Let $\mathcal{C}$ be a conic on the projective plane $PG(2,\mathbb{F})$ where $\mathrm{char}\mathbb{F}\neq 2$. Let $\ell$ be a line and let $N$ be a point on $\mathcal{C}\setminus \ell$. For $A,B\in \mathcal{C}$, let $L_{AB}=AB\cap \ell$. Prove that the following operation defines an abelian group $G$ on $\mathcal{C}\setminus\ell$:

$$A \circ B =\begin{cases} N, & \text{ if } L_{AB}N \text{ is the tangent at } N \text{ on }\mathcal{C} \\ M, & \text{ if } L_{AB}N = \{N,M\} \end{cases}.$$

Further show that if

  1. $\ell$ intersects the conic in one point (tangent) then $G$ is the additive group of $\mathbb{F}$
  2. $\ell$ intersects the conic in two points (secant) then $G$ is the multiplicative group of $\mathbb{F}$

Proof. Here is a figure where $A\circ B$ equals a point $M$ and $A\circ C$ equals the point $N$.

enter image description here

For the first part I can easily show closure, abelianity, identity (which is $N$) and inverses. I am a bit stuck on the associative part $(A\circ B)\circ C = A\circ (B\circ C)$. Let $A\circ B = M_1$ and $B\circ C = M_2$. I thought we can use Pascal's theorem to show that $M_1C$ is parallel to $AM_2$ but I am not sure how to continue. The second part of the exercise is not clear to me at all. Has anyone seen this type of group construction before? I've been looking at literature and cannot find anything else.

Update: The group part is done, and now we have to consider the cases when $\ell$ is a tangent and when $\ell$ is a secant. I think we can use the parabola and hyperbola for these cases and then use the fact that we can map conics to parabola and hyperbola - we usually do things like this but not sure how to continue.

  • $\begingroup$ What is $\mathbb K$? $\endgroup$ Dec 16, 2020 at 16:15
  • $\begingroup$ Oops I meant $\mathbb{F}$. $\endgroup$
    – mandella
    Dec 16, 2020 at 17:12
  • 1
    $\begingroup$ For a reference for this construction, there is this paper (arxiv.org/abs/math/0311306), but I don't think it will help you solve your exercise. $\endgroup$ Dec 16, 2020 at 17:51
  • $\begingroup$ This is similar to the way that the elliptic curve group operation is defined. $\endgroup$
    – Somos
    Jan 3, 2021 at 14:36
  • $\begingroup$ Any references? $\endgroup$
    – mandella
    Jan 3, 2021 at 14:37

1 Answer 1


Consider hexagon $ABCM_1NM_2$ inscribed in the conic. By construction, opposite sides $AB$, $M_1N$ meet on $\ell$, and so do opposite sides $BC$, $M_2N$. It follows by Pascal's Theorem that opposite sides $M_1C$ and $AM_2$ also meet on $\ell$, which implies $(A\circ B)\circ C=A\circ(B\circ C)$.


See this answer to a similar question for a proof that $G$ is the additive or multiplicative group of $𝔽$.


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