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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)

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  • $\begingroup$ Do you really mean that the $g_i$'s generate a cyclic group $C$? if so, why can't you take all of the $h_i$ to be the same element, a generator of the largest cyclic group containing $C$. My first thought is that this solution would satisfy (1), (2) and (3)... But I could be wrong. $\endgroup$
    – Nick Gill
    Commented Sep 21, 2012 at 13:04
  • $\begingroup$ No, I have edited my question. "The cyclic groupS that they generate" insted of "The cyclic group that they generate" $\endgroup$
    – calc
    Commented Sep 21, 2012 at 13:17
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    $\begingroup$ It should be obvious. Your $H$ is the pure subgroup generated by $g_i$-s, and in finite groups, pure subgroups should be the same as direct summands. $\endgroup$
    – Bugs Bunny
    Commented Sep 21, 2012 at 13:21
  • $\begingroup$ @Bugs Bunny Can you give a reference for the "pure subgroup generated by"? Also, how do you find the generators $h_i$ each dividing $g_i$? $\endgroup$
    – calc
    Commented Sep 21, 2012 at 13:23
  • $\begingroup$ en.wikipedia.org/wiki/Pure_subgroup $\endgroup$
    – Bugs Bunny
    Commented Sep 21, 2012 at 13:24

2 Answers 2

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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 $g_1 \not\in \langle h_1 \rangle$.

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  • $\begingroup$ I think you mean $8g = (8,0)$... $\endgroup$
    – Arturo Magidin
    Commented Sep 21, 2012 at 17:02
  • $\begingroup$ Thanks but there is something I am missing. Why do you not start with $g_1=(2,1)$? $\endgroup$
    – calc
    Commented Sep 21, 2012 at 17:03
  • $\begingroup$ Anyway I am still not convinced about the pure subgroup generated by the $g_i$. What is it? You generate a subgroup with the $g_i$. Then you take the roots. Then? Is this a subgroup at all? Or do you have to still take the subgroup generated and continue? $\endgroup$
    – calc
    Commented Sep 21, 2012 at 18:41
  • $\begingroup$ @calc: A pure subgroup $S$ of $G$ is a subgroup such that $nG\cap S=nS$ for all $n$. You may not need to take roots; for example, in $G=\mathbf{Z}_4\times\mathbf{Z}_2$, $S=\langle (1,1)\rangle$ is pure: $2G\cap S = \{(0,0),(2,0)\} = 2S$, and that's the only one you need to worry about. Even though $2(1,0)\in S$, you do not need $(1,0)$ in $S$ for $S$ to be pure, you just need $2(1,0)$ to be twice something in $S$. $\endgroup$
    – Arturo Magidin
    Commented Sep 21, 2012 at 21:27
  • $\begingroup$ @Calc: But if you have a subgroup $S$ of $G$, and you let $H=\{g\in G\mid ng\in S, ng\neq 0\}$ (which I suspect is what you mean; otherwise, in a torsion group, adding "all roots" would mean adding the entire thing), then $H$ need not be a subgroup; e.g., with $G$ and $S$ as above, $(1,0)\in H$ (since $2(1,0)=(2,0)\in S$ is nonzero), and $(1,1)\in H$ (since it is in $S$), but $(1,1)-(0,1)\notin H$. $\endgroup$
    – Arturo Magidin
    Commented Sep 21, 2012 at 21:31
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This old question just surfaced on MO for some reason, so I thought I'd add a comment to put Derek Holt's excellent answer in a wider context. Given a prime $p$, the problem of classifying pairs consisting of a finite abelian $p$-group of exponent $p^n$ together with a subgroup, up to isomorphism of pairs, is known to be of wild representation type for $n$ at least seven. See for example the paper of Ringel and Schmidmeier, "Submodule categories of wild representation type" (JPAA 2006). If what you're asking were true, it would imply a classification for such pairs. But more generally, trying to tame such pairs is destined to failure.

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