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

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？

• Note the abelianization of the free group $F_n$ is $\Bbb Z^n$.
– anon
Commented Aug 11, 2020 at 7:38

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. $$\Box$$

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 $$|X^1| = |X^2|$$.