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I've been pondering the concept of the direct sum of a finite number of groups, particularly when it becomes cyclic. I suspect that for the direct sum to be cyclic, some or perhaps all of the constituent groups must themselves be cyclic, but I'm not entirely certain. I'm approaching this inquiry from a general knowledge perspective.

Could anyone shed light on the conditions under which the direct sum of finite groups becomes cyclic? I would greatly appreciate any brief explanations, intuitions, or clues that could help me grasp this concept better.

Thank you for your insights and assistance in advance.

hadadrefael
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  • Well, every subgroup of a cyclic group is itself cyclic. Consider the subgroup ${e^G}\times H$ in a cyclic group $G\times H.$ So, that's necessary. If you ask about the sufficient condition, look at $|G|$ and $|H|$ and their $\gcd$? – PinkyWay Oct 20 '23 at 08:55
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    Suppose you have two finite cyclic groups generated by $x$ and $y$ respectively. What will be the order of $(x , y)$. That should answer your question. – Tony Pizza Oct 20 '23 at 09:07
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    See https://math.stackexchange.com/questions/4789873/prove-or-disprove-the-direct-sum-of-the-multiplicative-groups-u-n-and-u-m – Nicky Hekster Oct 20 '23 at 11:31
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    A direct sum of a finite number of finite groups is a cyclic group if and only if each summand is a cyclic group and their orders are pairwise cosimple. Try to prove it. – kabenyuk Oct 20 '23 at 12:50

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Thanks a lot, I understood it using these facts:

$\Bbb Z_n\oplus\Bbb Z_m$ is isomorphic to $\Bbb Z_{nm}$ if and only if $\gcd(n,m)=1$.

Cyclic groups of the same order are isomorphic.

So, suppose I have a direct sum of two cyclic groups which are isomorphic to $\Bbb Z_n$ and $\Bbb Z_m$ (while $n$ and $m$ is their order), their direct sum is isomorphic to $\Bbb Z_{nm}$, which is cyclic if and only if $(n, m) = 1$. Thank you for the clarification!

PinkyWay
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hadadrefael
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