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I have to prove this statement:

Let free abelian group $F_i$ with free basis $B_i$ for $i=1,2$ then $F_1\cong F_2$ iff $\lvert B_1 \rvert = \lvert B_2 \rvert$.

I prove it by the idea that element of basis is linear independent and they generate whole group. So we have injection $F_1= \langle B_1 \rangle \rightarrow \langle B_2\rangle=F_2$, WLOG $\lvert B_1 \rvert \leq \lvert B_2 \rvert$.

So now is trivial both implication.

My question is can I do use this idea for proving?

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If you accept that vector spaces over a field have a well-defined dimension, we can prove this using that result as follows:

Theorem. Let $F$ be a free abelian group, and let $B$ be a basis for $F$. Let $G=F/2F$; then $G$ is a vector space over $\mathbb{F}_2$, the image of $B$ is a basis for $G$ as a vector space, and the restriction of the canonical map $\pi\colon F\to G$ to $B$ is one-to-one. In particular, $|B|=\dim_{\mathbb{F}_2}(G)$.

Proof. $G$ is certainly an abelian group of exponent $2$; that means it is a vector space over $\mathbb{F}_2$. If $b_1,b_2\in B$ and $\pi(b_1)=\pi(b_2)$, then there exists $x\in G$ such that $b_1=b_2+2x$. Expressing $x$ in terms of the basis $B$ and using uniqueness of the representation we conclude that $x=0$ and $b_1=b_2$, so that $\pi|_{B}$ is injective.

Now, $B$ spans $F$, hence $\pi(B)$ spans $G$. To show that $\pi(B)$ is linearly independent over $\mathbb{F}_2$, let $b_1,\ldots,b_n\in B$ be pairwise distinct, and let $a_1,\ldots,a_n\in\mathbb{F}_2$ be such that $$a_1\beta_1+\cdots+a_n\beta_n = 0,$$ where $\beta_j=\pi(b_j)$. We want to show that $a_1=\cdots=a_n=0$. Let $r_i=0$ if $a_i=0$, and let $r_i=1$ if $a_i\neq 0$. Then $r_1b_1+\cdots+r_nb_n\in F$ is such that $\pi(r_1b_1+\cdots+r_nb_n)=a_1\beta_1+\cdots+a_n\beta_n$. Since the image is trivial, there exists $x\in F$ such that $r_1b_1+\cdots+r_nb_n=2x$. Again, expressing $x$ in terms of $B$ and using the uniqueness of the representation we conclude that all $r_i$ must be even, and hence equal to $0$. Thus, $a_i=0$ for all $i$, so $\beta_1,\ldots,\beta_n$ are linearly independent. This proves that $\pi(B)$ is linearly independent, and hence is a basis. $\Box$

Now the result follows easily: $F_1/2F_1\cong F_2/2F_2$ as abelian groups, hence as vector spaces. Thus, $|B_1|=\dim_{\mathbb{F}_2}(F_1/2F_1) = \dim_{\mathbb{F}_2}(F_2/2F_2) = |B_2|$. The converse is trivial.

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I believe this can be shown using the universal property of free objects. I am assuming that all bases of a free abelian group have the same cardinality.

Let $\phi: F_1 \to F_2$ be an isomorphism. Then, I claim that $\phi(B_1)$ is a basis for $F_2$. To show this, let $G$ be an abelian group and $f: \phi(B_1) \to G$ be a function. Then, $f \circ \phi: B_1 \to G$ is a function. By the universal property of free objects, there is some unique $g: F_1 \to G$ so that $g$ restricts to $f \circ \phi$ on $B_1$. Then, $\phi \circ g: F_2 \to G$ is the unique homomomorphism which restricts to $f$ on $\phi(B_1)$, proving that $\phi(B_1)$ is a basis of $F_1$. Since $\phi$ is an isomorphism, it is a bijection. Then, since all bases of free abelian groups have the same cardinality, $$|B_1| = |\phi(B_1)| = |B_2|.$$

EDIT: Here is a proof that bases of free modules over sufficiently nice rings have equal cardinality. This applies here since $\mathbb{Z}$ is a commutative ring with 1.

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  • $\begingroup$ I would say that invoking that two bases of the same free abelian group have the same cardinality makes your argument circular. If you know that, then this problem is trivial. Conversely, if you take this problem as a given, then it immediately follows that any two bases of the same free abelian group have the same cardinality. In other words, you are invoking something which is trivially equivalent to what you are trying to prove in order to prove it. $\endgroup$ – Arturo Magidin Dec 24 '20 at 22:39

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