Prove that a finite abelian p-group is generated by its elements of maximal order.
Let P be a finite abelian p-group. P is certainly finitely generated, so you can use the Fundamental Theorem of Finitely Generated Abelian Groups.
A p-group by definition has order $\displaystyle p^n$ where $\displaystyle n\in\mathbb{N}$.
Thus $\displaystyle P\cong\mathbb{Z}_{p^n}$ or $\displaystyle P\cong\mathbb{Z}_{p^{n-1}}\oplus\mathbb{Z}_{p}$ or $\displaystyle P\cong\mathbb{Z}_{p^{n-2}}\oplus\mathbb{Z}_{p^2}$ or $\displaystyle P\cong\mathbb{Z}_{p^{n-2}}\oplus\mathbb{Z}_{p}\oplus\mathbb{Z}_{p}$, etc.
Note that the order of an element is simply the lcm of the orders in each coordinate, generally you write these factors in descending order from left to right and since each factor is a power of a prime, you notice that the order of each element is really just the order of the component in the first coordinate (which is cyclic so there is a nonzero element here with maximal order that generates everything in the first coordinate). Then you could pick every other combination of elements for the rest of the positions and it would still have the same order. In this way I think this shows you why this is true then that this group would necessarily be generated by the elements of maximal order.
While I think there is probably a much easier way of proving this, it is not immediate to me anyway. Though this method is certainly a bit messy , I feel that it at least explains more constructively why this is actually the case.