Post
Topic
Board Development & Technical Discussion
Re: zero point of secp256k1
by
AliKhan189
on 20/12/2021, 17:45:53 UTC
Hello,

I need a math clarification please.

Since the points on the Elliptic Curve (secp256k1) form a group, we can apply the laws of addition, substruction and multiplication.

This require a zero.

I checked that -1*G+1*G=N*G (with N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141).

Therefore, could we consider N*G as the zero ?

If yes, then why -2*G+2*G = M such as there seems to be no link between M and N.

Generaly speaking, is there any link between (-1*G+1*G) and (-k*G+k*G) ?
Prove that if ϕ : G → H is a group homomorphism and G is cyclic, then the subgroup ϕ(G) is cyclic. Proof Suppose G is cyclic, so G = 〈a〉 = {ak : k ∈ Z} for some a ∈ G. ... Then, as a generates G, we have c = ak for some integer k. Thus b = ϕ(c) = ϕ(ak) = ϕ(a)k ∈ 〈ϕ(a)〉.