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.

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