# Why are there only 5 abelian groups of order 48?

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?

• I find this resource on the OEIS wiki quite helpful Commented Nov 29, 2022 at 2:51
• Thanks, but this resource gives me an answer, not an explanation.
– IAAW
Commented Nov 29, 2022 at 2:55
• 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. Commented Nov 29, 2022 at 2:58
• Is theorem 51 the structure theorem? You might want to see a few examples of how it's used Commented Nov 29, 2022 at 2:59
• Commented Nov 29, 2022 at 3:04

$$48=2^4×3$$. 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.

• 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.
– IAAW
Commented Nov 29, 2022 at 4:50
• @iateawalrus -- Yes, that's right. $\mathbb{Z} /8 \mathbb{Z} \times \mathbb{Z} /2 \mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z} \ \cong \ \mathbb{Z}/ 24 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} \ \cong \ \mathbb{Z} / 8 \mathbb{Z} \times \mathbb{Z} / 6 \mathbb{Z}$. Commented Nov 29, 2022 at 5:09