Rank of a Subgroup of a Free Abelian Group.

Question

Question:

Let $F$ be a free abelian group with countable rank. Let $G \leq F$. Is the rank of $G$ countable?

I know that this is true when $F$ has a finite rank.

Thanks

Answer 1

Let $A$ be an abelian free group and $X$ be a free basis of $A$. Then there is a bijection between $A$ and $\mathbb{Z}^{(X)}$ (that is, the set of functions from $X$ to $\mathbb{Z}$ with finite support), so $|A|=|X|$ if $X$ is infinite.

Consequently, if the rank of $B \leq A$ is $\kappa$, $\kappa \leq |B| \leq |A|=|X|$ the rank of $A$.

Answer 2

Every abelian group can be seen like a $\mathbb{Z}$-module and there is a theorem that claims you that if you have a free $R$-module $A$ and a $R$-submodule $B$ of $A$ then $B$ is free and the rank $B$ is minor that the rank of $A$ if $R$ is a PID. Therefore, since $\mathbb{Z}$ is a PID, your question is true.

Answer 3

Suppose $X = \{x_1, x_2, \ldots\} \subset F$ is a countable set that generates $F$. Let $M_n = \langle x_1, \ldots, x_n \rangle \cong \mathbb Z^n$. We see that $|M_n| = |\mathbb Z|$ for all $n$. Since $F = \bigcup_{n=1}^\infty M_n$, it follows that $|F| = |\mathbb Z|$.

Answer 4

The answer to the question as asked is, $G$ is not necessarily of countable rank.

The subgroup $G$ is necessarily free abelian (cf. F is a free abelian group on a set X , H⊆F is a free abelian group on Y, then |Y|≤|X| ), but the question seems to assume this fact and ask rather about the rank of $G$ being countable.

What can be said (see link) is that rank of $G$ is at most the rank of $F$. The latter is assumed countable, but the rank of $G$ could either be countable or finite (i.e. denumerable as explained in my comment). The simplest case is $G$ of rank $1$, that is the infinite cyclic subgroup generated by any nonidentity element of $F$.