Order of subgroups formed by Elliptic Curves with a Cofactor

Question

In this question:

Why are the lower 3 bits of curve25519/ed25519 secret keys cleared during creation?

The answer indicates that the order of all points on the curve over the finite field 2^255 - 19 is 8 times the size of the subgroup formed by G.

i.e. the subgroup size is 1=2^252+27742317777372353535851937790883648493 whereas the number of points in the curve itself is 8(1).

The answer then states: "This means that there are a few remaining points that have small order."

However, as stated in the answer the few remaining points are in fact 8 times the number of points in the cyclic subgroup G.

So how can one conclude that the remaining points form small order groups?

Isn't there scope for a group within the set of remaining points to be bigger than 1?

How do we know the other points not inside 1, form a variety of small order groups?

Answer 1

The answer indicates that the order of all points on the curve over the finite field $2^{255} - 19$ is 8 times the size of the subgroup formed by $G$.

Obviously, this is incorrect, and Samuel never claims it.

This curve defines a group with $8q$ elements (with $q = 2^{252} + 27742317777372353535851937790883648493$ prime), and the factorization of $8q = 2 \times 2 \times 2 \times q$. Hence, the possible orders of points are $1, 2, 4, 8, q, 2q, 4q, 8q$. In addition, this curve happens to be a cycle curve (not all elliptic curve groups are), and so for each possible order, there are in factor at least one group element with that order.

$G$ happens to be one of the points of order $q$ (actually, it didn't just 'happen', a point of that order was deliberately selected to be $G$).

How do we know the other points not inside 1, form a variety of small order groups?

Because we know the complete factorization of the number of points on the curve ($8q$). If there is a subgroup of size $\lambda$, that would imply that $\lambda$ was a factor of $8q$. We know all the values that are a factor of $8q$, and there are none between 8 and $q$; hence, there cannot be any subgroups with a size between 8 and $q$.

Group theory is your friend.