3

Given the binary operator $*$ of a finite (even small) commutative group, literally or as a table, how can I proceed to identify the name a mathematician knowing group classification would call it (short of learning group classification)?

If there's no silver bullet, what's a standard name for this particular one?

Let $p>2$ be a prime, and integer $a\in[2,p)$ such that $a^{(p-1)/2}\equiv-1\pmod p$. Let $S=\{\infty\}\cup\mathbb F_p$. Define the binary operation $*$ on $S$: $$u*v=\begin{cases} v&\text{if }u=\infty\\ u&\text{if }u\ne\infty\text{ and }v=\infty\\ \infty&\text{if }u\ne\infty\text{ and }v\ne\infty\text{ and }u+v\equiv0\pmod p\\ \displaystyle\frac{u\,v+a}{u+v}\bmod p&\text{otherwise} \end{cases}$$

$(S,*)$ is a commutative group of order $p+1$, with unity $\infty$, and a single other root of unity $0$.

Example for $p=7$, $a=3$ $$\begin{array}{c} &&&&&&&&\quad&\text{order}\\ \infty&0&1&2&3&4&5&6&&1\\ 0&\infty&3&5&1&6&2&4&&2\\ 1&3&2&4&5&0&6&\infty&&8\\ 2&5&4&0&6&3&\infty&1&&4\\ 3&1&5&6&2&\infty&4&0&&8\\ 4&6&0&3&\infty&5&1&2&&8\\ 5&2&6&\infty&4&1&0&3&&4\\ 6&4&\infty&1&0&2&3&5&&8\\ \end{array} $$ Update: Added the order of each element.

Update: The group can be constructed as an Elliptic Curve Group on $\mathbb F_p$ for the curve $y^2=x(x-a)^2$, where $(x,y)=(u^2,u(u^2-a))$. That curve, with the field $\mathbb R$. y^2=x(x-a)^2

Update: Made the unity $\infty$.

Derek Holt
  • 90,008
fgrieu
  • 1,758
  • 4
    I would start by counting the order of the elements. Also look at the structure theorem for finitely generated abelian groups. The example you give $1$ has order $8$ so the group will be $\mathbb{Z}/(8)$ – MIO May 25 '22 at 12:29
  • 2
    The element $1$ has order $8$, not $4$. – Derek Holt May 25 '22 at 12:54
  • 2
    Ok so you want to know the structure of the group. You already know that it's going to be direct sum of cyclic groups. So you look at the prime factors of the cardinality of the group. Then for each prime factor there will be a subgroup of the form $\mathbb{Z}/(p^{n_1})\times \mathbb{Z}/(p^{n_2})\dots \times \mathbb{Z}/(p^{n_m})$ that gives you all you need to know about the group structure – MIO May 25 '22 at 13:12
  • 3
    This group is cyclic for the first few primes. It might help if you gave some indication of where this operation comes from. – Derek Holt May 25 '22 at 14:19
  • 2
    Using $-1$ to stand for the point at infinity seems a bit clunky to me, especially since you then use an order $<$ in your definition which is not really appropriate for a finite group. I think it's fine to just use the symbol $\infty$ for that element in your finite group. – Jaap Scherphuis May 25 '22 at 15:03
  • 3
    There is a general result that an elliptic curve group over a finite field of prime order is either cyclic or a direct product of two cyclic groups. I don't know whether the second possibility can occur with this example, but it has not been ruled out. – Derek Holt May 25 '22 at 16:00
  • 1
    @Math_Images_Only: I now realize the answer might be in your first comment: If the group is cyclic (as I think it is) it's isomorphic to the group of integers modulo $p+1$ (the group order) under addition. Is it that simple? – fgrieu May 25 '22 at 16:17
  • 1
    @fgrieu Yes, any group of order $n$ that contains an element of order $n$ is cyclic, and in particular isomorphic to $\mathbb Z/n \mathbb Z$. – Ravi Fernando May 25 '22 at 18:50
  • 1
    Yes if there is an element of the order $n+1$ you are done. For the more general case you need to observe the order of all elements and use the structure of finitely generated abelian groups. There is most likely a clever way to deduce the order of the elements just from $a$ and $n$ but I cannot help you here – MIO May 26 '22 at 00:03

1 Answers1

2

Claim. For all $p$ and $a$ as stated above, the group $\newcommand{\P}{\mathbb{P}} \newcommand{\F}{\mathbb{F}} (\P^1(\F_p), *)$ is cyclic, hence is isomorphic to $C_{p+1}$, the cyclic group of order $p+1$.

First, note that we can write the group operation more uniformly using projective coordinates: \begin{align*} (b_0 : b_1) * (c_0 : c_1) = (b_0 c_0 + a b_1 c_1 : b_0 c_1 + b_1 c_0) \, . \end{align*} (Here we have identified $u$ with $(u:1)$ for all $u \in \F_p$ and $\infty$ with $(1:0)$.) Now the condition $a^{(p-1)/2}\equiv-1\pmod p$ means that $a$ is not a square mod $p$, so the field extension $\F_p(\sqrt{a}) \supseteq \F_p$ has degree $2$. We claim that the map \begin{align*} \varphi: \P^1(\F_p) &\overset{\sim}{\to} \F_p(\sqrt{a})^\times/\F_p^\times\\ (b_0 : b_1) &\mapsto \overline{b_0 + b_1 \sqrt{a}} \end{align*} is a group isomorphism. First, $\varphi$ is well-defined as we've quotiented out by $\F_p^\times$ in the codomain: \begin{align*} (\lambda b_0 : \lambda b_1) &\mapsto \overline{\lambda b_0 + \lambda b_1 \sqrt{a}} = \overline{\lambda (b_0 + b_1 \sqrt{a})} = \overline{b_0 + b_1 \sqrt{a}} \end{align*} for any $\lambda \in \F_p^\times$. Since \begin{align*} (b_0 + b_1 \sqrt{a})(c_0 + c_1 \sqrt{a}) = b_0 c_0 + a b_1 c_1 + (b_0 c_1 + b_1 c_0)\sqrt{a} \end{align*} for all $b_0 + b_1 \sqrt{a}, c_0 + c_1 \sqrt{a} \in \F_p(\sqrt{a})^\times$, then $\varphi$ is a group homomorphism. Since $\varphi$ is surjective and $$ \#(\F_p(\sqrt{a})^\times/\F_p^\times) = \frac{p^2-1}{p-1} = p+1 = \#\P^1(\F_p) \, , $$ then $\varphi$ is bijective, hence an isomorphism.

Since every finite subgroup of the group of units of a field is cyclic (see here, for example), then $\F_p(\sqrt{a})^\times$ is cyclic. Then the quotient $\F_p(\sqrt{a})^\times/\F_p^\times$ is also cyclic, and thus $\P^1(\F_p)$ is cyclic as well.

(This answer was inspired by $\S4.2$ of Lemmermeyer's Conics - A Poor Man's Elliptic Curves (Waybackup).)

Viktor Vaughn
  • 19,278