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I will ask a slightly more precise question then in the title.

Let $G$ be a finite abelian group, and $g_1, \ldots, g_n \in G$ such that the cyclic groups 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 pure subgroup of $G$?

(recall that a subgroup $H < G$ is pure if for all $h \in H$ and all $n \in \mathbb{N}$, if $h$ is $n$-divisible in $G$, it is also $n$-divisible in $H$).

Added: in this context, pure is a synonim for being a direct summand. In fact, the latter is the relevant property of the question.

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    I believe so yes. (Someone feel free to write a nice answer.) Consider only $n=1$, and let $G$ be bounded (so $mG=0$ for some positive integer $m$), then an $h_1$ exists as you want, and must be a direct summand of $G$. Now work on $g_2$, etc. by working in a direct complement of $\langle h_1 \rangle$ containing the other $g_i$. The $h_i$ you produce will work for the original claim.2012-09-20
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    Trivially, $h_i=g_i$, $a_i=1$ will do. But what doe sthe question have to do with "smallest pure subgroup" and "containing a fixed subgroup"?2012-09-20
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    @HagenvonEitzen the choice you made does not seem to respect property 3, the 'pureness'. For example $G= \mathbb Z / n \mathbb Z$ where $n=p^2$ and choose $g=[p]$, then $h=g$ is $p$-divisible in $G$ but not in $H$.2012-09-20
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    @Hagen von Eitzen, the intersection of pure subgroups is not pure, so the title is a bit misleading. What I mean more precisely with "smallest" is explained in the question.2012-09-20
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    @Jack Schmidt Can you say something more on how you find this $h_1$? Moreover, how do we know that $g_2 \notin \langle h_1 \rangle$ ?2012-09-21
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    @calc: If $G$ is a $p$-group (abelian of bounded exponent) then automatically $g_2 \notin \langle h_1\rangle$, but otherwise you have to choose $h_1$ correctly. My proof currently only works for $p$-groups. You might check if it is false for non-p-groups (if so, I can post the p-groups proof).2012-09-21
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    Cross-posted at http://mathoverflow.net/questions/107768/on-the-existence-of-a-direct-summand-containing-a-fixed-subgroup2012-09-21
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    @Niccolò: Oh, I see. I mistakenly assumed that $\langle g_1\rangle \oplus\ldot = G$ was assumed.2012-09-24

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