I need help trying to solve this problem. Let $A, B$ be sets, and $F(A), F(B)$ the corresponding free groups. Assume $F(A) \cong F(B)$. If $A$ is finite, prove that so is $B$, and $A \cong B$ as sets.
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Please explain the notation ~=. – vadim123 Apr 24 '13 at 02:39
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it means it is an isomorphism. – Apr 24 '13 at 02:42
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4The same question for arbitrary set (not necessarily finite) has been asked here: free groups: $F_X\cong F_Y\Rightarrow|X|=|Y|$ – Martin Sleziak Apr 24 '13 at 05:07
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Let $A$ and $B$ be any sets. If the free groups $F(A)$ and $F(B)$ are isomorphic, so are their abelianizations. The abelianizations are the free abelian groups on the sets $A$ and $B$. It is much easier to see that if these are isomorphic, then $\# A \cong \# B$: for instance tensor with $\mathbb{Q}$ to reduce to the fact that bases of isomorphic vector spaces have the same cardinality.
Pete L. Clark
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Someone has to upvote this answer at some point, +1. Interestingly, point 3 here claims that it follows readily from the definition. – Julien Apr 24 '13 at 12:04
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I'm having trouble understanding the downvote here: what seems to be amiss? – Pete L. Clark Apr 24 '13 at 17:12