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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!

citedcorpse
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  • I believe the standard way to do this, is (1) consider the abelianisation (2) tensor with a field (3) use linear algebra. But depending on your prerequisites, this may not be a helpful insight. – Myself Jun 20 '13 at 10:08
  • What was you idea? Sometimes it is a good idea to salvage (if possible). It does not matter so much it is inefficient, one can still learn quite a bit by trying. – Baby Dragon Jun 20 '13 at 13:38

2 Answers2

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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.)

Derek Holt
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    I liked this stunningly simple proof which doesn't use abelianizations. +1 – DonAntonio Jun 20 '13 at 12:25
  • This feels like cheating! – user1729 Jun 20 '13 at 12:43
  • This is just using the universal property of free groups. So is abelianizing. And in any case, any map from a free group to $C_2$ factors through its abelianization. – Qiaochu Yuan Jun 20 '13 at 19:48
  • (And when $A, B$ are both infinite the set of homomorphisms to $C_2$ naturally acquires the structure of an $\mathbb{F}_2$-vector space of dimensions $|A|, |B|$ respectively.) – Qiaochu Yuan Jun 20 '13 at 19:49
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If we were talking about free abelian groups, can you see why it's true? Can you see how this helps?

citedcorpse
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  • @Myself Yes unfortunately I haven't had much experience with what you mentioned except for introductory linear algebra.

    excitingcorpse No I'm not really sure how that would make it true.

     – Search deeper Jun 20 '13 at 10:12
  • ah, so basically what happens is you go from $F(A) \cong F(B)$ to $fab(A) \cong fab(B)$, the free abelian group on $A$ and $B$. it's basically the same thing, except the universal property is stated for abelian groups instead of arbitrary ones. so you end up with $\mathbb{Z}^{\oplus A} \cong \mathbb{Z}^{\oplus B}$. then when you tensor with $\mathbb{R}$, you get $\mathbb{R}^A \cong \mathbb{R}^B$, and from linear algebra you can deduce the result you want. i'd just take it as given for now if you don't follow the details of this argument – citedcorpse Jun 20 '13 at 10:30
  • What does tensor mean? I'm sorry but I'm unfamiliar with this concept. – Search deeper Jun 20 '13 at 12:20
  • @Searchdeeper to take the tensor product with. basically $\mathbb{Z}^n \otimes \mathbb{R} = \mathbb{R}^n$. so if you do it with both of (the abelianisations of) your free groups, you get $\mathbb{R}^n = \mathbb{R}^m$, and from linear algebra we know this is true only when $n = m$. – citedcorpse Jun 20 '13 at 12:30