Write down the complete list of abelian groups with 360 elements Here is what I got:
$C_{360}; C_{180}\times C_{2};C_{90}\times C_{2}\times C_{2};C_{120}\times C_{3};C_{60}\times C_{6}$
I thought I would get much more but I got these so far. Is there something I miss?To me it doesn't seem so though
I guess there is a misundertanding I have
I choose my abelian groups according to this: 
$C_{n_1}\times C_{n_2}\times C_{n_3}\times...\times C_{n_r} $
such that $n_1$ is divisible by $n_2$, $n_2$ is divisible by $n_3$, $n_3$ is divisible by $n_4$, $n_{r-1}$ is divisible by $n_r$ and so on..
 A: First, factorize $360 = 2^3\cdot 3^2 \cdot 5$. Recall that $C_{nm}\cong C_n\times C_m$ if $n$ and $m$ are coprime. This excludes a lot of possibilities. And looking at the partitions of the exponents, we have $$\begin{align} &\color{red}{C_2 \times C_2 \times C_2} \times \color{blue}{C_3 \times C_3} \times \color{darkgreen}{C_5} \cong C_{30} \times C_6 \times C_2\\ & \underline{\color{red}{C_2 \times C_2 \times C_2} \times \color{blue}{C_9} \times \color{darkgreen}{C_5}} \cong C_{90} \times C_2 \times C_2 \\ & \color{red}{C_2 \times C_4} \times \color{blue}{C_3 \times C_3} \times \color{darkgreen}{C_5} \cong C_{60}\times C_3 \times C_2 \cong C_{60}\times C_6 \\ &\underline{\color{red}{C_2 \times C_4}\times \color{blue}{C_9} \times \color{darkgreen}{C_5}} \cong C_{180} \times C_2 \\ & \color{red}{C_8} \times \color{blue}{C_3 \times C_3} \times \color{darkgreen}{C_5}\cong C_{120}\times C_3 \\ & \color{red}{C_8} \times \color{blue}{C_9} \times \color{darkgreen}{C_5}\cong C_{360}.\end{align}$$The equivalences in the right side are obtained gluing together the largest of each color group, one of each, and repeating until there are no factors left.
A: You have most, but not all. Note, as I think you have, that $360= 2^3*3^2*5$. 
I think one thing that you're missing is that $C_2 \times C_2$ is not isomorphic to $C_4$ -in the latter group we have an element of order 4, and no such elements in the former group. 
We thus have 3 non-isomorphic groups of order $8=2^3$: $C_2 \times C_2 \times C_2$, $C_2 \times C_4$, and $C_8$. We have 2 non-isomorphic groups of order $9=3^2$: $C_3 \times C_3$, and $C_9$. And we have a single group of order 5, $C_5$. 
Our different groups of order 360 then should be all of the possible combinations of the above groups, one from each order, giving us 6 groups:


*

*$C_2 \times C_2 \times C_2 \times C_3 \times C_3 \times C_5 \cong C_{30} \times C_6 \times C_2$

*$C_2 \times C_2 \times C_2 \times C_9 \times C_5 \cong C_{90} \times C_2 \times C_2$

*$C_2 \times C_4 \times C_3 \times C_3 \times C_5 \cong C_{60} \times C_6$

*$C_2 \times C_4 \times C_9 \times C_5 \cong C_{180} \times C_2$

*$C_8 \times C_3 \times C_3 \times C_5 \cong C_{120} \times C_3$

*$C_8 \times C_9 \times C_5 \cong C_{360}$

A: It is always better to write the prime factorization of the number given first. $$360=2^33^25^1$$
By using this factorization, we can list the abelian groups of order $360$ efficiently, that are
(i) $C_{2^3}\times C_{3^2}\times C_5 $
(ii) $C_{2^3}\times C_3\times C_3\times C_5$
(iii) $C_{2^2}\times C_2\times C_{3^2}\times C_5 $
(iv) $C_{2^2}\times C_2\times C_3\times C_3\times C_5$
(v) $C_2\times C_2\times C_2\times C_{3^2}\times C_5$
(vi) $C_2\times C_2\times C_2\times C_3\times C_3\times C_5$
You can check the number of abelian groups of order $360$ by calculating the product of number of partitions of number $3,2,1$ in the prime factorization. 
To write those abelian groups in the form that you desired, you can "group" those cyclic groups with different primes as follows:
$$C_2\times C_2\times C_2\times C_3\times C_3\times C_5 \cong (C_2\times C_3\times C_5)\times (C_2\times C_3)\times C_2\cong C_{30}\times C_6\times C_2$$
It would be a good practice for you to write the rest of those abelian groups in similar form.
