TITLE: Is there a slick proof of the classification of finitely generated abelian groups?
QUESTION [67 upvotes]: One the proofs that I've never felt very happy with is the classification of finitely generated abelian groups (which says an abelian group is basically uniquely the sum of cyclic groups of orders $a_i$ where $a_i|a_{i+1}$ and a free abelian group).
The proof that I know, and am not entirely happy with goes as follows: your group is finitely presented, so take a surjective map from a free abelian group.  The kernel is itself finitely generated (this takes a little argument in and of itself; note that adding a new generator to a subgroup of free abelian group either increases dimension after tensoring with $\mathbb{Q}$ or descreases the size of the torsion of the quotient), so our group is the cokernel of a map between finite rank free groups.  Now, (and here's the part I dislike) look at the matrix for this map, and remember that it has a Smith normal form.  Thus, our group is the quotient of a free group by a diagonal matrix where the non-zero entries are $a_i$ as above.
I really do not think I should have to algorithmically reduce to Smith normal form or anything like that, but know of no proof that doesn't do that.
By the way, if you're tempted to say "classification of finitely generated modules for PIDs!" make sure you know a proof of that that doesn't use Smith normal form first.

REPLY [3 votes]: Here is a proof of the finite case that I think is pretty different from the others. It is inspired by the proof of Lemma B.13 in Pete Clark's algebra notes.
Let $G$ be a finite group, written additively. Let $n$ be the highest order of any element of $G$ and let $g$ be an element of order $n$. We will show that $\langle g \rangle$ splits off as a direct summand of $G$; then we induct.
Let $H$ be a subgroup of $G$ which is maximal with respect to the condition that $\langle g \rangle \cap H = \{ 0 \}$. We can always take $H = \{ 0 \}$, so at least one such subgroup exists. We will show that $G = \langle g \rangle \oplus H$.
If $\langle g \rangle \oplus H$ is not all of $G$, then we can find some $u$ in $G$ not lying in $\langle g \rangle \oplus H$. Let $b$ be the lowest positive integer such that $bu \in \langle g \rangle \oplus H$; say $bu = ag+ h$. Write $a = qb+r$ with $0 \leq r < b$. Putting $v:=u-qg$, we have $bv = rg + h$, and $b$ is the lowest positive integer such that $bv \in \langle g \rangle \oplus H$.
Case 1: $r=0$. In this case, put $H' = H + \langle v \rangle$; we have $H' \supsetneq H$ and $\langle g \rangle \cap H' = \{ 0 \}$, contradicting the maximality of $H$.
Case 2: $r>0$. In this case, we claim that the order of $v$ is greater than $n$, contradicting the maximality of $n$. Indeed, the order of $v$ must be divisible by $b$, and the order of $bv$ is at least $\tfrac{n}{r} > \tfrac{n}{b}$. $\square$
I feel like this should be adaptable to cover the case of finitely generated abelian groups, but it keeps not coming out slick when I try. Maybe someone can show me how!