We state and prove a sufficient condition for finitely generated Abelian Groups to be the direct product of its generators, and state a counterexample to the conclusion when the condition is not satisfied.

## Theorem

Let be an abelian group and .

Suppose the generators are linearly independent over , that is, whenever for some integers , we have .

(Here we are using additive notation for , where the identity of is written as 0, the inverse of is written as ).

Then

## Proof

Define the following map by

We can check that is a group homomorphism.

We have that is surjective since any element is by definition a combination of finitely many elements of the generating set and their inverses. Since is abelian, for some .

Also, is injective since if , then all the coefficients are zero (by the linear independence condition). Thus is trivial.

Hence is an isomorphism.

## Remark

Note that without the linear independence condition, the conclusion may not be true. Consider which is abelian with order 30. Consider , .

We can see that , by observing that , , . However has order 90. Thus .