Monstrous Moonshine

Question

Wikipedia claims that the group of units of Z24 (1,5,7,11,13,17,19,23), which all have order 2, and are isomorphic to (Z/2Z)^3 have an important connection to Monstrous Moonshine theory, however, I cannot find any other reference besides Wikipedia that claims this --- It was recommended on sci.math that I pose this question here.

Perhaps it's a mistake? And he meant that the primes of the Monster, which continue to 71, are what are considered in Moonshine.

Paul Hjelmstad, B.M, B.A.

[Edit (PLC): Here is the relevant passage from wikipedia:]

24 is the highest number $n$ with the property that every element of the group of units $(\mathbb{Z}/n\mathbb{Z})^{\times}$ of the commutative ring $\mathbb{Z}/n\mathbb{Z}$, apart from the identity element, has order $2$; thus the multiplicative group $(\mathbb{Z}/24\mathbb{Z})^{\times} = \{1,5,7,11,13,17,19,23\}$ is isomorphic to the additive group $(\mathbb{Z}/2\mathbb{Z})^3$. This fact plays a role in monstrous moonshine.

Answer 1

$\newcommand{\Q}{\mathbf Q} \newcommand{\Z}{\mathbf Z}$

I don't know about monstrous moonshine, but $(\Z/24\Z)^\times$ is the group of automorphisms of the maximal elementary abelian $2$-extension $\Q_2\left(\root2\of{\Q_2^\times}\right)=\Q_2(\root2\of5, \root2\of3, \root2\of2)=\Q_2(\zeta_{24})$ of $\Q_2$. See for example Lemma 8 of Lecture 19 of my course on Local arithmetic.

Answer 2

I think the claim goes back to the 1979 paper "Monstrous Moonshine" by Conway and Norton, where they discuss the "defining property of 24": If $n$ is a positive integer such that $xy \equiv 1$ mod $n$ implies $x \equiv y$, then $n|24$. This fact is used in Atkin's determination of the normalizer of $\Gamma_0(N)$ in $SL_2(\mathbb{R})$. The number 24 plays a special role here, in the sense that the normalizer is $\Gamma_0(n|h)+$, where:

1. $h$ is the largest divisor of 24 such that $h^2|N$
2. $n = N/h$
3. $\Gamma_0(n|h) = \left\{ \begin{pmatrix} a & b/h \\ cN & d \end{pmatrix} \mid a,b,c,d \in \mathbb{Z}, ad-bcn/h = 1 \right\}$. The notation is meant to suggest that the group is conjugate to $\Gamma_0(n/h)$, and contains $\Gamma_0(nh)$.
4. The $+$ means we adjoin all possible Atkin-Lehner involutions. If $n/h$ has $k$ prime factors, then this extends $\Gamma_0(n|h)$ by an elementary abelian 2-group of rank $k$.

The connection between your observation and moonshine does not seem particularly strong to me, but that may be because I was too young to have experienced firsthand the heady days of numerical exploration. It involves the normalizers of $\Gamma_0(N)$ in the following way: There is a graded representation $V^\natural = \bigoplus V^\natural_m$ of the monster, such that for each element $g$ of the monster, the McKay-Thompson series $T_g(\tau) = \sum_{m \geq -1} Tr(g|V^\natural_m)q^m$ is the $q$-expansion of a modular function invariant under some genus zero group $\Gamma$ that contains and normalizes some $\Gamma_0(N)$, and therefore lies in some $\Gamma_0(n|h)+$ , where $n = |g|$. This fact was essentially the main conjecture in the Conway-Norton paper, although the paper enhances this claim with an explicit list of the candidate functions and their invariance groups. My understanding of the solution process is:

1. Atkin, Fong, and Smith gave a computational proof of existence (1980).
2. Frenkel, Lepowsky, and Meurman constructed a candidate representation $V^\natural$ (1984), and showed that it had a vertex operator algebra structure (1988).
3. Borcherds proved that the candidate representation was satisfactory (1992).

Wikipedia and sundry expository books by Gannon, du Sautoy, Ronan, and others can say more about the precise history than I can. I should mention that the number 24 is important as the central charge of $V^\natural$ in Borcherds's solution to the Monstrous Moonshine conjecture, but the paper does not make explicit use of the number 24 in the "group of units" role. One might reasonably argue (through a somewhat convoluted path) that these are the same 24, though.

There may be other connections, but I am unaware of them.