Let $G$ be a finite abelian group and suppose $x\in G$ has largest order. Then $\langle x\rangle$ is a direct summand of $G$.
This exercise is intended to give an alternative proof of the fundamental theorem. My proof is as follows.
Set $|x|=n$. Let $H\leq G$ be maximal with respect to the property that $H\cap\langle x\rangle=0$. Suppose $G\neq H\oplus\langle x\rangle$, then we show that $H$ cannot be maximal.
Take $g\in G-(H\oplus\langle x\rangle)$. Let $k$ be the order of the image of $g$ in the quotient group $G/(H\oplus\langle x\rangle)$ under the canonical projection, then $k\mid |g|$. But $|g|\mid|x|$ (this follows from the fact that, give any two elements $a,b$ in an abelian group $G$, $G$ has an element of order $\operatorname{lcm}(|a|,|b|)$), it follows that $k\mid n$.
Also we have $kg=h+lx$ for some $h\in H$ and $l\in\mathbb{Z}$. Multplying by $n/k$, we have
$$-(n/k)h=(nl/k)x=0,\;\text{ by}\; H\cap\langle x\rangle=0$$
hence $\;n\mid(nl/k)$, i.e., $k\mid l$.
Set $y=g-\frac lkx$, then $ky=h$. Clearly $y\notin H$. We claim that $(H+\langle y\rangle)\cap\langle x\rangle=0$, from it will show that $H$ is not maximal. Suppose $h'+ry=sx\neq0$ for some $r,s\in\mathbb{Z}$ and $h'\in H$. Now $rg\in H\oplus\langle x\rangle$, thus $k\mid r$. But $ry=h\in H$, hence $sx=h+h'\neq0$, contradicting the assumption that $H\cap\langle x\rangle=0$.
Is my proof correct?