# On the existence of a direct summand containing a fixed subgroup

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 $\langle g_1 \rangle \oplus \ldots \oplus \langle g_n \rangle$. 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 direct summand of $G$?

(I asked this question on math.stackexchange, see http://math.stackexchange.com/questions/199928/smallest-pure-subgroup-containing-a-fixed-subgroup)

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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. –  Nick Gill Sep 21 '12 at 13:04
No, I have edited my question. "The cyclic groupS that they generate" insted of "The cyclic group that they generate" –  calc Sep 21 '12 at 13:17
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. –  Bugs Bunny Sep 21 '12 at 13:21
@Bugs Bunny Can you give a reference for the "pure subgroup generated by"? Also, how do you find the generators $h_i$ each dividing $g_i$? –  calc Sep 21 '12 at 13:23

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 $g_1 \not\in \langle h_1 \rangle$.
I think you mean $8g = (8,0)$... –  Arturo Magidin Sep 21 '12 at 17:02
Thanks but there is something I am missing. Why do you not start with $g_1=(2,1)$? –  calc Sep 21 '12 at 17:03
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? –  calc Sep 21 '12 at 18:41
@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 $S$. –  Arturo Magidin Sep 21 '12 at 21:27
@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 $(1,1)-(0,1)\notin H$. –  Arturo Magidin Sep 21 '12 at 21:31