Number of elements in a finite abelian groups

Question

Is the following true?

Let $G$ be a finite abelian group with a minimal generating set $S$. By minimal generating set I mean we cannot reduce the cardinality further. Let $S=\{a_1,a_2,\ldots ,a_k\}$ with orders $r_1, r_2, \ldots, r_n$ respectively. Then the group $G$ has $r_1r_2\cdots r_n$ elements.

For example $G=\{e,a,b,c,ab,bc,ca,abc\}$ Take $S=\{a,b,c\}$ Then $|G|=2\times2\times2=8$.

Answer 1

No this is not true in general. Let $G = \langle a \rangle \times \langle b \rangle$ with $|a|=2$, $|b|=4$. Then $\{ ab,b \}$ is a minimal generating set (in any sense), but they both have order $4$, whereas the group has order $8$.

Of course by the Fundamental Theorem, there always exsists a minimal generating set with the property you describe, such as $\{ a, b \}$ in the above example.

Answer 2

Let $g_1,g_2\dots g_n$ be the generators for the group, then every element of the group can be seen as $g_1^{a_1}g_2^{a_2}\dots g_n^{a_n}$ where $0\leq a_i< |g_i|$. This tells us $|G|\leq\prod\limits_{i=1}^n|g_i|$.

The inequality can be sharp however, consider the group $\mathbb Z_4+\mathbb Z_6$ of order $24$. The set $\{(1,2),(1,1)\}$ is a minimal and minimum generating set of the group since the group is not cyclic. On the other hand the product of the orders of the elements is $144$.