# Find the smallest $n \in \mathbb{N}$ such that the group is isomorphic to the direct product of $n$ cyclic groups

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?

• 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 $n$). – Theo C. Apr 23 at 21:21
• 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. – Bernard Apr 23 at 21:24
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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.

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>