# [FOM] New Proof of Fundamental Theorem of Arithmetic

joeshipman@aol.com joeshipman at aol.com
Sun Sep 13 14:03:09 EDT 2009

```In the proof posted below, I left out a step -- the reason the
induction applies is that a quotient of a cyclic group must also be
cyclic (because quotients are generated by the image of any generating
set so can't require more generators than the original group). -- JS

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

Suppose n=p1* p2*...*p_k. The group Zn of integers mod n has an
additive subgroup {0, p2*p3*...*p_k, 2*p2*p3*...*p_k, 3*p2*p3*...*p_k,
..., (p1-1)*p2*p3*...*p_k} which has order p1; since Zn is abelian this
a normal subgroup and since p1 is prime it is a simple subgroup;
therefore there is a composition series for Zn which includes this
group, and by induction there is one which involves k groups of order
p1, p2, ..., p_k; by the Jordan-Holder theorem any composition series
for Zn involves the same groups up to order of factors so any other
prime factorization of n is the same up to order of factors.

Since this proof does not use division with remainder or GCD, and since
the proof of the Jordan-Holder theorem is completely abstract and
involves only the group operation, and doesn't depend on any properties
of rings which have two operations, should this be counted as
essentially different from the usual proofs?

-- JS

```