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Abelian Groups - László Fuchs [Springer]

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444 12 Torsion-Free <strong>Groups</strong><br />

A D B i ˚ C i with C i DhX i ˚ Y; p 1 .k i x i C y/i (12.11)<br />

where the C i ’s are non-isomorphic (and indecomposable). Then to ensure that k i x i C<br />

y D k i r i b 1 C .k i t i 1/x 1 C .x 1 C y/ will be divisible by p, we must select k i such<br />

that<br />

k i r i 0 and k i t i 1 mod p: (12.12)<br />

For any k i with 12m 1, then we guarantee that<br />

k j 6 ˙k i mod p for i ¤ j, and therefore, no two of the groups C 1 ;:::;C m are<br />

isomorphic.<br />

ut<br />

This theorem also shows that, for any rank r 3, and for every integer m, there<br />

exist torsion-free groups of rank r which have at least m pairwise non-isomorphic<br />

decompositions into indecomposable summands. However, Lady’s theorem 6.9 will<br />

show that no torsion-free group of finite rank admits infinitely many non-isomorphic<br />

decompositions.<br />

Possible Rank Distributions The question of rank distributions in direct<br />

sums of finite rank torsion-free groups is too important to leave out completely.<br />

The problem was completely solved by Yakovlev and Blagoveshchenskaya in a<br />

satisfactory way. The proof is too long, it requires further structural information<br />

to be determined, so here we cannot do more than just stating the theorem without<br />

proof. As expected, the rank 1 summands play a special role.<br />

Theorem 5.6 (Yakovlev [1], Blagoveshchenskaya–Yakovlev [1]). Let<br />

n D r 1 CCr k D s 1 CCs`<br />

be two partitions of the positive integer n 3.r i ; s j 2 N/. Let u and v denote the<br />

number of r i and s j , respectively, that are equal to 1.<br />

A necessary and sufficient condition for the existence of a torsion-free group of<br />

rank n to admit a direct decomposition into indecomposable summands of ranks r i ,<br />

and also one with indecomposable summands of ranks s j , is that<br />

(i) r i n v for all i, and s j n u for all j; and<br />

(ii) if r i D n v for some i, then exactly one s j ¤ 1.and is equal to n v/, and if<br />

s j D n u for some j, then exactly one r i ¤ 1.and is equal to n u/. ut<br />

The proof of sufficiency is based on a clever construction of special indecomposable<br />

groups, called flower groups.

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