In the book Rational Points on Elliptic Curves by Silverman/Tate one examines the elliptic curve $y^2 = x^3 + x + 1$ over $F_5$. One can then easily determine the group $$ C(F_5) = \lbrace \mathcal{O}, (0, \pm 1), (2, \pm 1), (3, \pm 1), (4, \pm 2)\rbrace $$ which is an abelian group of order 9. The book claims that $C(F_5)$ is either abelian of order 9 or two cyclic groups of order 3 which makes sense. However, i can't understand the reasoning behind the fact that it must be a abelian group of order 9. There is two points of order $3$ and the rest are of order $9$.
Can anyone assist me with this reasoning. Thanks!
It is known in general that the group $E(\mathbb{F}_q)$ for an elliptic curve $E$ over $\mathbb{F}_q$ is either cyclic or isomorphic to $\mathbb{Z}/k\mathbb{Z} \times \mathbb{Z}/\ell\mathbb{Z}$ for some $k,\ell$. For $y^2=x^3+x+1$ over $\mathbb{F}_5$ the group is isomorphic to $\mathbb{Z}/9\mathbb{Z}$, so it is cyclic. Indeed, $P=(0,1)$ is a generator of order $9$. Since $\langle P \rangle \subseteq E(\mathbb{F}_5)$, and both groups have $9$ elements, it follows that $\mathbb{Z}/9\mathbb{Z}\simeq \langle P \rangle\simeq E(\mathbb{F}_5)$.