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Abelian Groups - László Fuchs [Springer]

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7 Chains of Free <strong>Groups</strong> 111<br />

which is a countable reduced p-group. Therefore, we can apply the countable case<br />

to derive that A =A is a free group. Hence the chain of the A . < / has free<br />

factor groups, thus their union A is a free group.<br />

ut<br />

Example 7.11. Reduced totally projective p-groups (Sect. 6 in Chapter 11)admitanH.@ 0 /-family<br />

as stated in the theorem. However, no uncountable p-group with countable basic subgroups has<br />

such a family.<br />

The Summand Intersection Property We say that a group A has the summand<br />

intersection property if the intersection of two summands in A is likewise a<br />

summand of A. If the same holds for infinite intersections as well, then we refer<br />

to it as the strong summand intersection property. Needless to say, this property<br />

is shared by very special groups only. We are looking for free groups with this<br />

property.<br />

Proposition 7.12 (Kaplansky [K], Wilson [1]). All free groups have the summand<br />

intersection property. A free group has the strong summand intersection property if<br />

and only if it is countable.<br />

Proof. Let F be a free group, and F D B i ˚ C i .i D 1; 2/ direct decompositions.<br />

Then F=.B 1 \ B 2 / is isomorphic to a subgroup of the free group F=B 1 ˚ F=B 2 ,so<br />

is itself free. Hence B 1 \ B 2 is a summand of F.<br />

If F is a countable free group, then the same argument with countable summands<br />

leads to a countable factor group contained in a product of countable free groups<br />

F=B i . Theorem 8.2 below implies that this factor group is free, so the intersection<br />

of countably many summands is a summand.<br />

Finally, suppose jFj @ 1 .LetA be a torsion-free, non-free group of cardinality<br />

@ 1 ,and W F ! A a homomorphism that is a bijection between a basis fb j

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