MHB Direct sum of free abelian groups

[mathgirl1]

Show the direct sum of a family of free abelian groups is a free abelian group.

My first thought was to just say that since each group is free abelian we know it has a non empty basis. Then we can take the direct sum of the basis to be the basis of the direct sum of a family of free abelian groups. But not sure it makes sense to say direct sum of basis.

Is it just as simple that the direct sum of a family of free abelian groups is isomorphic to the direct sum of the additive group of integers?

I am probably making this way harder than it is but I am not sure how to state the obvious either. Any help is appreciated. Thanks!

 

[Fallen Angel]

Hi mathgirl,

Is it just as simple that the direct sum of a family of free abelian groups is isomorphic to the direct sum of the additive group of integers?

No, this is only true when your groups are finitely generated.

Try to prove that the disjoint union of the basis of your groups is a basis for the direct sum. (Probably you meant some like that when talking abaout "direct sum of basis")