# Up to isomorphism of $360$, why is $\Bbb Z_2 \times \Bbb Z_4 \times \Bbb Z_3 \times \Bbb Z_3 \times \Bbb Z_5$ one of them?

To save time from writing it all out yourselves, here is what the key says. The following uses the theorem of Fundamental Theorem of Finitely Generated Abelian Groups. Q1: I have a question about the phrase "Find all abelian groups, up to isomorphisms of order $360$". So far the solution only lists integer groups, but it doesn't list every possible group in the world that are abelian, have order $360$, and are isomorphic to one (actually all) of the integer product groups listed below. Am I wrong?

Q2: Why is the trivial $\Bbb Z_{360}$ not listed? It does not satisfy Fundamental Theorem of Finitely Generated Abelian Groups, I know, but the question doesn't even say I need to use it. It just says I need to find "abelian groups of order 360 + isomorphism".

Q3: Relating to above, doesn't this show that $\Bbb Z_2 \times \Bbb Z_4 \times \Bbb Z_3 \times \Bbb Z_3 \times \Bbb Z_5$ should be removed from the list since $gcd(4,2) = 2$ and therefore it the whole product can't be isomorphic to $\Bbb Z_{360}$

## 3 Answers

Q1: The statement is that every abelian group of order 360 is isomorphic to exactly one of these six groups. Whether those six groups are what you call "integer groups" or not is not objectionable, as long as the statement is true. Whether the list contains every possible abelian group of order 360 in the entire universe is also irrelevant, as long as every such group is isomorphic to one of the six in the list.

Q2: $\mathbb{Z}_{360}$ is isomorphic to $\mathbb{Z}_8 \times \mathbb{Z}_9 \times \mathbb{Z}_5$, so it is not necessary to list it separately.

Q3: $\mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_5$ is not isomorphic to any of the other five groups in the list. You are correct that it is not isomorphic to $\mathbb{Z}_{360}$, which is one of the other five, namely number 6 in the list as I just said in Q2.

• So relating to Q1, I could have replaced one of those groups with something completely different, so as long they are isomorphic? For example, I could have replaced $\mathbb{Z}_8 \times \mathbb{Z}_9 \times \mathbb{Z}_5$ with $\Bbb Z_{360}$? – Hawk Oct 1 '14 at 17:18
• @sidht: That is correct. – Lee Mosher Oct 1 '14 at 17:42
• so shouldn't Q1 be reformulated by asking "find all abelian groups of the form $\prod Z_{i}$, up to isomorpshim of order $360$" instead of just asking a vague list? – Hawk Oct 1 '14 at 19:09
• Oops I forgot one on the list between 4 and 5 if you can fill it in then you have understood how it works. – Marc Bogaerts Oct 1 '14 at 19:23
• I don't know what exactly the theorem you mention says but it is a fact that the subscipts are each a divisor of the next.This identifies the group in a unique manner. – Marc Bogaerts Oct 1 '14 at 19:35

Not all abelian groups of order $360$ are isomorphic to $\mathbb Z_{360}$. You know that such groups are uniquely determined by their invariant factors/elementary divisors from the fundamental theorem. Your answers are being given in terms of the elementary divisors, so we take that approach.

Every (finitely generated) abelian group $G$ can be realised as $$G\cong \mathbb Z^r\times\mathbb Z_{p_1^{\alpha_1}}\times\cdots\times\mathbb Z_{p_n^{\alpha_n}}$$ where $r$ is the rank of the group, $p_i$ are prime, and $\alpha_i>0$ are integers. If $G$ is finite, then $r=0$, and we have $|G|=\prod_{i=1}^n p_i^{\alpha_i}$ just by counting. The $p_i$ are allowed to repeat in the above expression. This is what the fundamental theorem tells us. Once you understand that, then you'll be able to ask your questions.

Q1 I covered above; Q2 relies on the fact that $\mathbb Z_n\times \mathbb Z_m\cong \mathbb Z_{n\cdot m}$ when $(n,m)=1$, i.e. the Chinese Remainder Theorem; Q3 I covered above as well.

The groups in the list can be (uniquely) written as :
$\Bbb{Z}_2\times \Bbb{Z}_6\times \Bbb{Z}_{30}$
$\Bbb{Z}_6\times \Bbb{Z}_{60}$
$\Bbb{Z}_2\times \Bbb{Z}_2\times \Bbb{Z}_{90}$
$\Bbb{Z}_2\times \Bbb{Z}_{180}$
$\Bbb{Z}_{360}$
Now look how one can obtain these from $360=2^33^25$
1: $360=2\times2.3\times2.3.5$
2: $360=2.3\times2.2.3.5$
3: $360=2\times2\times2.3.3.5$
4: $360=3\times2.2.2.3.5$
5: $360=2.3.2.2.3.5$
Each subscript is a divisor of the next.