Suppose $G$ is a finite abelian group, then $G$ is isomorphic to a product of cyclic groups.
Suppose $$G \cong C_2 \times C_2 \times C_2 .$$ where $ C_2 $ is cyclic of order 2. Then $\#G=2^3=8$.
What is the maximal order of an element of $G$?
If we use additive notation,the generators of $G$ are $(1,0,0),(0,1,0)$ and $(0,0,1)$, each of which has order $2$. Thus, it follows that the maximal order of an element of $G$ is 2. (since all other elements have order less that that of the generators).
Is this argument correct?
If we generalize this: Let $$ G \cong C_{p_1} \times C_{p_2}\times \cdot\cdot\cdot \times C_{p_l} ,$$ where the $p_i$ are primes (not necessarily distinct) dividing the order of $G$. Then $g=(1,1,\cdot \cdot \cdot, 1)$ is a (non-unique) maximal element of $G$ that has order equal to the product of distinct primes dividng $\#G$.
How do we know there isn't some other element of higher order? (intuitively, this is kind of obvious, I don't know how to explain it rigorously.)