Here is a simpler argument, combining 1--6 into one step.
Let $G$ be a countable abelian group generated by $x_1,x_2,\ldots$. Then a Følner sequence is given by taking $S_n$ to be the pyramid consisting of elements which can be written as
$a_1x_2+a_2x_2+\cdots+a_nx_n$ with $\lvert a_1\rvert\leq n,\lvert a_2\rvert\leq n-1,\ldots,\lvert a_n\rvert\leq 1$.
The invariant probability measure is then defined by $\mu(A)=\underset{\omega}{\lim}\lvert A\cap S_n\rvert / \lvert S_n\rvert$ as usual.
A more natural way to phrase this argument is:
- The countable group $\mathbb{Z}^\infty$ is amenable.
- All countable abelian groups are amenable, because amenability descends to quotients.
But I would like to emphasize that there is really only one step here, because the proof for $\mathbb{Z}^\infty$ automatically applies to any countable abelian group. This two-step approach is easier to remember, though. (The ideas here are the same as in my other answer, but I think this formulation is much cleaner.)
2016 Edit: Here is an argument to see that $S_n$ is a Følner sequence. It is quite pleasant to think about precisely where commutativity comes into play.
Fix $g\in G$ and any finite subset $S\subset G$. We first analyze the size of the symmetric difference $gS\bigtriangleup S$. Consider the equivalence relation on $S$ generated by the relation $x\sim y$ if $y=x+g$ (which is itself neither symmetric, reflexive, or transitive). We will call an equivalence class under this relation a "$g$-string". Every $g$-string consists of elements $x_1,\ldots,x_k\in S$ with $x_{j+1}=x_j+g$.
The first key observation is that $\lvert gS\bigtriangleup S\rvert$ is at most twice the number of $g$-strings. Indeed, if $z\in S$ belongs to $gS\bigtriangleup S$, then $z$ must be the "leftmost endpoint" of a $g$-string; if $z\notin S$ belongs to $gS\bigtriangleup S$, then $z-g$ must be the "rightmost endpoint" of a $g$-string; and each $g$-string has at most 2 such endpoints (it could have 1 if the endpoints coincide, or 0 if $g$ has finite order).
Our goal is to prove for all $g\in G$ that $\frac{\lvert gS_n \bigtriangleup S_n\rvert}{\lvert S_n\rvert}\to 0$ as $n\to \infty$. Since $\lvert abS\bigtriangleup S\rvert\leq\lvert abS\bigtriangleup bS\rvert+\lvert bS\bigtriangleup S\rvert= \lvert aS\bigtriangleup S\rvert+\lvert bS\bigtriangleup S\rvert$, it suffices to prove this for all $g_i$ in a generating set.
By the observation above, to prove that $\frac{\lvert g_iS_n \bigtriangleup S_n\rvert}{\lvert S_n\rvert}\to 0$, it suffices to prove that $\frac{\text{# of $g_i$-strings in $S_n$}}{\lvert S_n\rvert}\to 0$. Equivalently, we must prove that the reciprocal $\frac{\lvert S_n\rvert}{\#\text{ of $g_i$-strings in $S_n$}}$ diverges, or in other words that the average size of a $g_i$-string in $S_n$ diverges.
We now use the specific form of our sets $S_n=\{a_1g_1+\cdots+a_ng_n\,|\, \lvert a_i\rvert\leq n-i\}$. For any $i$ and any $n$, set $k=n-i$ (so that $\lvert a_i\rvert\leq k$ in $S_n$). The second key observation is that every $g_i$-string in $S_n$ has cardinality at least $2k+1$ unless $g_i$ has finite order. Indeed given $x\in S_n$, write it as $x=a_1g_1+\cdots+a_ig_i+\cdots+a_ng_n$; then the elements $a_1g_1+\cdots+bg_i+\cdots+a_ng_n\in S_n$ for $b=-k,\ldots,-1,0,1,\ldots,k$ belong to a single $g_i$-string containing $x$. If $g_i$ does not have finite order, these $2k+1$ elements must be distinct. This shows that the minimum size of a $g_i$-string in $S_n$ is $2n-2i+1$, so for fixed $g_i$ the average size diverges as $n\to \infty$.
When $g_i$ has finite order $N$ this argument does not work (a $g_i$-string has maximum size $N$, so the average size cannot diverge). However once $N<2k+1$, the subset containing the $2k+1$ elements above is closed under multiplication by $g_i$. In other words, once $n\geq i+N/2$ the set $S_n$ is $g_i$-invariant, so $\lvert g_iS_n\bigtriangleup S_n\rvert=0$.
I'm grateful to David Ullrich for pointing out that this claim is not obvious, since the quotient of a Følner sequence need not be a Følner sequence (Yves Cornulier gives an example here).
Best Answer
I reject the premise of the question. :-)
It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups. If $A$ is finite, then for every prime $p$ has a stable kernel $A_p$ and a stable image $A_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$. You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A_p$. A similar argument appears in Matthew Emerton's proof. As Terry says, this proof is nice because it works for finitely generated torsion modules over any PID. In particular, it establishes Jordan canonical form for finite-dimensional modules over $k[x]$, where $k$ is an algebraically closed field. My objection is that finite abelian groups look easier than finitely generated abelian groups in this question.
The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form. Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relations. That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices. This is one of the tricks in the proof that I give next.
Theorem. If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups.
Proof. By induction on $n$. If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done. Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$ that appears in some relation. Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation. (Pun intended. :-) )
The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$. By the same argument, we can assume that $a$ does not appear in any other relation.
The coefficient $z$ of another generator $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$. By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations. Thus $A \cong \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □
Compare the complexity of this argument to the other arguments supplied so far.
Minimizing the norm $|x|$ is a powerful step. With just a little more work, you can show that $x$ divides every coefficient in the presentation, and not just every coefficient in the same row and column. Thus, each modulus $x_k$ that you produce divides the next modulus $x_{k+1}$.
Another way to describe the argument is that Smith normal form is a matrix version of the Euclidean algorithm. If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated.
The proof immediately works for any Euclidean domain; in particular, it also implies the Jordan canonical form theorem. And it only needs minor changes to apply to general PIDs.