Modulo multiplication of cyclic groups that are divisors of 24

Question

I recently came across this reference in Wolfram Mathword regarding the modulo multiplication of cyclic groups ($M_n$)

The only ordered n for which the elements of $M_n$ are all self-conjugate are the divisors of $24: 1, 2, 3, 4, 6, 8, 12, 24 (OEIS > A018253; Eggar 2000)$. These correspond to the groups$ <e>, C_2, > C_2×C_2,$ and $C_2×C_2×C_2.$ This also means that no modulo multiplication group is isomorphic to a direct product of more than three copies of $C_2$.

The Eggar (2000) result is this:

"The divisors of 24 greater than 1 are the only positive integers n with the property $m^2 = 1 (mod n)$ for all integer m coprime to n."

Q1: What does self-congancy $m^2=1$ actually mean in this context? I get the math, for example in

$$Z_{12}^* = {1,5,7,11}$$

the square of all these numbers modulo-12 is $1$, but what implications or deductions follows from this?

Q2: More specifically, how does this self-conjugancy apply to models or theories built from cyclic groups? Naively, I'm thinking that any theory that needs inverses would be restricted by this since if the generators don't invert, then not all elements in the group invert.

Q3: More specific yet (or I may just be repeating myself), how should we view the limit $C_2×C_2×C_2$ and what deductions or implications can we make from this?

Any insights welcome and thank you in advance.

Answer 1

If the bounty description means you'd like to know why the "self-inverse" condition (or what I will call the "involutory" condition) forces $m$ to be a divisor of $24$, then that question is easy to answer.

The Chinese remainder theorem implies that if $m = p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k}$ is the prime factorization of $m$, then we have a canonical ring isomorphism

$$\mathbb{Z}/(m) \cong \mathbb{Z}/(p_1^{e_1}) \times \mathbb{Z}/(p_2^{e_2}) \times \ldots \times \mathbb{Z}/(p_k^{e_k})$$

which immediately implies an isomorphism of unit groups (groups of invertible elements)

$$(\mathbb{Z}/(m))^\ast \cong (\mathbb{Z}/(p_1^{e_1}))^\ast \times (\mathbb{Z}/(p_2^{e_2}))^\ast \times \ldots \times (\mathbb{Z}/(p_k^{e_k}))^\ast.$$

So the question boils down to: when is the group $(\mathbb{Z}/(p^e))^\ast$ ($p$ a prime) isomorphic to a product of copies of $C_2$?

There is a unique ring homomorphism $\phi: \mathbb{Z}/(p^e) \to \mathbb{Z}/(p)$; this map is surjective. It is well-known that the unit group $(\mathbb{Z}/(p))^\ast$ is cyclic of order $p-1$. (In fact any finite multiplicative group in a field is cyclic.) If $x \in \mathbb{Z}/(p^e)$ is any element such that $\phi(x)$ is a generator of $(\mathbb{Z}/(p))^\ast$, then $x$ is invertible and $p-1$ divides the order of $x$ -- and therefore the order of $x$ can't be $2$ if the prime $p$ is greater than $3$. Hence the only primes that can divide $m$ under the involutory condition are $2$ and $3$.

Let's now consider the case $p = 3$; what can the exponent $e$ be? Well, $4$ is invertible modulo $3^e$, and we can't have $4^2 \equiv 1 \; \text{mod}\; 3^e$ unless $3^e$ divides $15$. So the involutory condition would force $e = 0$ or $e = 1$.

Now consider the case $p = 2$. If $e \geq 4$, then we again have a surjective map $(\mathbb{Z}/2^e)^\ast \to (\mathbb{Z}/(16))^\ast$. The element $3 + (16) \in \mathbb{Z}/(16)$ has order $4$. If $x \in \mathbb{Z}/(2^e)$ is any element mapping to $3 + (16)$, then $x$ is invertible and $4$ must divide its order. So the involutory condition rules out $e \geq 4$ in the case $p = 2$.

Thus the involutory condition forces $m = 2^{e_1} \cdot 3^{e_2}$ where $e_1 \leq 3$ and $e_2 \leq 1$, so that $m$ divides $2^3 \cdot 3^1 = 24$.

(Incidentally: you can't actually mean $m^2 = 1$ in the OP and in the bounty description; you must have meant what I wrote in the opening paragraph. And you don't mean "groups that are divisors of 24"; you must have meant groups whose orders are divisors of 24. There are other abuses of language as well.)