From Dan Saracino, Abstract Algebra (Exercise 14.11)
Let $G$ be an abelian group of order $p^n$, where $p$ is a prime. An element $x \in G$ is said to be of maximal order if $o(x)\geq o(y)$ for all $y \in G$. Show that the only subgroup of $G$ that contains all the elements of maximal order is $G$ itself.
Based on the fundamental theorem of finite abelian groups, I can decompose $G$ into a direct product of cyclic groups of prime power order:
$$G \cong \mathbb{Z}_{p^{t_1}} \times \mathbb{Z}_{p^{t_2}} ... \times \mathbb{Z}_{p^{t_r}},$$ where $n = t_1 + t_2 + ... t_r$ is an arbitrary partition of $n$ with $t_1>t_2>...>t_r$. Then the set of elements
$$\{(1,0,0,0...,0), (1,1,0,0,0...,0), ...,(1,1,1,1...,1)\}$$
all have a maximal order of $t_1$.
I see that the elements in this set can generate $G$ together but how do I show this formally? Also, how to prove the "only" part about $G$?
Looking for some hints.