# Rank of a Subgroup of a Free Abelian Group.

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

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This has come up before, at least in greater generality (infinite rank as opposed to specifically countable rank). First however a comment on terminology. Countable means countably infinite. If you want a word that means either finite or countably infinite, the best choice is denumerable. – hardmath Jan 22 '13 at 12:21
@hardmath: you talk as though your use of countable is completely standard. It is not. Many mathematicians use it to mean finite or countably infinite. To me it seems very strange to say that a set with three elements is not countable. – Derek Holt Jan 22 '13 at 21:24
@DerekHolt: I posted to solicit clarification because the use seemed ambiguous. – hardmath Jan 22 '13 at 23:38
I tell my students that a set is countable if you can count the elements in it... Formally, $S$ is countable there exists a bijection from some subset of $\mathbb{N}$ to $S$. – user1729 Jan 23 '13 at 10:37

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 $A$.

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I am a bit confused. I've always thought it's $\mathbb Z^{\oplus X} \cong A$, where $\mathbb Z^{\oplus X}$ is a set of functions from $X$ to $\mathbb Z$ with finite support. (How did you define "support"?) – Tunococ Jan 22 '13 at 12:36
@Tunococ: You are completely right, I permuted $X$ and $\mathbb{Z}$. I corrected it, thank you. – Seirios Jan 22 '13 at 14:29

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.

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Is there an easier explanation. As I am still studying the group theory chapter. – Amr Jan 22 '13 at 11:23
Well, I'm a little low, but I think so. Maybe others can help you better than me, but let me think a second. – Diego Silvera Jan 22 '13 at 11:28

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 $|F| = |\mathbb Z|$.

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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 $F$.

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I think the OP means finite or countably infinite by countable.(i.e. There is an injection from the set to N) – Amr Jan 22 '13 at 14:28
@Amr: Perhaps. I put a comment on the question itself for clarification. In any case a couple of the answers point out that $F$ itself is a countable set, so any basis for $G$ is at most countable. – hardmath Jan 22 '13 at 14:44