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

  • $\begingroup$ I've added the commutative algebra tag, as this might be a standard facts of f.g. modules over PIDs. $\endgroup$
    – YCor
    Jul 20 at 11:34
  • 3
    $\begingroup$ What about the subgroup of ${\mathbb Z}/2 \oplus {\mathbb Z}/8$ generated by $(1,2)$? $\endgroup$
    – Derek Holt
    Jul 20 at 11:50
  • $\begingroup$ @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 . $\endgroup$ Jul 20 at 13:09
  • $\begingroup$ Holt's example is a counterexample. Then where was Robinson confused? Is it related to the condition of generating set that I made? $\endgroup$
    – YJ Kim
    Jul 20 at 13:11
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    $\begingroup$ 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! $\endgroup$
    – Derek Holt
    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.

  • $\begingroup$ After first seeing the question, I thought that surely the Smith normal form somehow gives you a positive answer. Which is wrong, as this answer shows. $\endgroup$
    – spin
    Jul 20 at 14:19
  • $\begingroup$ @spin : I made that error in my (now deleted) comment. $\endgroup$ Jul 20 at 14:23
  • $\begingroup$ @DerekHolt Thanks for your easy and clear example! $\endgroup$
    – YJ Kim
    Jul 21 at 1:34

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