Free abelian group

Expressive definition with explicit generating set
This is not the rigorous definition, but is the one that is most useful for explicit expressions and computations.

Suppose $$A$$ is a set (with no additional structure necessary). The free abelian group on $$A$$ (or the free abelian group with generating set $$A$$) is a group whose elements are defined as formal finite $$\mathbb{Z}$$-linear combinations of elements of $$A$$, i.e., finite sums of the form:

$$m = \sum_{a \in A} m_aa$$

where $$m_a \in \mathbb{Z}$$ for all $$a$$ and $$m_a \ne 0$$ for only finitely many $$a \in A$$. The addition rule is as follows: we group together coefficients of the same element of $$A$$. Thus, if $$m = \sum_{a \in A} m_aa$$ and $$n = \sum_{a \in A} n_aa$$, then $$m + n = \sum_{a \in A} (m_a + n_a)a$$.

When writing a formal sum, we ignore all the terms with zero coefficient, and use subtraction to denote the addition of something with a negative coefficient (so $$a + (-1)b$$ is written as $$a - b$$).

Note that if $$A$$ already had an additive structure, that structure is ignored for our purposes. A more precise formulation of this would be to not use the elements of $$A$$ themselves but a generating set equipped with a bijection to $$A$$; however, this extra layer of formality is often unnecessary.

Definition in terms of rank
Let $$\alpha$$ be a cardinal (a nonnegative integer if finite, otherwise an infinite cardinal). The free abelian group of rank $$\alpha$$ is defined as the free abelian group on any set of size $$\alpha$$. This is unique up to isomorphism: any bijection between two sets induces an isomorphism of the corresponding free abelian groups. In fact, something stronger is true: free abelian groups arising from sets of different cardinalities are not isomorphic, so the rank of a free abelian group is a unique cardinal.