Free groups on finite sets are essentially unique up to rank

Let F_1 and F_2 be free groups of finite rank. Prove that F_1 \cong F_2 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 A and B be sets and \theta : A \rightarrow B a bijection. Now by the universal property for free groups, there exists a unique group homomorphism \Phi : F(A) \rightarrow F(B) such that \Phi(a) = \theta(a) for all a \in A. Similarly, there exists a unique group homomorphism \Psi : F(B) \rightarrow F(A) such that \Psi(b) = \theta^{-1}(b) for all b \in B.

We claim that \Phi \circ \Psi = 1 and \Psi \circ \Phi = 1. To see this, note for instance that if w = a_1^{e_1} \cdots a_k^{e_k} \in F(A), then (\Psi \circ \Phi)(w) = (\Psi \circ \Phi)(a_1)^{e_1} \cdots (\Psi \circ \Phi)(a_k)^{e_k} = a_1^{e_1} \cdots a_k^{e_k} = w. Since \Phi has a two-sided inverse, it is an isomorphism.

Thus if F_1 and F_2 are free groups of the same rank, then F_1 \cong F_2. (This part of the proof does not require that the rank be finite.)

For the next direction, we need a lemma.

Lemma: Suppose F is a free group of finite rank and that F = \langle S \rangle, where S is finite. Then there is a subset T \subseteq S such that F \cong F(T). Proof: We proceed by induction on |S|. Suppose |S| = 1. Since F is cyclic, we have F \cong F(S). Suppose now that the result holds for all free groups of finite rank and all generating sets of cardinality at most k. Let F be a free group of finite rank with F = \langle S \rangle and |S| = k+1. If F \cong F(S), then we’re done. Suppose not; then there exists an element x \in S such that x \in \langle S \setminus \{x\} \rangle. In particular, F = \langle S \setminus \{x\} \rangle. By the induction hypothesis, there is a subset T \subseteq S such that F \cong F(T). \square

Suppose now that F_1 and F_2 are free groups of finite rank and that \Phi : F_1 \rightarrow F_2 is an isomorphism. Suppose further that F_1 = F(S). Then F_2 = \Phi[F_1] = \Phi[\langle S \rangle] = \langle \Phi[S] \rangle. Since \Phi is a bijection, |\Phi[S]| = |S|. By the lemma, then, F_2 = F(T) for some subset T \subseteq S, and we have \mathsf{rank}(F_2) \leq \mathsf{rank}(F_1). Likewise using \Phi^{-1}, \mathsf{rank}(F_1) \leq \mathsf{rank}(F_2). Thus F_1 and F_2 have the same rank.

This proof depends on the lemma, which only works due to induction on the natural numbers.

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