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Abelian Groups - László Fuchs [Springer]

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156 5 Purity and Basic Subgroups<br />

Proof. If the group contains a quasi-cyclic group Z.p 1 / for some prime p, then this<br />

is a direct summand. If it does not contain any quasi-cyclic subgroup, but it contains<br />

elements of order p, then it must also contain one of finite height; cf. (E). The claim<br />

follows from Corollary 2.2.<br />

ut<br />

Another corollary worthwhile recording is as follows.<br />

Corollary 2.4 (Kulikov [1]). A directly indecomposable group is either cocyclic<br />

or torsion-free.<br />

Proof. By Corollary 2.3 there are no indecomposable mixed groups, and the only<br />

indecomposable torsion groups are the cocyclic groups.<br />

ut<br />

Bounded Pure Subgroups We now proceed to show that certain pure subgroups<br />

are necessarily summands.<br />

Theorem 2.5 (Prüfer [2], Kulikov [1]). A bounded pure subgroup is a direct<br />

summand.<br />

Proof. If B is a bounded group, then Lemma 2.1 allows us to decompose B D B 1˚C<br />

where B 1 is a direct sum of cyclic groups of the same order p k , and the bound for C<br />

is less than the bound for B. IfB is pure in A, thensoisB 1 , and Lemma 2.1 implies<br />

that A D B 1 ˚ A 1 for some A 1 A. ThenB D B 1 ˚ C 1 with C 1 D B \ A 1 Š C.<br />

Here C 1 is pure in A 1 , and by induction, C 1 is a summand of A 1 , and hence B is<br />

one of A.<br />

ut<br />

Corollary 2.6 (Khabbaz [2]). p n A-high subgroups .n 2 N/ of a group A are<br />

summands.<br />

Proof. We show that a p n A-high subgroup B is a bounded pure subgroup. Since<br />

p n B B\p n A D 0, B is a bounded p-group. By induction, we prove that B\p k A <br />

p k B for integers k with 0 k n. This being trivially true for k D 0, assume it is<br />

true for some k where 0 k < n. Letb D p kC1 a ¤ 0.b 2 B; a 2 A/. Ifp k a 2 B,<br />

then by induction hypothesis p k a D p k b 0 for some b 0 2 B, andthenb D p kC1 b 0 .If<br />

p k a … B, thenhB; p k ai contains a non-zero element b 0 C p k a 2 p n A,whereb 0 2 B<br />

is also in p k A,sob 0 2 p k B as well. Now b 0 D p n a 0 p k a for an a 0 2 A, and<br />

hence b D p kC1 a D p nC1 a 0 pb 0 ; where the term p nC1 a 0 has to vanish because of<br />

B \ p n A D 0. Thus b D pb 0 2 p kC1 B: ut<br />

Corollary 2.7 (Erdélyi [1]). A p-subgroup embeds in a bounded summand if and<br />

only if the heights of its elements .computed in the containing group/ are bounded.<br />

Proof. The necessity being obvious, suppose that p m is an upper bound for the<br />

heights in the subgroup B of A.Thereisap mC1 A-high subgroup C such that B C.<br />

Invoke the preceding corollary to conclude that C is a summand of A. ut<br />

Cosets of Pure Subgroups Here is another characterization of purity.<br />

Lemma 2.8 (Prüfer [1]). A subgroup B of A is pure if and only if every coset of A<br />

modulo B can be represented by an element that has the same order as the coset.

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