Smallest pure subgroup containing a fixed subgroup I will ask a slightly more precise question then in the title.
Let $G$ be a finite abelian group, and $g_1, \ldots, g_n \in G$ such that the cyclic groups they generate are in direct sum $\langle g_1 \rangle \oplus \ldots \oplus \langle g_n \rangle$. Is it always possible to find elements $h_1, \ldots, h_n \in G$ and integers $a_1, \ldots, a_n$ such that the following three facts hold
1) $g_i= a_i h_i$, for all $1 \leq i \leq n$,
2) the cyclic subgroups generated by the $h_i$ are in direct sum, $H:=\langle h_1 \rangle \oplus \ldots \langle h_n \rangle$.
3) $H$ is a pure subgroup of $G$?
(recall that a subgroup $H < G$ is pure if for all $h \in H$ and all $n \in \mathbb{N}$, if $h$ is $n$-divisible in $G$, it is also $n$-divisible in $H$).
Added: in this context, pure is a synonim for being a direct summand. In fact, the latter is the relevant property of the question.
 A: Incomplete answer:
Suppose $G$ is an abelian group and $m$ is a positive integer such that $mG=0$. For instance, if $G$ is finite, then one could take $m=|G|$.
Let $g$ be an element of $G$ and let $H_0$ be a subgroup of $G$ that intersects $\langle g\rangle$ only in 0. Consider the collection of all subgroups of $G$ containing $H_0$ that intersect $\langle g\rangle$ trivially. It is non-empty (containing $H_0$) and closed under unions of chains, so Zorn's lemma guarantees a maximal element, $H$.
Now consider the collection of subgroups of $G$ that contain $g$ but intersect $H$ only in 0. The collection is non-empty, containing $\langle g\rangle$, and closed under unions of chains, so Zorn's lemma guarantees a maximal element, $K$.
Clearly $H+K$ is a direct sum. Let $x \in G$ have order $a$. If $\langle x \rangle \cap \langle g \rangle = 0$, then clearly $H+\langle x \rangle$ is a larger element of the first collection, contradicting the definition of $H$ unless $x$ is already in $H$. So we may assume $\langle bx \rangle \cap \langle g \rangle \neq 0$, in particular, we can choose $b$ so that $a/b$ is prime. Then if $a \neq 1$, $(\langle x \rangle +K)\cap H \neq 0$, so $0 \neq cx \in H$ for some $c$ dividing $a$. If $a$ is a prime power, then $\langle bx \rangle \leq \langle cx \rangle$, a contradiction.
In general (non $p$-group), I'm not sure why this is a contradiction, but I'm fairly sure that $H$ and $K$ are the summands if they exist.
At any rate: Take $g$ to be $g_1$, $H_0 = \langle g_2, \ldots, g_n\rangle$, then $h_1$ is a generator of $K$ (which is cyclic since $G$ is bounded) and $G=H\oplus K$.
I suggest trying to prove $H$ and/or $K$ is pure more directly. A bounded pure submodule is a direct summand, so everything works.
