of defining relations relative
to these generators (i.e., no relation can be omitted). [Hint: Theorem 1.6.]
(7) Let 0 ! A 1 ! A 2
˛!A3 ! 0 be an exact sequence, and i W F i ! A i .i D 1; 3/
epimorphisms where F i are free. If W F 3 ! A 2 is such that ˛ D 3 ,then
1 ˚ W F 1 ˚ F 3 ! A 2 is epic, and its kernel is Ker 1 ˚ Ker 3 .
(8) Let 0 ! F 1 ! F 2 !!F n ! 0 be an exact sequence of finitely generated
free groups. Prove the equality P n
kD1 . 1/k rk F k D 0.
(9) Assume fA n j n 2 Zg is a set of groups. Verify the existence of free groups
F n .n 2 Z/ and a long sequence
:::
˛n 2
! F n 1
˛n 1
! F n
˛n
! F nC1
˛nC1
! :::
such that ˛n 1˛n D 0 and Ker ˛n= Im ˛n 1 Š A n for every n 2 Z.
2 Finite and Finitely Generated Groups
We turn our attention to groups with a finite number of generators. First, we discuss
finite groups separately. Though this is a special case of the general theory of finitely
generated groups (to be developed independently), a short, direct approach to the
theory of finite groups is not without merit.
Finite Groups We start with a simple lemma.
Lemma 2.1. Let A be a p-group that contains an element g of maximal order p k for
an integer k >0.Thenhgi is a direct summand of A.
Proof. If A is infinite, then use Zorn’s lemma to argue that there is a subgroup B of A
maximal with respect to the property B\hgi D0. To show that A Dhgi˚B equals
A, by way of contradiction assume that some a 2 A does not belong to A . Replacing
a by p i a if necessary, we may also suppose that pa 2 A ,i.e.pa D mg C b for some
m 2 Z; b 2 B. By the maximality of the order of g, wehavep k 1 mg C p k 1 b D
p k a D 0. Hence p k 1 mg D 0, som must be divisible by p, say,m D pm 0 .Then
a 0 D a m 0 g … A satisfies pa 0 D b. By the maximal choice of B, hB; a 0 i\hgi ¤0,