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Abelian Groups - László Fuchs [Springer]

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2 Slender <strong>Groups</strong> 491<br />

Example 2.5. The following groups are slender: Z, and more generally, every proper subgroup of<br />

Q, and direct sums of these groups: the reduced completely decomposable groups. Also, direct<br />

sums of any number of reduced countable torsion-free groups A i . On the other hand, Q; P are not<br />

slender.<br />

Monotone Subgroups Though the group P is not slender, it contains a lot of<br />

uncountable slender subgroups, as is shown by the next theorem.<br />

A subgroup M of P D Q n2N he ni is called a monotone subgroup (Specker [1])<br />

if it satisfies:<br />

(i) P n2N k ne n 2 M implies that also P n2N m ne n 2 M where m n D<br />

max.1; jk 1 j;::::;jk n j/I<br />

(ii) P n2N k ne n 2 M and j`nj jk n j .n 2 N/ imply P n2N `ne n 2 M.<br />

Example 2.6.<br />

(a) PEvery monotone subgroup contains the smallest monotone subgroup B consisting of all<br />

n2N k ne n with jk n j bounded.<br />

(b) The infinite sums P n2N k ne n , for which there is a c 2 N such that the coefficients satisfy<br />

jk n jcnŠ, form a monotone subgroup.<br />

Interestingly, monotone subgroups are slender with a single exception: the<br />

obvious one.<br />

Theorem 2.7 (G. Reid [1]). All monotone subgroups of the product P D Q n2N he ni<br />

are slender, with the exception of P itself.<br />

Proof. Let M be a monotone subgroup of P, properly contained in P, and assume<br />

there is a homomorphism W P ! M such that e n ¤ 0 for infinitely many indices<br />

n. We may assume that e n ¤ 0 for all n. Infinite cyclic groups are slender, so<br />

for each k, there are but finitely many indices n for which the coordinate of e n<br />

in he k i is non-zero. In view of this, we may even drop more summands from P<br />

to have e n D .0;:::;0;t n;in ; t n;in C1;:::/ with t n;in ¤ 0 for a strictly increasing<br />

sequence i 1 < < i n < of indices. Pick an x D .`1;:::;`n;:::/ 2 P n M with<br />

1

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