How to find all abelian groups of order 180? How would I find, up to isomorphism, a list of all abelian groups of order 180?  Would I simply use the fundamental theorem of finitely generated abelian groups by breaking 180 into prime factors, $180 = 2^2 3^2 5$?  
Also, how would I find a group in the list that had an element of order $18$?  I was thinking of using a theorem that states that if $G$ is abelian and $|G| = pk$, where $p$ is a prime, then $G$ has an element of order $p$.  However, $18$ is not a prime.
Edit: I now know that I can use the fundamental theorem of finitely generated abelian groups to the first part of my question, but the second part to my question still holds.
 A: There is a theorem that is seldomly mentionned in notebooks but is sometimes quite useful:Every finite Abelian group can uniquely be written as $\Bbb Z_{n_1} \oplus \ldots \oplus \Bbb Z_{n_k}$ where $n_1|\ldots|n_{k}$.
 In our case this gives us:


*

*$\Bbb Z_{180}$

*$\Bbb Z_2 \times \Bbb Z_{90}$ 

*$\Bbb Z_3 \times \Bbb Z_{60}$ 

*$\Bbb Z_6 \times \Bbb Z_{30}$ 

A: Since you know how to apply the the fundamental theorem of finitely generated abelian groups for the first part, I will only answer the second part.
For the second part of your question you need that one of the $\mathbb{Z}_{d_i}$ factors to be divisible by $9$. If you only have only one $d_i$ (in this case $d_1$), the only case you can have is $\mathbb{Z}_{180}$, so this case lets you a generated group of order 180, and you can consider the element $10$, that is of order $18$.
If you have $d_1$ and $d_2$, (that are the same that I told in the answer of your question before,) it must happen that $d_1=2$ and $d_2=2\cdot 9\cdot 5$ (because $d_1$ must divide $d_2$, and this is the only group of order $180$ that can have an element of order $18$.
The fundamental theorem of finitely generated abelian groups states that your group will descompose into $\mathbb{Z}_{d_1}\oplus\dots \mathbb{Z}_{d_m}$, where $d_i\mid d_{i+1}$ and 
$$d_1\cdot \dots d_m=\vert G \vert$$
So if you have that $5$ divides $d_1$, then divides $d_2$ too, so you must have that $25\mid 180$, but that's not true. 
If you have that $3\mid d_1$, then $d_1=3$ or $2\cdot3$, so $d_2=5\cdot 4\cdot 3$ or $3\cdot 5$, but this implies that your group has structure
$$\mathbb{Z}_{3}\oplus \mathbb{Z}_{60} \text{ or } \mathbb{Z}_{6}\oplus \mathbb{Z}_{15}$$
But is clear that any of them has an element of order $18$ since any of them has an element of order $9$ because of their structure.
