if they have the same rank, then using the definition of a free group it's obvious that they must be isomorphic. the converse is not trivial. to prove it we'll first need a nice

lemma that extends a well-known fact about vector spaces to free modules over commutative rings with identity:

Lemma:(IBN for commutative rings) if is a commutative ring with identity and a free -module, then every two basis of have the same cardinality.

Proof:let be a fixed maximal ideal of and any basis for clearly is a vector space over let if

for some in then for some and which is impossible since is basis for and thus a similar argument

shows that is a basis for -vector space thus which completes the proof because does not depend on

Corollary:let be a commutative ring with identity and and be two free -modules with basis and respectively. if then

Proof:let be an isomorphism. then and both are basis for thus by the Lemma

now suppose and are two free groups on and respectively. suppose let and be their commutator subgroups. then clearly

it's easy to see, from the definition of free groups, that and are free (obviously abelian) on and respectively. so and can be considered

as two isomorphic -modules. the above corollary now shows that