Let and be free groups of finite rank. Prove that if and only if they have the same rank. What facts do you need in order to extend this proof to infinite ranks (where the result is also true)?
First let and be sets and a bijection. Now by the universal property for free groups, there exists a unique group homomorphism such that for all . Similarly, there exists a unique group homomorphism such that for all .
We claim that and . To see this, note for instance that if , then . Since has a two-sided inverse, it is an isomorphism.
Thus if and are free groups of the same rank, then . (This part of the proof does not require that the rank be finite.)
For the next direction, we need a lemma.
Lemma: Suppose is a free group of finite rank and that , where is finite. Then there is a subset such that . Proof: We proceed by induction on . Suppose . Since is cyclic, we have . Suppose now that the result holds for all free groups of finite rank and all generating sets of cardinality at most . Let be a free group of finite rank with and . If , then we’re done. Suppose not; then there exists an element such that . In particular, . By the induction hypothesis, there is a subset such that .
Suppose now that and are free groups of finite rank and that is an isomorphism. Suppose further that . Then . Since is a bijection, . By the lemma, then, for some subset , and we have . Likewise using , . Thus and have the same rank.
This proof depends on the lemma, which only works due to induction on the natural numbers.