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 \(2 + 2 \sqrt 3\).