## 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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