## 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 $2 + 2 \sqrt 3$.

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 11 Responses to A strange representation of Z6

1. Yitz says:

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).

• Yitz says:

Prime p of course.

• Brent says:

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

• Yitz says:

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

• Yitz says:

• Brent says:

Aha, now I see, thanks!

• Brent says:

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 $a + b \sqrt p$ modulo $p$, then you always get something isomorphic to $\mathbb{Z}(p^2)$. But you can also do the same construction on $a + b \sqrt{p}$ modulo $q$ where $p \neq q$ (they do not even have to be prime). Call the resulting group $G(p,q)$. Then, for example, after some fiddling with GAP I have determined that $G(3,5)$ and $G(3,7)$ are both cyclic (of order 24 and 48, respectively), but $G(3,11)$ is not cyclic.

• Yitz says:

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.

2. Ingo Blechschmidt says:

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.

3. Edward Kmett says:

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.

• Brent says:

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