# Prove that a free abelian group $A_{n}$ is finitely generated if and only if its rank $n$ is finite

I need to prove that a free abelian group $A_{n}$ is finitely generated if and only if its rank $n$ is finite.

The $(\Longrightarrow)$ direction is easy: Suppose that $A_{n}$ has finite rank. Then, since $A_{n}$ is free, it must have a basis $B$, and since it has finite rank $n$, $|B|=n$. Now, $B$ is a basis for $A_{n}$ if and only if $B$ is linearly independent and generates $A_{n}$, so $A_{n}$ is finitely generated.

The $(\Longleftarrow)$ direction is what I'm having trouble with: I was given the following hint:

Assume that $X$ is a finite generating set for $A_{\infty}$, a free abelian group of (countably) infinite rank. Show that not all elements of $A_{\infty}$ can be expressed in terms of $X$ by writing expressions of elements of $X$ in terms of a basis $B$ of $A_{\infty}$.

I'm at a loss as to how to do this, however. Something's telling me that the following theorem could be useful:

Theorem: Let $H$ be a subgroup of a free abelian group $A$. Then, there exists a basis $B$ for $A$ such that for some (finite or infinite) subset $\{b_{1},b_{2},\cdots, b_{i}, \cdots\}$ of $B$, a generating set $X$ for $H$ is of the form $X=\{d_{1}b_{1}, d_{2}b_{2}, \cdots , d_{i}b_{i}, \cdots \}$ where, for all $i$, $d_{i}>0$ and $d_{i}$ is a divisor of $d_{i+1}$.

Although at second glance, I'm not sure how, since it doesn't say anything about finite generating sets...

Now, if $X$ is a finite generating set for $A_{\infty}$, then $X \subset A_{\infty}$, so if $B = \{b_{1}, b_{2}, \cdots, b_{i}, \cdots \}$ is a basis for $A_{\infty}$, then $\forall x_{j} \in X$, $x_{j} = c_{1}b_{1} + c_{2}b_{2} + \cdots + c_{i}b_{i} + \cdots$. But then, how does doing this help show that not all elements of $A_{\infty}$ can be expressed in terms of $X$?

1. Tell me if my $(\Longrightarrow)$ direction is correct.
2. Help me prove the $(\Longleftarrow)$ direction.

Thank you.

• How are you defining the rank of an abelian group? – anomaly Feb 6 '17 at 4:04
• @anomaly from the typed up lecture notes my prof hands out, it would appear that we are defining the rank of a free abelian group $A$ as the number of copies of $\mathbb{Z}$ that $A$ is isomorphic to. I.e., $A_{n} \simeq \mathbb{Z}^{n}$ and $A_{\infty}$ is the direct sum of an infinite countable number of copies of $\mathbb{Z}$. I know it seems kind of weird, and most other places I've looked at have used the number of elements in the basis (i.e., the number of elements in a generating set) to define the rank, but that's not the way we're doing it in my class, so I have to prove it as written. – ALannister Feb 6 '17 at 4:09

Notice that when you find expressions for all the finitely many elements in terms of the basis, you've only used finitely many basis elements. Pick a basis element $b_i$ that you didn't use. It's a direct application of the definition of a basis that no integer linear combination of the elements in the finite set will give you $b_i$.
• how is it that you only use finitely many of the basis elements? When I wrote out an expression for $x_{j}$, I used all of them. I must have not written it correctly. Could you elaborate on your answer a bit and show me the correct way to express $x_{j}$ in terms of the basis? – ALannister Feb 6 '17 at 4:11
• So, let's say for some $x_{j}$ in the finite generating set $X$, I express it as $x_{j} = c_{j1}b_{1} + c_{j2}b_{2}+\cdots + c_{ji}b_{i} + \cdots c_{jn}b_{n} + 0 + 0 + \cdots + 0 + \cdots$. – ALannister Feb 6 '17 at 4:15
• then can I have all the $x_{1}$ to $x_{n}$ be WLOG a linear combination of $b_{1},b_{2}, \cdots, b_{n}$? Then, I could choose, say $b_{n+1}$, and how is it a direct application of the basis that no integer linear combination of these will give me $b_{n+1}$? – ALannister Feb 6 '17 at 4:21