Free abelian group $F_i$ with free basis $B_i$

Question

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?

Answer 1

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.

Answer 2

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.