Abelian group with order product of primes = cyclic?

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SUMMARY

An abelian group G with order equal to the product of distinct primes, expressed as \# G = p_1 p_2 \cdots p_n, is definitively cyclic. This conclusion is supported by Cauchy's theorem, which states that for each prime order element g_i, their product must equal the order of G. The result can also be confirmed through invariant factor decomposition, indicating that G is isomorphic to a direct product of cyclic groups, specifically G ≅ Z_{k_1} × Z_{k_2} × ... × Z_{k_n}, where all k_i are coprime.

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  • Understanding of abelian groups
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It seems rather straight forward that if you have an abelian group G with [itex]\# G = p_1 p_2 \cdots p_n[/itex] (these being different primes), that it is cyclic. The reason being that you have elements [itex]g_1, g_2, \cdots g_n[/itex] with the respective prime order (Cauchy's theorem) and their product will have to have the order of G. Rather simple, but I wanted to check that I'm not overlooking something simple because I find the result rather interesting although I was never told this in any of my algebra classes, which strikes me as strange.
 
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I think this can be confirmed by invariant factor decomposition (http://en.wikipedia.org/wiki/Finitely-generated_abelian_group) although really Cauchy's theorem should be sufficient for a proof. [itex]G \cong Z_{k_1} \times Z_{k_2} \times \ldots \times Z_{k_n}[/itex] such that [itex]k_1 \vert k_2 \vert \ldots k_{n-1} \vert k_n[/itex]. All [itex]k_i[/itex] are coprime by hypothesis so there can only be one factor.
 

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