Find the complete list of abelian groups of order $n$, up to isomorphism

Question

Here is the question:

Find the complete list of abelian groups of order $n$, up to isomorphism

I don't understand what this is asking me. I've checked online, and I understand it's about the factors of $n$; however, I'm confused on what an isomorphic abelian group of order $n$ means.

For example the question is asking about $\Bbb Z_{30},$ which can be broken down to $\Bbb Z_2, \Bbb Z_3, \Bbb Z_5$ and their products. What properties of the group am I looking for so that it satisfies being both isomorphism and abelian?

Answer 1

For example, for $n=8,\,$ $\mathbb Z_2\times \mathbb Z_4 $ is isomorphic to $\mathbb Z_4\times \mathbb Z_2$,

so the question is asking you not to list those separately.

On the other hand, those are not isomorphic to $\mathbb Z_{8}$, so $\mathbb Z_{8}$ should be listed separately.

For $n=30$, there is only one Abelian group, up to isomorphism.

Answer 2

There are as many finitely many abelian groups of order $n$, but you you need to compute only those which are isormorphic.

Definition: Let $m_1, m_2, \dots, m_n$ be relatively prime positive integers then $$\mathbb{Z}_{m_1} \cdot \mathbb{Z}_{m_2} \cdot \mathbb{Z}_{m_3} \cdots \mathbb{Z}_{m_n} \cong \mathbb{Z}_{m_1m_2m_3 \cdots m_n}$$

for $n=30$ there is only abelian group under upto isomorphism.

Answer 3

Take $n=32=2^5$, a slightly non-trivial example.

For each partition of the exponent $5$, we get a different abelian group.

It works thusly:

the partitions are $1+1+1+1+1,2+1+1+1,3+1+1,4+1,5,2+3,2+2+1$.

There's $7$.

Then we get $$\Bbb Z_2×\Bbb Z_2×\Bbb Z_2×\Bbb Z_2×\Bbb Z_2\\ \Bbb Z_4×\Bbb Z_2×\Bbb Z_2×\Bbb Z_2\\\Bbb Z_8×\Bbb Z_2×\Bbb Z_2\\\Bbb Z_{16}×\Bbb Z_2\\\Bbb Z_{32}\\\Bbb Z_4×\Bbb Z_8\\\Bbb Z_4×\Bbb Z_4×\Bbb Z_2$$

This idea completely generalizes, and all we are using is the structure theorem (and,CRT, if you want).