What are the invariant factors of $G = (\Bbb Z/12\Bbb Z) \times (\Bbb Z/25 \Bbb Z) \times (\Bbb Z/45\Bbb Z)$?

Question

What are the invariant factors of $G = (\Bbb Z/12\Bbb Z) \times (\Bbb Z/25 \Bbb Z)\times(\Bbb Z/45\Bbb Z)$ ?

My way of solving it: $$G \simeq (\Bbb Z/2^2\Bbb Z \times \Bbb Z/3\Bbb Z)\times(\Bbb Z/5^2\Bbb Z)\times(\Bbb Z/3^2\Bbb Z \times \Bbb Z/5\Bbb Z)$$

And because it is easy to "see": $$G \simeq (\Bbb Z/(3*5)\Bbb Z)\times (\Bbb Z/(3^2*5^2*4)\Bbb Z)=(\Bbb Z/15\Bbb Z)\times (\Bbb Z/900\Bbb Z)$$ as when writing $G \simeq (\Bbb Z/n_1\Bbb Z) \times (\Bbb Z/n_2 \Bbb Z)$ we need $n_1$ to divide $n_2$.

But:

1. What is the general method to find the invariant factors?
2. How many $n_i$ should we use?

Answer 1

Ok, I finally figured it out. The proper method is the following:

1. Decomposition of G in a product of $\Bbb Z/p_i^k\Bbb Z$ where $p_i$ is prime ($i \in \{0,1,...,r\}$) and $k$ an integer.

2. We order the p_i as follows:

   p_1^k_11 p_2^k_21 p_3^k_21 ...

   p_1^k_12 p_2^k_22

   ...

   with $k_i1 < k_i2 < ... < k_ir$ $\forall i \in \{0,1,...,r\}$

   ie starting with the higher exponents, on each row we sort the primes by the increasing order and on each column we sort the prime by exponents.

   3. Finally make the product of each row to get the invariant factors $n_k$.
   4. Check that they are such that $n_k | n_{k+1}$

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Here it would yield: $$2^0\ 3\ 5$$ $$2^2\ 3^2\ 5^2$$

and the invariant factors are 15 and 900 and 15|900.