I found the following algorithm to find the generator of an elliptic curve:

  1. Find the order of the curve - N.
  2. Choose any random point on the curve - P.
  3. Find the order of that point - n.
  4. Calculate co-factor - h=N/n.
  5. Find the generator - G=h x P.

Here, let us assume the order of the generator G is n_g. It accounts to say n_g x G=O. On replacing G from step 5, we can observe that n_g x (h x P)=O. From this we see that n_g x h >= n_g which makes P a stronger generator with an higher order.

How should I counter this argument?


1 Answer 1


The fact that $n_g\cdot h\cdot P=\mathcal{O}$, does not mean that the order of $P$ is $n_g\cdot h$. It means that the order of $P$ divides $n_g\cdot h$, so it could simply be $n_g$.


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