[FOM] New Proof of Fundamental Theorem of Arithmetic

joeshipman@aol.com joeshipman at aol.com
Thu Sep 24 17:20:26 EDT 2009

Here's a simpler proof that avoids quoting Jordan-Holder:

Let n=ab be divisible by a prime p. Consider the cyclic group Z_n of 
order n and its natural subgroups Z_p (of order p, generated by n/p) 
and Z_a (of order a, generated by b=n/a). Suppose p does not divide a. 
Then Z_p is not a subgroup of  Z_a, and since Z_p has no nontrivial 
proper subgroups the intersection of Z_p and Z_a is the trivial group. 
Therefore the internal direct product Z_p X Z_a has order pa, and so pa 
divides n=ab, so p divides b. We've just proved Euclid's lemma (if a 
prime divides a product it divides one of the factors) and the 
Fundamental Theorem of Arithmetic follows in the usual way, without 
having used GCDs or division with remainder.

-- JS

-----Original Message-----
From: joeshipman at aol.com

The way you can tell it is a truly different proof is that it doesn't 
apply to other Euclidean Domains like the GCD proof does...The proof 
only works for Z because the primes in Z do double duty as 
cardinalities of sets so their irreducibility implies that the 
corresponding groups have no proper subgroups and so are aleady simple. 
-- JS 

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