[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 

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