A friend recently showed me some nice machinery from Representation Theory that allows you to show relatively simply that if $G$ is abelian, then it has $|G|$ distinct characters (homomorphisms into $\mathbb{C}$). Thinking about this led me to the conclusion that the product of the periods of the elements that make up any generating set of $G$ is always $|G|$:
Consider a character $f$ of $G$ - we know by definition that there is an element $a$ of $G$ such that $f(a) \neq 0$ and that $f(ab) = f(a)f(b)$, so for an arbitrary element $x$ of $G$ $$f(x) = f(1*x) = f(1)*f(x)$$ so $f(1) = 1$. Then, consider that the period of $x$ is $n$ (i.e. $x^n = 1$), so $f(1) = f(x^n) = f(x)^n =1$ so $f(x)$ is a root of unity!
Now, consider a given element $a_i$ from a generating set $\{a_0,a_1,...a_n\}$ of $G$ with period $k_i$. If we suppose that we wish to construct a character of $G$, since $a_i$ satisfies the relation $f(a_i)^{k_i}=1$, we can assign any of the $k_i$th roots of unity to $f(a_i)$ and, further, if we assign values to $f(a_j)$ for all $j$ then we have uniquely determined $f$, as all other elements' outputs in $f$ can be determined by products of the generating set's outputs.
An example would be the residue class modulo 12, for which $\{5,7\}$ is a generating set: both 5 and 7 have period 2, so in constructing a character for the group we can assign the values 1 or -1 for each of $f(5)$ and $f(7)$, uniquely determining the values of $f$ for the rest of the group through products.
So, if each of the generating set's elements can be assigned each of the $k_i$th roots of unity as outputs, then the number of characters of the group is the product of their periods. In the above case, for example, the elements of the generating set have period 2 and so the number of characters of that group is $2\times2 = 4$, which can be checked quickly. Now, since for an abelian group $|G|$ is the both the number of elements of the group and the number of characters of the group, it follows that the product of the periods of the elements in any given generating set of $G$ is $|G|$.
Is this correct? I realise now that I am assuming (I think!) that the generating sets in question are, in some sense, "minimal" in that $\{5,7,11\}$ would have also been a generating set for the above example but the inclusion of 11 is entirely unnecessary.