Abelian Groups - László Fuchs [Springer]

446 12 Torsion-Free Groups
(5) For every integer n 1, there is an indecomposable group A of rank n such that
A ˚ B Š A ˚ C for suitable non-isomorphic indecomposable groups B; C of
finite rank. [Hint: Theorem 5.5.]
(6) (Sa¸siada) If X D h5 1 x;3 1 y;2 1 .x C y/i, then every indecomposable
summand of X .n/ for any n >1is isomorphic to X. [Hint: any such summand
must have a subgroup Šh5 1 x;3 1 yi of index 2.]
(7) (a) (Corner) Let A be torsion-free of rank n, and B; C isomorphic pure
subgroups of A of rank n 1. ThenA=B Š A=C. [Hint: reduce to
n D 2; B ¤ C, and calculate the torsion subgroup of A=.B C C/ in two
ways.]
(b) Cancellation by a finite rank torsion-free group is permitted if the complements
are of rank 1.
6 Direct Decompositions of Finite Rank Groups:
Positive Results
We have seen that a torsion-free group of finite rank may have numerous nonisomorphic
direct decompositions. Is there anything positive which can be stated
about such decompositions? In particular, must such a group have but finitely many
non-isomorphic such decompositions? Lady [3] has answered this question in the
positive.
This is an important result, therefore we give a detailed proof, though from here
on nearly everything is ring theory. The proof of the main Theorem 6.9 relies on
several not so well-known facts from ring theory, in particular, on a deep result by
Jordan–Zassenhaus that will be quoted without proof.
Lemmas on Rings and Modules In the next lemma, E stands for the endomorphism
ring of A.
Lemma 6.1. Let B and B 0 be direct summands of a group A, and let ; 0 denote
the projections onto them. Then B Š B 0 if and only if the right ideals E and 0 E
are isomorphic as right E-modules.
Proof. See the proof of the more complete Lemma 1.6 in Chapter 16. ut
We break down the discussion of the required ring-theoretical background into
a series of easy lemmas. The first lemma can be found in most text books on rings
and modules. R will denote a ring with 1, and J its Jacobson radical.
Lemma 6.2 (Nakayama’s Lemma). If M is a finitely generated right R-module
such that M D MJ, thenMD 0.
Proof. Let fx 1 ;:::;x k g with k 1 be a minimal generating system for M. By
hypothesis, we have x 1 D x 1 r 1 CCx k r k with r i 2 J. Since1 r 1 is invertible in
R, x 1 can be expressed as a linear combination of the other generators, contrary to
the minimality of the generating system. Thus k D 0, and hence M D 0. ut