I Fundamental theorem of arithmetic from Jordan-Hölder theorem

eoghan
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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?
 
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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.
 
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?
 
eoghan said:
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.
You could take an element of lowest highest order. That also proves ##p_1\,|\,n## (Lagrange). Since primality and irreducibility are exchangeable, this should prove primality.
eoghan said:
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).
Yes, seems convincing. A note that the group orders decline with each step without repetitions would help to guarantee that it comes to an end. ##(\mathbb{Z},+) ## has no composition series.
eoghan said:
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?
I mean that every statement about prime numbers is always up to units, and the units in ##\mathbb{Z}## are ##\pm 1.## ##-5## is as prime as ##5## is. Mention that we only consider positive numbers and done. But I mainly meant that sets are equal or not, containing each other or not. They are not isomorphic since there is no "morph" on sets.
 
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It makes sense!

Thank you for checking it out and for the hints :)
 
eoghan said:
It makes sense!

Thank you for checking it out and for the hints :)
I had to correct myself: take an element of highest order, not lowest for irreducibility. High order means low prime, I confused them.
 
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