From my linear algebra notes(http://alistairsavage.ca/mat3141/notes/MAT%203141%20-%20Linear%20Algebra%20II.pdf , p.99 if anyone is interested).
(The structure theorem). Let $M$ be a module of finite type over a euclidean domain $A$. Then $M$ is a direct sum of cyclic submodules $M = Au_1 ⊕ Au_2 ⊕ · · · ⊕ Au_s$ with $A \supsetneq Ann(u_1) ⊇ Ann(u_2) ⊇ · · · ⊇ Ann(u_s)$.
The next theorem states that: In the case where $A = \Bbb Z$, an $A$-module is simply an abelian group. Let $G$ be a finite abelian group. Then $G$ is a direct sum of cyclic subgroups $G = Zg_1 ⊕ Zg_2 ⊕ · · · ⊕ Zg_s$ with $o(g_1) \neq 1$ and $o(g_1)| o(g_2)| . . . | o(g_s)$. (the orders)
Furthermore, the exponent of $G$ is $o(g_s)$.
Thoughts: I am unsure why $o(g_1)| o(g_2)| . . . | o(g_s)$. Any clarification or insights much appreciated.
Let $\langle g\rangle$ be a cyclic (additive) group with order $n$. Then $n\mathbb{Z}=\operatorname{Ann}(g)$.
Indeed, $n\mathbb{Z}$ is the kernel of the surjective homomorphism $z\mapsto zg$, just by computing cardinalities after applying the homomorphism theorem.
Also $n\mathbb{Z}\supseteq m\mathbb{Z}$ if and only if $n\mid m$.
So the statement about abelian groups is exactly the same as the general statement over Euclidean domains.
The statement about the exponent should be easy: if $(x_1,x_2,\dots,x_s)\in\mathbb{Z}g_1\oplus\dots\oplus\mathbb{Z}g_s$, then the order of $x_i$ divides $o(g_s)$ by what seen before. Therefore any element of $G$ has order a divisor of $o(g_s)$. Since $g_s$ has that order, the exponent is $o(g_s)$.