$g\in G$ maximal order in $G$ abelian then $G=\left\oplus H$ If $g\in G$ is an element of maximal order in a finite abelian group $G$ then exists $H\leq G$ such that $G=\left<g\right>\oplus H$
Attempt: Using fundamental theorem I know that $G=C_{n_1}\times\cdots\times C_{n_k}$ where $n_i|n_{i+1}$. With some work I proved $|\left<g\right>|=n_k$, so $C_{n_k}\cong \left<g\right>$. Then $G\cong H\times \left<g\right>$, but this is not an equality.

Here is a proof using the structure theorem. By taking a primary decomposition of $G$ we can assume that there is a prime $p$ and exponents $a_i \in \mathbb{N}$ such that $n_i = p^{a_i}$ for each $i$. Let 
$$ G = \langle g_1 \rangle \oplus \ldots \oplus \langle g_k \rangle $$
where $g_i$ has order $p^{a_i}$ for each $i$ and $a_1 \le \ldots \le a_k$. Let
$$g = c_1g_1 + \cdots + c_{k-1}g_{k-1} + c_kg_k$$ 
where $c_i \in \mathbb{N}$ for each $i$.
If every $c_ig_i$ has order dividing $p^{a_k-1}$ then $g$ has order dividing $p^{a_k-1} = n_k/p_k$, a contradiction. Therefore there exists $i$ such that $a_i = a_k$ and $p$ does not divide $c_i$. By reordering the $g_i$, we may assume that $i = k$. Take $r$ such that $rc_k \equiv 1$ mod $p^{a_k}$. Then $rg = (rc_1)g_1 + \cdots + (rc_{k-1})g_{k-1} + g_k$ and it is straightforward to show that
$ G = \langle g_1, \ldots, g_{k-1} \rangle \oplus \langle g \rangle. $
Remark. The reduction to the primary case is needed to make this argument work. For example, if $G = \langle g_1 \rangle \oplus \langle g_2 \rangle $ where $g_1, g_2$ both have order $6$ then $2g_1 + 3g_2$ is an element of order $6$, but neither $2g_1$ nor $3g_2$ has maximal order.
