# 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.

• You asked a question equal to this one, but using $p^5$ instead of $180$. math.stackexchange.com/questions/2022131/… – MonsieurGalois Nov 20 '16 at 3:49
• I asked this question before I realized the theorem also applies to a product of prime factors. Let me fix this question. – MomoTheSir Nov 20 '16 at 3:52

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:

1. $\Bbb Z_{180}$
2. $\Bbb Z_2 \times \Bbb Z_{90}$
3. $\Bbb Z_3 \times \Bbb Z_{60}$
4. $\Bbb Z_6 \times \Bbb Z_{30}$

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.

• Do you mean that the order of $\mathbb{Z}_{d_i}$ be divisible by 9? Also, why must it be divisible by 9 to have an element of order 18? – MomoTheSir Nov 20 '16 at 4:12
• Because the elements of the group are determined by the structure of the group. If you only have a sum of subgroups of order not divisible by $9$, then it can't happen to you to have an element of order $9$. I tried to explain in the answer again. – MonsieurGalois Nov 20 '16 at 4:20
• @MonsieurGalois, how about the group $Z_4\oplus Z_{45} = Z_4 \oplus Z_9 \oplus Z_5$ and the element $(2, 1, 0)$? :) – Sasha Mayer Nov 20 '16 at 4:30
• @SashaMayer I forgot that case, but, that's the case when $d_1=180$ – MonsieurGalois Nov 20 '16 at 4:32
• @MonsieurGalois, remaining group with such element is $V_4\oplus Z_{45}$, btw – Sasha Mayer Nov 20 '16 at 4:33