(Z/10557Z)* as Abelian Groups using Chinese Remainder Theorem

[RVP91]

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?

 

[spamiam]

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?

Good start. You're looking for the invariant factor decomposition of the group--let's start out by computing its elementary divisor decomposition. Use the Chinese remainder theorem again on $\mathbb{Z}/18 \mathbb{Z} \times \mathbb{Z}/ 16 \mathbb{Z} \times \mathbb{Z}/ 22 \mathbb{Z}$ to write them as a product of cyclic groups with prime power order. This gives you the elementary divisors of the group.

From there, take a look at this link to convert to the invariant factor decomposition. Basically you find the largest factor first by choosing the largest prime power for each of the distinct primes and then multiplying them together. You then proceed similarly until you've used up all the elementary divisors. The link I gave has some examples, too.

Hope that helps!