Here is a simple problem from abstract algebra.

Prove that a group with exactly five elements is commutative.

(In reply to Counting elements by vswitchs)

My earlier post overlooked that a or b may be their own inverses.  If this is the case only for one of them, we still have at least six elements by my previous arguments. If both equal their inverses, I show that the products aba and bab will be yet different elements:

Take a b a.  It cannot equal e, as then both ab and ba would equal a_ (the inverse of a).  It cannot equal a, because then ba would equal e (and therefore ab, see my last post). And it cannot equal b, because then (by multiplication with a) ab=ba.  It also cannot equal ab or ba because then a would be the unit element.

Likewise for b a b.  So this again gives us at least six elements (six because aba might equal bab).

Posted by vswitchs on 2006-09-18 13:28:40