Let E be defined over Fq and let n ≥ 1. Show that E(Fq)[n] and E(Fq)/nE(Fq) have the same order.

Question

Let E be an Elliptic Curve defined over $F_q$ and let n ≥ 1. Show that $E(F_q)[n]$ and $E(F_q)$/$nE(F_q)$ have the same order.

I feel like this is obvious. The n-th torsion group $E(F_q)[n]$ by definition has $n^{th}$ order. Then E/nE = E mod n should also have n elements. This is similar to $Z_n$ $\cong Z/nZ$.

The last part of the question I had proved that if $\Phi$ is a homomorphism from a finite group G to itself then $|Ker(\Phi)$| = $|G:\Phi(G)|$ using Lagrange's Theorem and the first Isomorphism theorem. I would say $E(F_q)[n]$ $\cong$ $Z_n$ and then maybe prove $E(F_q)$/$nE(F_q)$ $\cong$ Z/nZ?

Answer 1

This is a property of abelian groups, and the fact that the group is coming from elliptic curves has little to do with it. Let $G$ be a finite abelian group and let $\phi:G\to G$ be a homomorphism. Then, $$|\ker(\phi)|=|\operatorname{coker}(\phi)|.$$ This follows from the first isomorphism theorem, in particular from the fact that $G/\ker(\phi) \cong \text{image}(\phi)$, and by definition $\operatorname{coker}(\phi)\cong G/\text{image}(\phi)$.

Now replace $G=E(\mathbb{F}_q)$ and $\phi$ by the multiplication-by-$n$ map.