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If $F_1$ and $F_2$ are free groups, they are isomorphic iff they have the same rank.

I have prove it constructing an injective homomorphism from $F_1$ to $F_2$ and viceversa. My question is about when the rank is not finite. In the proof, what I actually use is the equivalency of cardinality (not finiteness). Can I use Schröder-Bernstein theorem? If not, what's the answer?

Thanks!

Asaf Karagila
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Peter
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1 Answers1

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First of all, to show that two free groups are isomorphic, it's not enough to construct injective homomorphisms in both directions. You have to show that there is a homomorphism with an inverse, i.e. a bijective homomorphism.

For example, the free group on three generators $\langle a,b,c\rangle$ maps into the free group on two generators $\langle x,y\rangle$ by $a\mapsto x^2,b\mapsto y^2,c\mapsto xy$.

But the hard part here is showing that two isomorphic free groups must have independent generating sets of the same cardinality. I don't think this is trivial, even for finite generating sets, but here is a proof that works for sets of any cardinality:

Suppose that $F$ and $G$ are free groups on the sets $S$ and $T$ respectively.

If $F$ is isomorphic to $G$, then their abelianizations $F^{ab}$ and $G^{ab}$, are isomorphic. These are free abelian groups on the sets $S$ and $T$.

Taking tensor products, we have $F^{ab} \otimes_\mathbb{Z} \mathbb{Q} \cong G^{ab} \otimes_\mathbb{Z} \mathbb{Q}$. These are vector spaces over $\mathbb{Q}$ with bases indexed by $S$ and $T$. Since every basis of a vector space has the same cardinality, $|S| = |T|$.

Andrew Dudzik
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  • Thank you! So how do you define the isomorphism (bijective homomorphism) if they have the same rank? – Peter Nov 19 '15 at 23:20
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    @Peter That depends on exactly how you define free groups. If you think of $F$ and $G$ as "words in $S$" and "words in $T$" then just substitute the letters of $S$ for the letters of $T$. – Andrew Dudzik Nov 19 '15 at 23:29
  • Well if you again consider the underlying set of the free groups. And forget that the map is an homomorphism you still have a bijection dont you? – Abellan Nov 20 '15 at 00:11
  • @Abellan If you're talking about a way to go from $F\cong G$ to $|S|=|T|$, this doesn't work, because there is no reason that an isomorphism between $F$ and $G$ needs to restrict to a function $S\to T$. – Andrew Dudzik Nov 20 '15 at 02:37