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Let $p$ be an odd prime, and $a$ be an element of field $\Bbb Z_p$.

Define $l$ as the Legendre symbol $\displaystyle\biggl(\frac ap\biggr)$.

When $l=+1$, define $b$ as a particular solution of $b^2=a$ in $\Bbb Z_p$, e.g. the odd one in range $[1,p)$.

Define the set $\mathcal S$ as: $$\mathcal S=\begin{cases} \{\infty\}\cup\Bbb Z_p&\text{when }l=-1\\ \{\infty\}\cup\Bbb Z_p-\{0\}&\text{when }l=0\\ \{\infty\}\cup\Bbb Z_p-\{b,\,p-b\}&\text{when }l=+1 \end{cases}$$

Define internal law $\boxplus$ in $\mathcal S$ as: $$x\boxplus y=\begin{cases} y&\text{when }x=\infty\\ x&\text{when }x\ne\infty\text{ and }y=\infty\\ \infty&\text{when }x\ne\infty\text{ and }y=-x\\ (x+y)^{-1}(x\,y+a)&\text{(computed in $\Bbb Z_p$) otherwise} \end{cases}$$

$(S,\boxplus)$ is a finite Abelian group¹ of order $p-l$. What's its name in standard literature?

¹ The Abelian group axioms hold:

  • Closure of $\boxplus$ follows from definitions, trivially for the case $l=-1$, and with some level of care in the other ones.
  • The neutral is $\infty$. This follows from the first three cases of the definition of $\boxplus$.
  • The opposite $-x$ is computed as in $\Bbb Z_p$ when $x\ne\infty$, and we define $-\infty$ as $\infty$. That $x\boxplus(-x)=\infty=(-x)\boxplus x$ follows from the first and third cases of the definition.

  • Associativity $(x\boxplus y)\boxplus z=x\boxplus(y\boxplus z)$ needs to be proved in multiple cases:

    • When any of $x$, $y$ or $z$ is $\infty$, we use the first two cases of the definition.
    • Otherwise, when $y$ is $-x$ or $-z$, we use the first three cases of the definition.
    • In the remaining general case, it boils down to basic algebra. The exclusions in the definition of $\mathcal S$ prevent degenerate cases.
  • Commutativity is easy.
fgrieu
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  • It is called $S$. – Eduardo Longa Feb 19 '20 at 21:27
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    It seems we need to assume $a\ne0$ for this to be a group. – Greg Martin Feb 19 '20 at 21:57
  • @AdrienZ that operation falls into the third case of the definition, not the last. – Greg Martin Feb 19 '20 at 22:01
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    Interestingly, if the ground field were $\Bbb R$, this group law seems to be something like a change of variables $x\mapsto\arctan\frac1x$, which would make the group isomorphic to $(\Bbb R/\pi,{+})$. – Greg Martin Feb 19 '20 at 22:04
  • I think it should be $\mathbb{F}{p^2}^* / \mathbb{F}_p^$. If instead we take $\mathbb{R}$ and $a = -1$ we can think of this as embedding $\mathbb{R}$ into $\mathbb{C}^$ by $x \to x+i$ and then doing multiplication in $\mathbb{C}^* / \mathbb{R}^*$. The point is $(x + i)(y+i) = (xy-1) + (x+y)i$ which is equivalent to $\frac{xy-1}{x+y} + i$ modulo scalars. My guess is there is some way to normalize things to do the same construction in $\mathbb{F}{p^2}$ with arbitrary $a$, but I haven't worked out the details. – Nate Feb 19 '20 at 22:21
  • @Greg Martin: thanks for pointing that in the initial statement, the structure was not always a group. Associativity, failed for some arguments when $\bigl(\frac ap\bigr)\ne-1$. That's fixed by excluding these. – fgrieu Feb 20 '20 at 07:05

1 Answers1

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For $k$ a field and $a\in k^*$ let $$H_{a,p} = \{ \pmatrix{r &x\\ -ax&r}\in GL_2(k), r^2+ax^2\ne 0\}$$ It is a subgroup of $GL_2(k)$ $$\pmatrix{r &x\\ -ax&r}\pmatrix{s&y\\ -ay&s}=\pmatrix{ -axy + rs & ry + sx\\-a( ry + sx)& -axy + rs }$$

Then $H_{a,p}/k^*$ is your subgroup of $PGL_2(k)$, which is isomorphic to the group of fractional linear transformations $$\pmatrix{r &x\\ -ax&r}\to \frac{tr+x}{-axt+r}= \cases{\frac{t+x/r}{-a\frac{x}rt+1}\ \text{if} \ r\ne 0\\ \frac{-1}{at}\text{ otherwise}}$$ That $a$ is a square or not tells if the subgroup of $GL_2(k)$ is a split or non-split torus. It would suggest the name "projective non-split torus" but it doesn't give any hit.

The case $a=0$ is degenerate and your group takes a completely different form, it is isomorphic to $(k,+)$ through $x\to \frac1x$.

reuns
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