$\langle x \rangle$ is a direct summand of a finite abelian group where $x$ is maximal order Let $x$ be an element of a finite abelian group $G$ where $x$ has maximal order. Then I want to show that $\langle x\rangle$ is a direct summand of $G$. Note that I do not want to use finite abelian group classification theorem, because I want to use this fact to prove it. 
My strategy is to find a subgroup $H\le G$ where $\langle x\rangle\cap H = \{e\}$ and $\langle x,H\rangle = G$. But I'm unsuccessful in finding such $H$. 
 A: We may of course assume that $\langle x \rangle$ is a proper subgroup of $G$.
Suppose there is an element of prime order $y \notin \langle x \rangle$. Then consider $G / \langle y \rangle$ and proceed by induction.
If there is no such $y$, take any $z \in G \setminus \langle x \rangle$. We may assume that $z$ has order a power of a prime $p$, and that $z^{p} \in \langle x \rangle$. If $z^{p} = u^{p}$ for some $u \in \langle x \rangle$, then $z u^{-1}$ is an element of order $p$ not in $\langle x \rangle$, a contradiction.
Thus the $p$-part of the order of $x$ equals $\lvert z^{p} \rvert$, so that $x = a b$, with $a$ of order prime to $p$, and $\lvert b \rvert = \lvert z^{p} \rvert$. One sees now that $a z$ has order $\lvert x \rvert \cdot p$, a contradiction.
A: Assume $G$ is $p$-primary. Suppose $x$ is of maximal order, a power of $p$. We may assume $G$ is not cyclic. Thus, there exist two proper subgroups of the same order $p$, $H,K$. Let $H$ be the subgroup that is not contained in $\langle x\rangle$ ($\langle x\rangle$ has a unique subgroup of order $p$). It follows $H\cap \langle x\rangle$ is a proper subgroup of $H$, so it is trivial. Hence
$$\overline X=\frac{\langle x\rangle+H}H\simeq{\langle x\rangle}$$
Thus $\overline X$ is a cyclic subgroup of $G/H$, and of maximal order (the same order as $x$). Now choose a direct summand by induction on $|G|$ and lift it to a direct summand in $G$, and conclude using that $G=\bigoplus G(p_i)$ for some primes $p_1,\ldots,p_n$, by taking $x=x_1+\cdots+x_n$ where the $x_i$ are the components of $x$ in the sum. 
Claim An abelian $p$-group $G$ is cyclic if it has a unique subgroup of order $p$.
Proof Let $a\in G$ be of largest order, say $|a|=\exp G=p^k$. Let $H$ be the unique subgroup of order $p$. Then $H\leqslant \langle a \rangle$. Suppose there is $x\in G\smallsetminus \langle a\rangle$. We may assume that $x^p\in\langle a\rangle$, since iterating the situation $x^p\notin\langle a\rangle$, we may look at $x^{p^2}$ and so on. In some moment we will reach $x^{p^k}=1\in \langle a \rangle$. Then $x^p=a^m$. If $k=1$ then $x^p=1$ so $x\in H$ which cannot be. Suppose $k>1$. Then $1=x^{p^k}=a^{mp^{k-1}}$, so $p\mid m$. But then $x^p=a^{pl}$, so $x^{-1}a^l$ has order $p$, and is in $H$; which gives $x^{-1}$ is in $\langle a\rangle$, which cannot be. $\blacktriangleleft$
