TITLE: On the existence of a direct summand containing a fixed subgroup
QUESTION [8 upvotes]: 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 $\langle g_1 \rangle \oplus \ldots \oplus \langle g_n \rangle$. 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 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)

REPLY [10 votes]: 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 $g_1 \not\in \langle h_1 \rangle$.