[FOM] New Proof of Fundamental Theorem of Arithmetic

joeshipman@aol.com joeshipman at aol.com
Wed Sep 16 19:53:51 EDT 2009

Because since it has NO proper subgroups it has no proper subgroups and 
therefore no proper normal subgroups. This depends on the theorem that 
the order of a subgroup divides the order of the group, but that 
doesn't require GCD, you get it from a coset argument; and it depends 
on p1 having no proper divisors but that is a definition and doesn't 
require proof.

-----Original Message-----
From: Franklin Vera Pacheco <franklin.vp at gmail.com>

> a normal subgroup and since p1 is prime it is a simple subgroup;

How do you prove this part without Euclid's division with remainder or 

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