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.

• What's your background? What book(s) did you read to learn about elliptic curves? – Álvaro Lozano-Robledo Jan 31 '17 at 20:57
• Maths undergrad, writing a thesis on ECC. I've read some of Hoffstein's An Introduction to Cryptography and some of Koblitz' A Course in Number Theory and Cryptography. I've found a statement of the theorem in Silverman's An Introduction to the Theory of Elliptic Curves found here math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf on page 29, but it provides no insight as to why. – user406579 Feb 1 '17 at 0:03
• Of course it is not surprising to me that an elliptic curve over a finite field is isomorphic to the product of two cyclic groups, after all an elliptic curve over $\mathbb{C}$ is homeomorphic to a torus, which itself is homeomorphic to the product of two circles. But I would like some more insight as to why this is the case, particularly for finite fields. – user406579 Feb 1 '17 at 0:20
• – Jyrki Lahtonen Feb 2 '17 at 18:11

Let $n=\#E(\mathbb{F}_q)$, which is finite because $\mathbb{F}_q$ is. Then $E(\mathbb{F}_q)\subset E[n]\cong \mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$.
• Where $E[n]$ is the n-torsion points on $E$? I'm not especially familiar with the notion of n-torsion, would you care to explain why $E[n] \cong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n \mathbb{Z}$ please? – user406579 Feb 1 '17 at 20:48
• Indeed $E[n]$ is the $n$-torsion subgroup. That material should be available in many places, I don't think I can do better. Perhaps the easiest is to look at division polynomials, which are polynomials whose roots correspond exactly to $n$-torsion points. These polynomials have degree $n^2$, which is a big step towards proving the statement. For example, try Silverman's "Arithmetic of Elliptic Curves" exercise III.3.7. – CurveEnthusiast Feb 2 '17 at 7:22