Why is the abelian group of points on an elliptic curve over a finite field isomorphic to the product of at most two cyclic groups? Could anyone explain why the abelian group of points on an elliptic curve over a finite field is isomorphic to at most two cyclic groups? Why is it that it cannot be the product of more than two cyclic groups? 
I have tried searching for an answer but the best I could find is that it might be because of Lefschetz principle since an elliptic curve is an abelian variety with dimension 2. But I am not very familiar with algebraic geometry so I don't really understand this answer, let alone know if it is correct.
Any help much appreciated. 
 A: *

*For an elliptic curve $\widetilde{E}/\Bbb{F}_q$ - given by a Weierstrass equation - over a finite field you can construct an elliptic curve $E/\Bbb{Q}(\zeta_{q-1})$ whose it is a reduction, then $E[n]\to \widetilde{E}[n]$ is surjective under adequate definition of the reduction map (not sure if it is possible to give a two lines proof of this claim).


*$E[n]$ has $n^2$ elements, this follows from the isomorphism from $E/\Bbb{C}$ to a complex torus.
This implies that for all $n$, $\widetilde{E}[n]$ has at most $n^2$ elements, which itself implies that $\widetilde{E}[n]\cong \Bbb{Z}/n_1\Bbb{Z}\times \Bbb{Z}/n_2\Bbb{Z}$ for some $n_1,n_2$ dividing $n$.


*Finally $\widetilde{E}(\Bbb{F}_q)$ is a subgroup of $\widetilde{E}[n]$ with $n=|\widetilde{E}(\Bbb{F}_q)|$, and a subgroup of a product of two cyclic groups is a product of two cyclic groups itself.
A: Let $n=\#E(\mathbb{F}_q)$, which is finite because $\mathbb{F}_q$ is. Then $E(\mathbb{F}_q)\subset E[n]\cong \mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$.
