Let be a set. The group presented by where is called the free abelian group on . Denote it by . Prove that has the following universal property: If is any abelian group and is a set function, then there is a unique group homomorphism such that . Deduce that if is a free abelian group on a set of (finite) cardinality , then .
Recall from the universal property of free groups that there is a unique group homomorphism such that . Moreover, clearly . Then there is a unique group homomorphism such that . Identifying in with in , we have . This gives existence.
To see uniqueness, note that any homomorphism extending is uniquely determined by , which is uniquely determined by .
Now let and let . If , then is a quotient of , so that in fact . Suppose , and for each , define to be the -tuple with 1 in the th coordinate and 0 in all other coordinates. By the universal property, there exists a unique group homomorphism extending ; is surjective since generates . Now if , note that since is abelian, we can collect terms so that . Now , and in fact . Thus is injective, and we have .