$\newcommand\la{\langle}\newcommand\ra{\rangle}$Let $G=\mathbb{Z}/p^{i\_1}\times\cdots\times\mathbb{Z}/p^{i\_r}$ with $i\_1\leq\ldots\leq i\_r$ be a finite abelian $p$-group. Then there can be many choices of generators $\{x\_1,\ldots,x\_r\}$ such that the order of $x\_j$ is $p^{i\_j}$ and $G=\la x\_1\ra\times\cdots\times \la x\_r\ra$.


Let $H$ be a subgroup of $G$. Then $H$ is of the same form with less or equal number of factors.


Does there exist a choice of generators $\{x\_1,\ldots,x\_r\}$ of $G$ as above such that $H$ is a product of subgroups of $\la x\_j\ra$?


If it is not true, is there an easy counterexample?



Structures of subgroups of a finite abelian p-group
Let $G$ = ${\mathbb Z}/2 \oplus {\mathbb Z}/8$, and let $H$ be the cyclic subgroup of order $4$ generated by the element $h = (\bar{1},\bar{2})$.


There is no element $g \in G$ with $2g = h$, and so $H$ cannot be a subgroup of a cyclic direct summand of $G$ of order $8$. And clearly it cannout be a subgroup of a summand of order $2$, so the answer to the question is no, and this is a counterexample.


