# Structures of subgroups of a finite abelian p-group

$$\newcommand\la{\langle}\newcommand\ra{\rangle}$$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?

• I've added the commutative algebra tag, as this might be a standard facts of f.g. modules over PIDs.
– YCor
Jul 20 at 11:34
• What about the subgroup of ${\mathbb Z}/2 \oplus {\mathbb Z}/8$ generated by $(1,2)$? Jul 20 at 11:50
• @Derek Holt: You are right, that is a counterexample. I must be getting old(er): I was thinking of the structure theorem for (quotients by) subgroups of finitely generated free Abelian groups . Jul 20 at 13:09
• Holt's example is a counterexample. Then where was Robinson confused? Is it related to the condition of generating set that I made? Jul 20 at 13:11
• I have to confess that when I was very much younger than I am now I made the same mistake and asserted somewhere (I think in a lecture to students) that the answer to this question is yes and easily proved. Fortunately a bright student found a counterexample, which explains why I remember it! Jul 20 at 13:13

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.