Can (Z/10557Z)* be written as Cn1 x Cn2 x Cn3 with n1 dividing n2 dividing n3?

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If I was to try to work this out I would use the Chinese Remainder Theorem and since 10557 = 3^3 . 17 . 23
end up with (Z/10557Z)* isomorphic to (Z/27Z)* x (Z/17Z)* x (Z/23Z)* isomorphic to C18 x C16 x C22 where Cn represents the Cyclic group order n.

How would I then write this as Cn1 x Cn2 x Cn3 s.t. n1 divides n2 divides n3?
 
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this is explained in section 17 of these free notes:

http://www.math.uga.edu/%7Eroy/844-2.pdf and probably in many other algebra books.
 
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(Z/10557Z)* as Abelian Groups using Chinese Remainder Theorem
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