A strange representation of Z6

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 $M$ be a monoid, and let $M^*$ denote the subset of elements of $M$ which actually have an inverse. Then it is not hard to show that $M^*$ is a group: the identity is its own inverse and hence is in $M^*$; it is closed under the monoid operation since if $a$ and $b$ have inverses then so does $ab$ (namely, $b^{-1}a^{-1}$); and clearly the inverse of every element in $M^*$ is also in $M^*$, because being an inverse also implies having one.

Now let $M = \{a + b\sqrt{3} \mid 0 \leq a,b < 3\}$, where the operation is multiplication, but the coefficients $a$ and $b$ are reduced modulo 3. For example, $(2 + \sqrt 3)(2 + 2 \sqrt 3) = (4 + 6) + (4 + 2) \sqrt 3 = 1 + 0 \sqrt 3$. This does turn out to be associative, and is clearly commutative; and $1 = 1 + 0\sqrt 3$ is the identity. I wrote a little program to see which elements have inverses, and it turns out that the three elements with $a = 0$ 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 $\mathbb{Z}_6$. And, sure enough, $M^*$ turns out to be generated by $2 + \sqrt 3$ and $2 + 2 \sqrt 3$.