TITLE: Let A,B be sets and F(A), F(B) their respective free groups. Assume F(A) ≅ F(B). If A is finite, prove that so is B, and A ≅ B.
QUESTION [2 upvotes]: I thought I had this but apparently my idea wasn't even close. I'm not sure what I'm missing or need to consider. Thanks for any help!

REPLY [2 votes]: Let $C_2$ be the cyclic group of order 2. By definition of a free group, each map $A \to C$ uniquely determines a homomorphism $F(A) \to C_2$, so there are excatly  $2^{|A|}$ such homomorphisms. Hence if $F(A) \cong F(B)$ then $2^{|A|} = 2^{|B|}$.
This is enough to solve your problem. But if $A$ and $B$ are infinite, then I believe that the statement $2^{|A|} = 2^{|B|}$ implies $|A|=|B|$ is independent of ZFC. (But there are other ways of proving the result when $A,B$ are both infinite.)
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