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Find the smallest $n \in \mathbb{N}$ such that the group $\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$ is isomorphic to the direct product of $n$ cyclic groups.

I'm not sure but if I understand correctly,

$\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$ is isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{4} \times \mathbb{Z}_{5} \times \mathbb{Z}_{5} \times \mathbb{Z}_{9}$,

$\mathbb{Z}_{2} \times \mathbb{Z}_{5} \times \mathbb{Z}_{9}$ is isomorphic to $\mathbb{Z}_{90}$,

$\mathbb{Z}_{3} \times \mathbb{Z}_{4} \times \mathbb{Z}_{5}$ is isomorphic to $\mathbb{Z}_{60}$,

and therefore, $\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$ is isomorphic to $\mathbb{Z}_{60} \times \mathbb{Z}_{90}$ and the answer is $n = 2$. Is this a correct solution?

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    $\begingroup$ I agree that the answer is $n=2$. For the sake of being thorough, I would probably explain why $n=1$ is impossible (just because you've found one situation where $n=2$ doesn't necessarily mean it's the smallest $n$). $\endgroup$ – Theo C. Apr 23 at 21:21
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    $\begingroup$ Yes, it is. You just have to find the smallest number of groups of moduli such that the moduli in each group are pairwise coprime. $\endgroup$ – Bernard Apr 23 at 21:24
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$ – Shaun 2 days ago
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You may find the invariant factors.

Actually $\mathbb{Z}_6\times \mathbb{Z}_{20}\times \mathbb{Z}_{45}\cong (\mathbb{Z}_2\times\mathbb{Z}_4)\times(\mathbb{Z}_3\times\mathbb{Z}_9)\times(\mathbb{Z}_5\times\mathbb{Z}_5)$. Now we can pick the largest ones in each bracket to form $\mathbb{Z}_4\times\mathbb{Z}_9\times\mathbb{Z}_5\cong\mathbb{Z}_{180}$. Then we pick the largest ones of the remaining, which is $\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_5\cong\mathbb{Z}_{30}$, and nothing remain. Hence the group is isomorphic to $\mathbb{Z}_{180}\times\mathbb{Z}_{30}$ and so $n\le 2$. Since the group is obviously not cyclic, we have $n = 2$.

I think we can always get $n$ for any abelian group by finding the invariant factors.

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According to GAP, the group is $$\Bbb Z_{180}\times \Bbb Z_{30}$$ and, of course, not cyclic.

gap> G:=DirectProduct(CyclicGroup(6) , DirectProduct(CyclicGroup(20), CyclicGroup(45)));
<pc group of size 5400 with 8 generators>
gap> StructureDescription(G);
"C180 x C30"
gap> IsCyclic(G);
false
gap>
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