# All Isomorphic Classes of Abelian Groups of Order $n$

I know that each finite abelian group is isomorphic to a direct product of cyclic groups of prime orders $> 1$. This means taking a finite abelian group of order $n$, I can find the prime factorization of $n$, and the group will be isomorphic to this direct product. Are these the isomorphic classes of abelian groups of order $n$?

For example, for $n = 200$, I know the prime factorization is $2\times2\times2\times5\times5$. How would I know how many isomorphic classes of abelian groups of order $200$ there are? Am I looking for cyclic subgroups of the abelian group of order $200$, with order $2$ or order $5$?

Thanks.

EDIT: I think I figured out where my misunderstanding of the fundamental theorem of abelian groups was. Basically, since $200 = 2^3 \cdot 5^2$, I am looking for all abelian groups of order $8$ and all abelian groups of order $25$, and then multiply the two numbers together to get the number of abelian groups isomorphic to order $200$.

• Hey I think you are either misunderstanding or misstating the Fundamental Theorem of Abelian Groups. From what I read it looks like you aren't thinking about $\mathbb{Z}_{p^n}$ as a possibility in a direct product only $\mathbb{Z}_p$ – user171177 Oct 29 '14 at 0:55
• Hmm I see what you mean. Could you explain how you would find the isomorphic classes in this case then? – jstnchng Oct 29 '14 at 0:58
• So the prime factorization of n is 2x2x2x5x5. Am I looking for groups of order 8 or order 25? – jstnchng Oct 29 '14 at 1:00
• Yes you are looking for groups of those orders (and then also the different ways to choose one from each of those sets) – user171177 Oct 29 '14 at 1:12

So the fundamental theorem of abelian groups says that if $G$ is a finite abelian group then $G\cong \Pi \mathbb{Z}_{p_i^{n_i}}$ with $\vert G \vert = \Pi p_i^{n_i}$ where the product is taken over the index $i$. Overall it is clear that we need to have $p$ show up as many times as it does in the prime factorization of the size of $G$. As a more concrete example all of the abelian groups of order 4 are $\mathbb{Z}_{4}$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$. So in your case you are looking for the different ways to split 2 and 5 up so that the total number of them appearing is 3. The easiest 2 ways would be $\mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_5 \times\mathbb{Z}_5$ and $\mathbb{Z}_8 \times \mathbb{Z}_{25}$ thought this isn't an exhaustive list.