Let $A=\{a_1,a_2,...,a_k\}$ be a subset of an abelian group $G$. Let $\langle{A}\rangle$ be a subgroup generated by $A$. Suppose that each $a_i\in A$ has finite order $d_i$. Then there are exactly $d_i$ distinct powers of $a_i$ for each $i$. Observe that each element of $\langle{A}\rangle$ is of the form $a_1^{\alpha_1}a_2^{\alpha_2}...a_n^{\alpha_n}$ with $\alpha_i\in\mathbb{Z}$.
Claim: The total number of distinct products of the form $a_1^{\alpha_1}a_2^{\alpha_2}...a_n^{\alpha_n}$ is at most $d_1d_2...d_k$
That is,
$$| \langle A \rangle |\leq d_1d_2...d_k$$
My Question
I am confused why the order of the generated group is bounded above by the product. In what case might it be less than the product and in what case is it exactly the product of the orders of the generating elements. I am assuming that the product itself follows by a counting argument.
Thanks.