Q: On the order of elements of the generating set? My question is only for the abelian case. The following example is to clarify the idea in the non-abelian case.
The dihedral group has the following presentation $$D_{2i}=\left<s,r/s^2=r^i=e,sr=r^{-1}s \right>$$
Order of $D_{2i}$ is $2i$. If we take $S=\{s,r\}$ and $H=\{s,sr\}$, then  $D_{2i}=\left<S \right>=\left<H\right>$, but sum order of elements of $S$ is $2+i$ and of $H$ is $4$. From this, we deduce that there is no relation between the order of the non-abelian group and the orders of elements in the generating set.
My question is there any relation between the order of abelian group and the order of elements in its  generating sets?
I guess the following is true $S=\{s_1,\ldots,s_k\}$ is a generating set of abelian group $G$ then $\Pi_{i=1}^k order(s_i) \geq order(G)$. 

Is this equation true, and are there any other relations?


A: The equation 
$$ \tag 1 \def\ord{\operatorname{ord}}\prod_i \ord s_i \ge \ord G$$ 
is true, for a proof note that given $g \in G = \def\<#1>{\left<#1\right>}\<s_1, \ldots, s_n>$, we can write 
$$ g = \prod_i s_i^{a_i(g)} $$
for some $a_i(g) \in \{0, \ldots, \ord s_i - 1\}$. This gives a one-to-one map $a = (a_1, \ldots, a_n) : G \to \prod_i \{0,\ldots, \ord s_i - 1\}$, hence we have (1). 
