How do you determine if an elliptic curve over a finite field is cyclic? I know the group order and the points of the elliptic curve $y^2 = x^3 + Ax + B$, but I am confused on how to determine if they from a cyclic group
The curve $y^2 = x^3 + 2x +2$ in $\Bbb F_{11}$ which has $9$ points: $(1,4),$ $(1,7),$ $(2,5),$ $(2,6),$ $(5,4),$ $(5,7),$ $(9,1),$ $(9,10)$ and the point of infinity, $(0,1)$. 
 A: Well, you're looking at an abelian group (say with operation $\oplus$) of order $9,$ correct? Then either the group will be cyclic, or it will be isomorphic to $C_3\times C_3$. Let's rewrite the non-identity elements of the group as: $$(1,4),\ominus(1,4),(2,5),\ominus(2,5),(5,4),\ominus(5,4),(9,1),\ominus(9,1)$$ (these are written in the same order as you had them, but making  the inverse elements clear). Since elements have the same order as their inverses, then we need consider only $$(1,4),(2,5),(5,4),(9,1).$$ Either all four of them have order $3$, or three of the four of them have order $9$. In the latter case, we'll have a cyclic group. Regardless, you'll only need to check two of them, and if you're checking $(p,q)$, then you'll only need to check if $(p,q)\oplus(p,q)=\ominus(p,q)$. If not, then $(p,q)$ has order $9$ and we're done.
A: A shorter solution depending on a bit more machinery is the following.
As others have said, an abelian group of order $9$ is either cyclic or isomorphic to $C_3\times C_3$.
It is known (see e.g. Silverman) that the $3$-torsion of this curve (over an algebraic closure of $\Bbb{F}_{11}$ is isomorphic to $C_3\times C_3$. Furthermore, the Weil pairing on that $3$-torsion is non-degenerate. So if all  of the $3$-torsion points have coordinates in a field $K$, then the multiplicative group $K^*$ contains a primitive third root of unity.
But $\Bbb{F}_{11}^*$ is cyclic of order ten, so there are no elements of order three there. The conclusion is that $K$ must be a proper extension of $\Bbb{F}_{11}$.
This leaves the cyclic group as the only alternative.
A: Use the following code in Magma:

F:=GF(11);
E:=EllipticCurve([0,0,0,2*F.1,2*F.1]);
P:=E![1,4];
[2*P,3*P,4*P,5*P,6*P,7*P,8*P,9*P];

The result is:

[ (2 : 5 : 1), (9 : 10 : 1), (5 : 4 : 1), (5 : 7 : 1), (9 : 1 : 1), (2 : 6 : 1),
  (1 : 7 : 1), (0 : 1 : 0)]

Hence $P=(1,4)$ has order $9$.
