Here is a simple problem from abstract algebra.

Prove that a group with exactly five elements is commutative.

Let's assume that the group is not commutative and count elements:

We have to have a neutral unit element e. Besides we have to have two elements a and b which do not commute:

a b != b a

Now as they do not commute, a cannot equal b.  Also, a cannot equal the inverse of b, since every element commutes with its inverse, even in non-commutative groups.  For the same reason, neither a nor b can be the unit element.

This gives us already five different elements:

e, a, b, a_, b_

(a_ and b_ are my notation for the inverses of a and b.)
If our group is to have only five elements in total, ab and ba have to equal one (a different one) of the five above.  By the above arguments, they cannot equal e, a or b.  So let's assume ab=a_.  Multiplying with a from the right and then with a_ from the left (left and right multiplication being different in non-commutative groups), we get:

aba = e
ba = a_

So ba would equal ab, contrary to our starting-point.  The case ab=b_ can be disproved analogously.  So ab and ba equal neither of our elements above, and the group has at least seven different elements.

Posted by vswitchs on 2006-09-18 13:00:05