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 $X$?
Could someone please:
- Tell me if my $(\Longrightarrow)$ direction is correct.
- Help me prove the $(\Longleftarrow)$ direction.
Thank you.
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 $b_i$.