Then

where

is the subgroup of all non-torsion elements (along with the identity).

is a subgroup because

is abelian. We get that

by inducting on the number of generators of the group,

. Clearly the result holds for a 1-generated group.

Assume the result holds for all groups generated by less than or equal to

elements. Let

be a subgroup generated by

elements. Then,

is of finite index as

contains only torsion elements. Thus,

is generated by less than or equal to

elements (this can be seen by viewing

as an

-dimensional vector space). Is it of the form we wish? Let

. Then as we are in an Abelian group these subgroup as normal and so we can apply the correspondence theorem to get that they are normal subgroups of

:

. Then,

, and so they are the same subgroups. Thus,

is of the correct form. As it consists only torsion elements it is cyclic.

Viewing our group as a vector space we can glue these two bits together - their intersection is trivial and

and so we have a direct sum, and so a direct product. Thus,

.

I think we cannot do anything with

. Clearly it is just

, but

conforms to the conditions stipulated.