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Find a subgroup of $\Bbb Z_{12} \oplus \Bbb Z_{4} \oplus \Bbb Z_{15}$ of order $9$.

I know that this group is non-cyclic, but I can't figure a way to even begin finding a subgroup of this order since $9$ doesn't divide $12, 4,$ or $15$ which implies there are no elements in any of the $\Bbb Z$'s whose order is $9$.

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  • $\begingroup$ It's true 9 doesn't divide 12, 4, or 15. But what is the order of $\mathbb{Z}_{12} \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_{15}$? It's not “12, 4, or 15”. :-) $\endgroup$ – Matthew Leingang Aug 4 '16 at 16:52
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    $\begingroup$ Can you find two subgroups of order $3$? If so, try to see how to "combine" them to find a subgroup of order $9$. $\endgroup$ – Bungo Aug 4 '16 at 16:55
  • $\begingroup$ @Bungo Interesting, I see that $\langle (4,0,0) \rangle \cup \langle (0,0,5) \rangle$ has $9$ elements. Does this type of method usually work for finding subgroups or is it in this case simpler because of the small amount of combinations? $\endgroup$ – Oliver G Aug 4 '16 at 17:02
  • $\begingroup$ It generally works for subgroups of finitely generated abelian groups; for others, it's a trickier business. $\endgroup$ – John Hughes Aug 4 '16 at 17:08
  • $\begingroup$ @OliverG The key fact here is that in an abelian group, all subgroups are normal. Consequently the product $HK$ of any two subgroups $H$ and $K$ is itself a subgroup. So finding any two subgroups of order $3$ and forming their product gives you a subgroup of order $9$. $\endgroup$ – Bungo Aug 4 '16 at 17:20
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How about $4\mathbb{Z}_{12}\oplus \{0\}\oplus 5\mathbb{Z}_{15}$? This group has $9$ elements. It doesn't matter that $9$ doesn't divide $12$, $4$, or $15$. $9$ must devide the order of the group, which is $12\cdot 4\cdot 15=720$.

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    $\begingroup$ Do you perhaps mean $4\Bbb Z_{12} \oplus \{0\} \oplus 5\Bbb Z_{15}$? $\endgroup$ – David Wheeler Aug 6 '16 at 2:51

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