On my other blog I am writing about a proof of the Lucas-Lehmer test, and today in the course of working up some examples I stumbled across this little gem.
Let be a monoid, and let denote the subset of elements of which actually have an inverse. Then it is not hard to show that is a group: the identity is its own inverse and hence is in ; it is closed under the monoid operation since if and have inverses then so does (namely, ); and clearly the inverse of every element in is also in , because being an inverse also implies having one.
Now let , where the operation is multiplication, but the coefficients and are reduced modulo 3. For example, . This does turn out to be associative, and is clearly commutative; and is the identity. I wrote a little program to see which elements have inverses, and it turns out that the three elements with do not, but the other six do. So this is an Abelian group of order 6; but there’s only one such group, namely, the cyclic group . And, sure enough, turns out to be generated by and .