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 .

## About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

Well, sure, what’s so strange about that? For any p, that is isomorphic to the group of invertible elements of Z(p^2), which is cyclic of order p(p-1).

Prime p of course.

Well, strange is in the eye of the beholder, of course! Could you elaborate on your comment a bit? What is Z(p ^ 2)?

I mean the integers modulo p^2. Your multiplication law is (a,b)(c,d)=(ac,ad+bc). Note that for a prime p, (a+bp)(c+dp)=ac+(ad+bd) mod p^2. So your group is just the group of invertible elements of the integers modulo p^2. That is a cyclic group of order p(p-1). See, for example:

Weisstein, Eric W. “Modulo Multiplication Group.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/ModuloMultiplicationGroup.html

Sorry. (a+bp)(c+dp)=ac+(ad+bc)p mod p^2

Aha, now I see, thanks!

So if I’m understanding this correctly, the example I picked is special, because the number under the square root and the modulus are the same. If you consider invertible modulo , then you always get something isomorphic to . But you can also do the same construction on modulo where (they do not even have to be prime). Call the resulting group . Then, for example, after some fiddling with GAP I have determined that and are both cyclic (of order 24 and 48, respectively), but is not cyclic.

Correct. Not only that, you can generalize the square root to a linear combination of elements of an algebraic field extension over the rational numbers. The set of all such integral linear combinations for a given set of extension elements is a module over Z. If the module contains a set of linearly independent elements equal to the degree of the field extension, it is called a full module, and if it forms a ring and contains the number 1 then it is called an order of the extension. The group of units of an order is cyclic. That is actually a geometric fact; it is related to the theory of lattices. See, for example, chapter 2, “Representation of Numbers by Decomposable Forms,” in Borevich and Shafarevich, “Number Theory”. Your non-cyclic examples are non-full modules.

The modulus part of your version is a bit of a red herring in this setting. First determine the structure of the full group of units, which is a finitely-generated abelian group. In general, it will have some infinite components and some finite components. Then apply your modulus, with the obvious effects on the components.

Your ring turns out to be isomorphic to the ring of *dual numbers* over the field with three elements:

F_3[sqrt(3)] = F_3[X]/(X^2 – 3) = (Z/(3))[X]/(X^2 – 3) = (Z[X]/(3))/(X^2 – 3) = Z[X]/(3, X^2 – 3) = Z[X]/(3, X^2) = … = F_3[X]/(X^2) = F_3[eps]/(eps^2).

Here “A/I” means the factor ring of a ring A modulo the ideal I. “(3)” means the ideal generated by 3.

I came here to drop in more or less the same comment that Ingo made.

The dual number thing clicks when you see that you have numbers of the form a + bi, with i^2 = 0, but i /= 0.

Ah, yes, that makes sense, thanks! And now I understand Ingo’s comment too.