Fundamental theorem of arithmetic from Jordan-Hölder theorem

[eoghan]

TL;DR Summary

I propose a proof of the fundamental theorem of arithmetic based on the Jordan-Hölder theorem. I would like to know if the proof makes sense.

I was intrigued by a comment in Brilliant.org:

In fact, it is not hard to show that the fundamental theorem of arithmetic follows from Jordan–Hölder

Besides the proof provided by Brilliant, I also found a couple of other websites. But none of these proofs were entirely clear to me. So I tried to come up with my own proof. Since I am not a group theorist, I wanted to ask if the proof makes sense, or if there are some mistakes.

Let $n$ be a positive integer. Let's consider $\mathbb{Z}_n$, I know that
the quotient $\mathbb{Z}_n/\mathbb{Z}_d$ is defined and is simple if and only if $n/d=p$ with $p$ prime.

This allows us to create a composition series for $\mathbb{Z}_n$. Let's take $p_1$ be a prime number dividing $n$ and consider $\mathbb{Z}_\frac{n}{p_1}$. Because of what we said, $\mathbb{Z}_n/\mathbb{Z}_\frac{n}{p1}$ is simple. Let's now take another prime $p_2$ that divides $\frac{n}{p_1}$ and consider $\mathbb{Z}_\frac{n}{p_1p_2}$. Then $\mathbb{Z}_\frac{n}{p_1}/\mathbb{Z}_\frac{n}{p_1p_2}=p_2$ and is therefore simple. We can then build
$$
\mathbb{Z}_n \triangleright \mathbb{Z}_\frac{n}{p_1} \triangleright \mathbb{Z}_\frac{n}{p_1p_2}
$$
We can continue this process until $\frac{n}{p_1...p_k}$ is prime. Then we take as next subset $\mathbb{Z}_1=\{0\}$ and we have $\mathbb{Z}_\frac{n}{p_1...p_k}/\mathbb{Z}_1=\frac{n}{p_1...p_k}$ prime. So we found the composition series

$$
\mathbb{Z}_n \triangleright \mathbb{Z}_\frac{n}{p_1} \triangleright \mathbb{Z}_\frac{n}{p_1p_2}\triangleright...\triangleright \mathbb{Z}_1 = \{0\}
$$

But since $n=p_1...p_k$, I can rewrite the composition series as

$$
\{0\} \triangleleft \mathbb{Z}_{p_1} \triangleleft \mathbb{Z}_{p_1p_2}\triangleleft...\triangleleft \mathbb{Z}_{p_1...p_k}=\mathbb{Z}_n
$$

I could have also chosen another decomposition of $n$ to obtain the series
$$
\{0\} \triangleleft \mathbb{Z}_{q_1} \triangleleft \mathbb{Z}_{q_1q_2}\triangleleft...\triangleleft \mathbb{Z}_{q1...q_l}=\mathbb{Z}_n
$$

By the Jordan-Hölder theorem, the two series must be equivalent, i.e. the set
$\{p_1,...,p_k\}$ must be isomorphic to the set $\{q_1,...,q_l\}$ and it also proves that $n$ can be decomposed into prime factors. Therefore, this proves the fundamental theorem of arithmetic.


Do you think this proof is correct, or am I overlooking something?

 

[fresh_42]

Looks ok.

I have some minor remarks:
1.) $\mathbb{Z}_n / \mathbb{Z}_d$ is only defined if $d\,|\,n,$ not always.
2.) Where do you take the existence of $p_1$ from?
3.) Why does a composition series exist? Not that you have hidden a circular reasoning here!
4.) $\{p_1,\ldots,p_k\} = \{q_1,\ldots,q_l\}$ (up to $\pm 1$) not isomorphic.

 

[eoghan]

Thank you for checking.

1. Correct, $d$ must divide $n$. I need to specify this better.
2. Yes, this needs an explanation. I was simply supposing that if $n$ is not prime, then there exists a prime number that divides $n$. But this is a consequence of the FTOA, that I am trying to prove here. What I am doing here is just proving the uniqueness of the factorisation. I need to think if there is a group theory tool to argue that $p_1$ exists.
3. If we assume that a non prime number is divisible by a prime number, see point above, then the composition series exists because during the proof I build it (i.e, take one prime number $p_1$ that divides $n$ and consider $\mathbb{Z}_{n/p_1}\triangleleft \mathbb{Z}_n$ and so on).
4. I am not sure I understand what you mean here. I guess with (up to $\pm 1$) you mean that the sets do not contain $\pm 1$, because $1$ is not prime. But why are they not isomorphic?