In this post, we provide an example of an abelian group which is not free, i.e. does not have a basis. Namely, we show that the additive group formed by the countably infinite direct product of the integer group , which we denote by is not free. Here, is the set of natural numbers, including zero.
Definition 1. An abelian group is said to be free abelian if is isomorphic to the direct sum (note, not direct product) of infinite cyclic groups. Explicitly, there exists a subset consisting of elements of infinite order, which we call a basis of , with , where denotes the subgroup generated by .
We allow the possibility that , in which case is the trivial group.
An equivalent characterization of a free abelian group is that has a basis such that every element can be uniquely written as an expression of the form
where almost all (i.e. all but finitely many) .
Let denote the countably infinite direct product (note, not direct sum) of .
Theorem 2. Let be a homomorphism. Then there exists such that
Proof. If , then we’re done; so assume otherwise. For , define a subset by
Let be an enumeration of . Set and . Suppose we have chosen and such that
If , then we’re done. Otherwise, we can choose
Choose such that and therefore . Hence, for any ,
which implies that . Either this process terminates after finitely many steps, or by induction, we obtain an element . Observe that by construction,
But since , we conclude that , which is a contradiction.
The reader can verify that the following result is a ready consequence of the preceding lemma.
Proposition 3. For , let be the inclusion
If we have a countable collection of homomorphisms, which contains an infinite subet of nonzero maps, then there does not exist a homomorphism such that
As an application of the preceeding proposition, taking to be the identity mapping for all , we see that there does not exist a homomorphism such that
Theorem 4. is not a free abelian group.
Proof. Suppose is free abelian and therefore there exists a -module isomorphism
where consists of elements of infinite order. Let be an at most countable subset of , and for each , let be the projection map
By composition, we obtain a homomorphism . By Lemma 2, there exists such that , where is defined as above.