Is this a valid "easy" proof that two free groups are isomorphic if and only if their rank is the same?

Question

I have read on different sources that it is not possible to give a simple proof that "two free groups are isomorphic if and only if they have the same rank" using only what "a student who has just read the definition of free group as a set of words over an alphabet" would know. See for example the answers to this question Is there a simple proof of the fact that if free groups $F(S)$ and $F(S')$ are isomorphic, then $\operatorname{card}(S)=\operatorname{card}(S')?$.

I think I have come with such a proof, but I would like to know if it is valid. The proof goes as follows.

If A and B have the same cardinality, we can define a bijection between letters on A alphabet and letters on B alphabet. This establishes a bijection between (reduced) words on A and (reduced) words on B, and the isomorphism between the free groups F(A) and F(B). This proves the "if".

Now suppose that |A|<|B|. We can define a bijection between letters on A and a subset of the letters on B. Put differently, we can “relabel” the letters on A and say that B contains all the letters on A plus, at least, one letter that is not in A. Let x be this “extra” letter. Let w be any reduced word on A, that is, any element of F(A). Then wx is a reduced word on B, but not a valid word on A. That is, there is at least one element of F(B) that is not on F(A). Therefore, F(A) and F(B) cannot be isomorphic. This proves the "only if".

Answer 1

Let me write out, in full, what the "only if" direction says:

If there exists an isomorphism between the free group $F(A)$ and the free group $F(B)$ then $A$ and $B$ have the same cardinality.

Let's write the contrapositive:

If $A$ and $B$ have different cardinality then no isomorphism exists between the free group $F(A)$ and the free group $F(B)$.

Your proof amounts to the statement that one particular homomorphism from $F(A)$ to $F(B)$, namely the homomorphism induced by a certain choice of injection $A \hookrightarrow B$, is not an isomorphism. Your proof that this one particular homomorphism is not an isomorphism is correct. But that does not amount to a proof that no isomorphism exists. And continuing in that vein is really not going to work, because you cannot possibly go through the list of all possible homomorphisms, testing them one at a time to be sure that none of them is an isomorphism.

The standard proof of the "only if" direction is to show that the tensor product $\mathbb R \otimes_{\mathbb Z} F(A)_{\text{ab}}$ --- where $G_{\text{ab}}$ denotes "abelianization" --- is a vector space whose dimension over $\mathbb R$ is equal to $|A|$. So assuming that $F(A)$ and $F(B)$ are isomorphic, one can use the theorem on well-definedness of vector space dimension to conclude that $|A|=|B|$ (and one must throw in some "functor" language as well to make the argument work).

Answer 2

Your proof is not valid. A valid proof must show that no map between them is an isomorphism; you've just shown that this particular map you've described is not an isomorphism.

Note that there are maps $\Bbb{N} \rightarrow \Bbb{N}$ which are injective but not surjective. Your proof would then purport to show that $F(\Bbb{N}) \not\cong F(\Bbb{N})$, which is of course untrue.

Answer 3

I think you have proven that there can't be any isomorphism between $F(A)$ and $F(B)$ that maps elements of $A$ onto elements of $B$. It's natural to think that any isomorphism must be like this, but it isn't true.

You may have the intuition from just looking at elements of the free group that you can tell from the group structure alone which elements are singleton letters, and you might try to say something like "the singleton letters are those elements which aren't the product of any other elements". If true, then isomorphisms, since they preserve the group structure, would have to map letters to letters, and your proof would work. (As an aside, this is true for e.g. free monoids, and I think your argument works there.)

But of course it isn't true, e.g. in $F(\{a,b\})$, $a$ is the product of $ab$ and $b^{-1}$, so you can't identify letters this way. Indeed, $ab$ and $b^{-1}$ also generate the free group on two elements, and the map from $F(\{a,b\})$ to itself that sends $a$ to $ab$ and $b$ to $b^{-1}$ is an isomorphism, as is the map that sends $a$ to $bab$ and $b$ to $bab^3ab$, so the property of being a letter can't be in the group structure, since these isomorphic groups have different letters.

Perhaps more strikingly, there are subgroups of $F(\{a,b\})$ isomorphic to $F(\{a,b,c\})$, or even to $F(\mathbb N)$, so it's not true that "smaller alphabet" means "smaller group", in the sense of can-be-embedded (really, it means that can-be-embedded isn't really a "smallness" relation for groups).