Consider the theorem
Let $G$ be a finite Abelian group with order $|G|=p^n$ and $a$ an element of maximal order in $G$, then there is a subgroup $H$ of $G$ such that $G\cong |a|\times H$.
I'm interested only in the begin of the proof (I've seen some online and all go a similar way through). Consider $H$ the maximal subgroup of $G$ with $H\cap \langle a\rangle =\{e\}$.
No one says in his proof why this subgroup exist. I must miss something very elementary. I undestand the following: Lets take $b\in G-\langle a\rangle$, then $\langle a\rangle\cap \langle b\rangle=\{e\}$. Now the problem arises from the maximality. Let's take $c\in G-\langle a\rangle\langle b\rangle$ (we consider that $b$ and $c$ exist otherwise we are already done).
What can I say about $\langle b\rangle\langle c\rangle$? Could we build $H$ iteratively $H=\langle b\rangle\langle c\rangle\cdots$?
$G$ is finite, and you know $H=\{e\}$ satisfies your condition, so you can just take $H$ of maximum order that satisfies this condition. You don't actually have to construct it.