[FOM] New Proof of Fundamental Theorem of Arithmetic
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.
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.
More information about the FOM