# 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? Feb 6, 2017 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.
– user100463
Feb 6, 2017 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?
– user100463
Feb 6, 2017 at 4:11
• @Jessy You don't know what the finitely many elements are, so you don't know specifically which basis elements to use, so in a literal sense your abstract expression will use all of them. Feb 6, 2017 at 4:13
• @Jessy But all but finitely many of the coefficients must be zero because summing infinitely many elements is undefined. So there are only finitely many different basis elements with nonzero coefficients collectively in the expressions. Feb 6, 2017 at 4:15
• So, let's say for some $x_{j}$ in the finite generating set $X$, I express it as $x_{j} = c_{j1}b_{1} + c_{j2}b_{2}+\cdots + c_{ji}b_{i} + \cdots c_{jn}b_{n} + 0 + 0 + \cdots + 0 + \cdots$.
– user100463
Feb 6, 2017 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 $b_{n+1}$?
– user100463
Feb 6, 2017 at 4:21