Theorem. Let $G$ be a finite abelian group and $g \in G$. Then $G$ contains a subgroup $H$ such that $G/H$ is cyclic and the order of $gH$ in $G/H$ is the same as the order of $g$ in $G$.
Sketch of the proof. Write $G \simeq \oplus_p G_p$ where $G_p$ is a $p$-Sylow of $G$ and write $G_p \simeq \oplus_r \mathbb{Z}/p^{i_r}\mathbb{Z}$. So, define $H_p \leq G_p$ as direct sum of some $\mathbb{Z}/p^{i_r}\mathbb{Z}$ (there is some important details to do that) and take $H = \oplus_p H_p$.
That is the idea of the book "Class Field Theory" by Nancy Childress. I don't know if it is possible, but I would like to know if there is some different proof. Maybe it is not the best way to describe, but I would like to see a "less constructive" proof. Anyway, I appreciate any references.
Here’s an inductive proof.
It suffices to prove this for abelian $p$-groups (do each $p$-part separately, then take their direct sums).
Let $G$ be abelian of order $p^n$, $g\in G$, and assume the result holds for any abelian groups of order $p^k$, $k\lt n$. If $G$ is cyclic, take $H=\{e\}$ and we are done.
If $G$ is not cyclic, then it has more than one subgroup of order $p$. As $\langle g\rangle$ contains at most one subgroup of order $p$, let $H$ be a subgroup of order $p$ with $H\cap\langle g\rangle = \{e\}$. Then $gH$ has the same order in $G/H$ as $g$ has in $G$, and $|G/H|\lt |G|$. Inductively, $G/H$ has a subgroup $M/H$ such that $G/M \cong (G/H)/(M/H)$ is cyclic, and the order of $(gH)M = gM$ in $G/M$ is the same as the order of $gH$ in $G/H$, which is the same as the order of $g\in G$. $\Box$