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)

When I first looked at this, I was confused, because nowhere in the definition of a group does it say that each element must have only one inverse, but you assumed it to be true.  I didn't think it was immediately obvious, so here's a proof.

To be proved: For every elements a and b in the group, there exists an element c in the group such that ac = b, and an element d in the group such that da = b.

Proof:
Let a' be an inverse of a.
(aa')b = a(a'b)
b = a(a'b)
a'b is the element c that we needed
Similarly, ba' is the element d that we need.

It follows that if b != c, then ab != ac and ba != ca, and therefore, no element has two inverses.

Posted by Tristan on 2006-09-19 14:59:20