The question in question

I am trying to understand exactly how this works. I am following this procedure up until it shows the second equivalences. For example, why is $\mathbb{Z}_{2^3}\times \mathbb{Z}_{2} \times \mathbb{Z}_{3} \cong \mathbb{Z}_2 \times \mathbb{Z}_{24}$? Furthermore, why isn't it isomorphic to $\mathbb{Z}_{48}$, or any of the other products on the RHS of the second equivalence?

  • $\begingroup$ I find this resource on the OEIS wiki quite helpful $\endgroup$
    – Graviton
    Commented Nov 29, 2022 at 2:51
  • $\begingroup$ Thanks, but this resource gives me an answer, not an explanation. $\endgroup$
    – IAAW
    Commented Nov 29, 2022 at 2:55
  • 1
    $\begingroup$ That is true, as I realize now your question is more about why those groups are isomorphic. In that case, a better resource would probably be this Wikipedia article on Direct Products. It doesn't give any specific examples, but the discourse on isomorphisms throughout might yield better understanding. $\endgroup$
    – Graviton
    Commented Nov 29, 2022 at 2:58
  • $\begingroup$ Is theorem 51 the structure theorem? You might want to see a few examples of how it's used $\endgroup$ Commented Nov 29, 2022 at 2:59
  • 2
    $\begingroup$ This may help: Product of two cyclic groups is cyclc iff their orders are co-prime $\endgroup$ Commented Nov 29, 2022 at 3:04

3 Answers 3


$48=2^4×3$. The number of partitions of $4$ is $5$. They're $$1+1+1+1, 1+1+2,2+2,1+3,4$$.

Now apply the structure theorem. It implies (along with the Chinese remainder theorem) the general result that if $n$ has prime factorization $p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, then the number of abelian groups of order $n$ is $p(a_1)p(a_2)\cdots p(a_k)$, where $p(x)$ is the number of partitions of $x$.


I have to assume that "Theorem 51" is the classification of finite abelian groups. If we recall, this says that

Every finite abelian group can be written as a direct sum of cyclic groups of prime power order

$$ \bigoplus_\alpha \mathbb{Z} \big / p_\alpha^{k_\alpha} \mathbb{Z}$$

Moreover, this decomposition is unique up to (noncanonical) isomorphism

So, let's start small. If we want to understand all abelian groups of size $12$, we can factor $12 = 2^2 \cdot 3$. Now there are two ways to get all of these prime powers:

  • $\mathbb{Z} / 2 \mathbb{Z} \oplus \mathbb{Z} / 2 \mathbb{Z} \oplus \mathbb{Z} / 3 \mathbb{Z}$
  • $\mathbb{Z} / 2^2 \mathbb{Z} \oplus \mathbb{Z} / 3 \mathbb{Z}$

The "uniqueness" clause of the classification tells us that these decompositions are not isomorphic. Of course, we can also see this directly: the latter has an element of order $4$ while the former doesn't. So they cannot be isomorphic!

Let's see another example. What are the possible abelian groups of order $2^3 \cdot 3^2 \cdot 7$?

Well, we have to have $3$ copies of $\mathbb{Z}/2\mathbb{Z}$ somehow. We can do this as either

  • $\mathbb{Z} / 2\mathbb{Z} \oplus \mathbb{Z} / 2 \mathbb{Z} \oplus \mathbb{Z} / 2 \mathbb{Z}$
  • $\mathbb{Z} / 2 \mathbb{Z} \oplus \mathbb{Z} / 4 \mathbb{Z}$
  • $\mathbb{Z} / 8 \mathbb{Z}$

(do you see why?)

Then we have to have $2$ copies of $\mathbb{Z} / 3 \mathbb{Z}$. We can do this is either

  • $\mathbb{Z} / 3 \mathbb{Z} \oplus \mathbb{Z} / 3 \mathbb{Z}$
  • $\mathbb{Z} / 9 \mathbb{Z}$

Lastly, we have to have $1$ copy of $\mathbb{Z} / 7 \mathbb{Z}$, and there's only one way to do this

  • $\mathbb{Z} / 7 \mathbb{Z}$

All in all, this gives us $6$ groups of this order (again, do you see why?)

In general, then, how do we figure out all the possible abelian groups of order $n$?

  1. First we factor $n$ into prime powers.
  2. Then we list all the ways of partitioning each prime exponent into a sum of integer pieces.
  3. Lastly we have to choose one of these partitions for each prime, which completes the classification.

As for your followup question, "why is $\mathbb{Z} / 8 \mathbb{Z} \oplus \mathbb{Z} / 3 \mathbb{Z} \cong \mathbb{Z} / 24 \mathbb{Z}$". The answer is the Chinese Remainder Theorem.

I hope this helps ^_^


The fact you need is the Chinese Remainder Theorem: if $(m,n)=1$, then $\mathbb{Z}/mn\mathbb{Z}\cong \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$. Here's a quick proof: we have a map $\pi: \mathbb{Z}\rightarrow \mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$ given by $\pi(a)=(\bar{a}, \bar{b})$. Clearly $mn\mathbb{Z}\subset \ker\pi$. Conversely, if $x\in\ker\pi$, then $x$ is divisible by both $m$ and $n$. Since $m$ and $n$ are coprime, $x$ must be divisible by their product $mn$. Thus $mn\mathbb{Z}=\ker\pi$. Now it suffices to show that $\pi$ is surjective. Let $(\bar{r},\bar{s})\in \mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$. Since $m$ and $n$ are coprime, we can use Bezout's identity to find $p,q\in\mathbb{Z}$ so that $pm+qn=1$.Then $\pi(spm+rqn)=(\bar{r},\bar{s})$.

We can use the Chinese Remainder Theorem to establish some of these isomorphisms you're seeing. For instance $\mathbb{Z}/8\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/24\mathbb{Z}$ since 8 and 3 are coprime.

  • $\begingroup$ In this case, can we say that $\mathbb{Z}/8\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/8\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$, since 2 and 3 are coprime? Sorry, I'm really trying to grasp this. $\endgroup$
    – IAAW
    Commented Nov 29, 2022 at 4:50
  • 1
    $\begingroup$ @iateawalrus -- Yes, that's right. $\mathbb{Z} /8 \mathbb{Z} \times \mathbb{Z} /2 \mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z} \ \cong \ \mathbb{Z}/ 24 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} \ \cong \ \mathbb{Z} / 8 \mathbb{Z} \times \mathbb{Z} / 6 \mathbb{Z}$. $\endgroup$ Commented Nov 29, 2022 at 5:09

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