# order of elliptic curve subgroup when curve has point (0,0)

I'm a beginner. But I understand that the order of a subgroup is a divisor of the group order. The curve $$y^2=x^3+7$$ over $$\mathbb{Z}_7$$ has eight points (7 points and the point at infinity). The order of the point (0,0) is 2 (?), but the order of all the other subgroups is 7, not 8. This seems to violate LaGrange's Theorem.

I did the same thing for $$y^2=x^3+7$$ over $$\mathbb{Z}_{11}$$, and the subgroup orders were all divisors of 12. That is what I expected. Why doesn't it work over $$\mathbb{Z}_7$$?

I hope I've explained OK. I am not into the "weeds" of higher math.

Thayer

• When $p=7$ the discrimant is zero: $2A^3-27B^2 = 0.$ ($y^2 = x^3 + Ax +B$) and the reason is the cusp that I've used as an example for cusps. Dec 17, 2021 at 17:27

The problem is that when we reduce the curve equation $$y^2=x^3+7$$ modulo 7, we get the equation $$y^2=x^3$$ which is not counted as an elliptic curve. The technical term for this is that the rational curve $$y^2=x^3+7$$ has "bad reduction" at the prime 7.
The reason that curves of the form $$y^2=x^3$$ are not elliptic curves is because they are not "smooth". This means that they have a special singular point that does not behave well. Roughly speaking this means that tangent lines at that point are not well-defined (which in particular means the doubling rule on the curve does not make sense at that point). In this case the singular point is $$(0,0)$$ which is a cusp. (Bad) reduction to this sort of curve is called additive reduction because there is a group law on the non-singular points, but it is the same as the additive group of the finite field. In this case, the group is the same as addition modulo 7. The isomorphism between the groups is easy: an $$t\neq 0$$ integer mod 7 goes to the point $$(t^{-2},t^{-3})\mod 7$$ and 0 goes to the point at infinity. Likewise the inverse map sends the point $$(x,y)$$ to the integer $$x/y\mod 7$$.