If the order of elliptic group is prime then every point is a generator of that group. I tested the above statement on some elliptic curves and found it true. Does that really work on all curves?

Is there any lemma or theorem which states that?

  • $\begingroup$ Actually, it's not true - the identity element (neutral element, point at infinity) is a point on the curve, and it doesn't generate the entire curve. It's true that every other point on the curve does. $\endgroup$
    – poncho
    Commented Mar 13 at 2:10

1 Answer 1


This is true of any group of prime order, over elliptic curves or not. This is due to Lagrange's Theorem which states that the order of a subgroup $H$ of group $G$ divides the order of $G$.

Since orders are elements of the ring of integers and since this is a principal ideal domain, unique factorization exists and primes make sense. Or put another way, primes behave how you've been taught they do. Therefore the only possible subgroups of a group of order $p$ are the trivial group of order $1$ and the group of order $p$, i.e. the group itself.

This has a consequence, usually presented as a lemma of Lagrange's theorem - no element can generate a subgroup, so it must generate the full group.

For elliptic curve groups, there is a valid proper subgroup, the trivial subgroup consisting only of the identity, in this case $\mathcal{O}$.


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