# order of group of points of secp256k1

points on curve secp256k1 form a group E(Fp) over field Fp.

p = 2^256 - 2^32 - 2^9 - 2^8 - 2^7 - 2^6 - 2^4 - 1 is prime.

n is the order of group E, n=115792089237316195423570985008687907852837564279074904382605163141518161494337

Is n prime too?

Is E(Fp) a cyclic group?

Theorem. Working over a finite field, the group of points E(Fp) is always either a cyclic group or the product of two cyclic groups.

• `n` has to be a prime such that `(Gx, Gy) * n = slope is infinity`. Hopefully someone can answer the other part – Wizard Of Ozzie Jun 21 '15 at 0:34

The comment stating `n` “has to be prime” is a bit confusing.
The order of base point “has” to be prime in the sense that this is a requirement in the particular documents defining standard curves—for example, in SECG, which includes secp256k1. Bitcoin's base point order `r` is prime.
In SECG, it is also stated that cofactor of secp256k1 curve is 1, which makes `n = r × 1`, again prime. A group of prime order is obviously cyclic.
• @arulbero re SEC defined G: great question. Anyone know why? Why not start at `x=0`? – Wizard Of Ozzie Jun 21 '15 at 14:17