What is the Sufficient and necessary condition of isomorphism of free groups

Question

Let $F(X_1)$ and $F(X_2)$ be free groups of free basis are $X_1$ and $X_2$. The rank of them are finite (i.e.The cardinality of $X_1$ and $X_2$ are finite). Then $F(X_1)\cong F(X_2)$ if and only if they have the same rank.

I have proved the Sufficient condition, but struggling with the Necessary condition (I only know the image of Generate set is also Generate set when i prove the necessary condition).

If the free basis is Infinite then Is this proposition still correct？

Answer 1

Lemma. $\mathbb{F}(X)\simeq\mathbb{F}(Y)$ if and only if $|X|=|Y|$ in the sense of cardinal numbers.

Proof.

"$\Leftarrow$" Consider bijection $f:X\to Y$ and its inverse $g:Y\to X$. Compose them with inclusions to obtain $f':X\to\mathbb{F}(Y)$ and $g':Y\to\mathbb{F}(X)$. By the universal property of free groups these induce group homomorphisms $F:\mathbb{F}(X)\to\mathbb{F}(Y)$ and $G:\mathbb{F}(Y)\to\mathbb{F}(X)$. It is not hard to see that these are inverses of each other. Hence $\mathbb{F}(X)\simeq\mathbb{F}(Y)$.

"$\Rightarrow$" Since $\mathbb{F}(X)\simeq\mathbb{F}(Y)$ then $\mathbb{F}(X)_{ab}\simeq\mathbb{F}(Y)_{ab}$ where $G_{ab}:=G/[G,G]$ is group abelianization. It is well known that $\mathbb{F}(X)_{ab}\simeq\bigoplus_{x\in X}\mathbb{Z}$, see here. And so we reduced the claim to: if $\bigoplus_{x\in X}\mathbb{Z}\simeq \bigoplus_{y\in Y}\mathbb{Z}$ then $|X|=|Y|$.

Note that we now deal with $\mathbb{Z}$-modules, and group homomorphism $\bigoplus_{x\in X}\mathbb{Z}\to\bigoplus_{x\in X}\mathbb{Z}$ are actually $\mathbb{Z}$-homomorphisms. We further reduce the problem by tensoring both sides with $\mathbb{Q}$ to obtain $\bigoplus_{x\in X}\mathbb{Q}\simeq \bigoplus_{y\in Y}\mathbb{Q}$.

Again $\mathbb{Z}$-homomorphisms $\bigoplus_{x\in X}\mathbb{Q}\to\bigoplus_{y\in Y}\mathbb{Q}$ are actually $\mathbb{Q}$-linear maps. Now the standard linear algebra applies: vector spaces are uniquely determined (up to isomorphism) by the dimension. $\Box$

Answer 2

This can be solved by considering $\mathbb{Q}$ free vector spaces. Consider a set $X$ and the free vector space $G(X)$. There is a natural group homomorphism $F(X) \to G(X)$ wuch that the image of the morphism spans $G(X)$.

Now suppose we have $F(X^1) \simeq F(X^2)$. Then there is a group map $f : F(X^2) \to G(X^1)$ such that the span of the image of this map is $G(X^1)$. Thus, the span of $\{f(x) : x \in X^2\}$ is $G(X^1)$. This implies that $|X^2| \geq |X_1|$. That is, $F(X^1) \simeq F(X^2)$ implies $|X^1| \leq |X^2|$. Therefore, $F(X^1) \simeq F(X^2)$ implies that $|X^1| = |X^2|$.