Let $G$ be a finite abelian group, and $g\_1, \ldots, g\_n \in G$ such that the cyclic groups that they generate are in direct sum $\langle g\_1 \rangle \oplus \cdots \oplus \langle g\_n \rangle$. 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. $g\_i = a\_i \cdot h\_i$, for all $1 \leq i \leq n$,
2. the cyclic subgroups generated by the $h\_i$ are in direct sum, $H:=\langle h\_1 \rangle \oplus \cdots \oplus \langle h\_n \rangle$.
3. $H$ is a direct summand of $G$?


(I asked this question on math.stackexchange, see <https://math.stackexchange.com/questions/199928/smallest-pure-subgroup-containing-a-fixed-subgroup>)



On the existence of a direct summand containing a fixed subgroup
I am not convinced that this is true, because the pure subgroup generated by the $g\_i$ might
not have the stipulated form as a direct sum of the $h\_i$.


Let $G = {\mathbb Z}/16{\mathbb Z} \oplus {\mathbb Z}/4{\mathbb Z}$, $n=1$, and $g\_1=(4,2)$.
What could $h\_1$ be?


We can prove that there is no such $h\_1$ as follows. For any $g \in G$ of order 16, we have $8g = (8,0)$. Since $(8,0) \in \langle h\_1 \rangle$, $\langle h\_1 \rangle$ cannot be a direct summand of $G$ of order 4. So $\langle h\_1 \rangle$ would have to be a direct summand of order 16. But then $4h\_1 = (4,0)$ or $(12,0)$ and so $g\_1 \not\in \langle h\_1 \rangle$.


