How many 2-torsion points in an elliptic curve?

Question

N torsion points have the structure ker([n]) ≅ Zn×Zn , so ker([2]) ≅ Z2×Z2 , gives us 3 2-torsion points.2-torsion points

but ker([4]) ≅ Z4×Z4 ,this means we have 5 subgroup of order 4 . In each subgroup , there is a generator Pi, my question is , these 5 [2]pi are all different order 2 points right? Does not this means we have 5 4-torsion points ??

What's wrong here?

Answer 1

The question, as i understand it, is not really a question about elliptic curves, we can start in general with the group $$ (G,+) = (\Bbb Z/4\times \Bbb Z/4,\ +)\ , $$ which has $16$ elements. (Well, this comes in the question from an elliptic curve, but we replace it for the purpose of the question with the above isomorphic choice, and typing becomes easier.)

Its elements are $$ \begin{array}{|c|c|c|c|} \hline (0,0) & \color{blue}{(1,0)} & (2,0) & \color{blue}{(3,0)} \\\hline \color{blue}{(0,1)} & \color{blue}{(1,1)} & \color{blue}{(2,1)} & \color{blue}{(3,1)} \\\hline (0,2) & \color{blue}{(1,2)} & (2,2) & \color{blue}{(3,2)} \\\hline \color{blue}{(0,3)} & \color{blue}{(1,3)} & \color{blue}{(2,3)} & \color{blue}{(3,3)} \\\hline \end{array} $$ and the blue entries are the elements of order $4$ in $G$.

Here are the subgroups of order 4, listed (for easy typing) by sage, and it is easy to check this also with bare hands.

sage: G.<P,Q> = AbelianGroup(2, [4, 4])                                                                                       
sage: print(f"G = {G}\nwith generators {H.gens()}\n")                                                                         
G = Multiplicative Abelian group isomorphic to C4 x C4
with generators (P, Q)

sage: for H in G.subgroups(): 
....:     if H.order() == 4: 
....:         print(f"H = {H} generated by {H.gens()}") 
....:                                                                                                                         
H = Multiplicative Abelian subgroup isomorphic to C4 generated by {P} generated by (P,)
H = Multiplicative Abelian subgroup isomorphic to C4 generated by {P*Q} generated by (P*Q,)
H = Multiplicative Abelian subgroup isomorphic to C4 generated by {P*Q^2} generated by (P*Q^2,)
H = Multiplicative Abelian subgroup isomorphic to C4 generated by {P*Q^3} generated by (P*Q^3,)
H = Multiplicative Abelian subgroup isomorphic to C2 x C2 generated by {Q^2, P^2} generated by (Q^2, P^2)
H = Multiplicative Abelian subgroup isomorphic to C4 generated by {P^2*Q^3} generated by (P^2*Q^3,)
H = Multiplicative Abelian subgroup isomorphic to C4 generated by {Q} generated by (Q,)
sage:  

(Sage uses a multiplicatively written group operation.)

So there are six cyclic subgroups of order four isomorphic to C4$\cong\Bbb Z/4$, each one generated by the one or the other blue element. Two "blue" elements $(a,b)\in G$ and $-(a,b)=(-a,-b)=(4-a,4-b)$ are generating the same group.

And there is one subgroup of order $4$ isomorphic to C2 x C2$\cong\Bbb Z/2\times \Bbb Z/2$, just to have the full list and a correct statement regarding the subgroups of order four.

Now let us consider the C4 subgroups one by one, the possible choices for the generator $\pi$, and the value of $2\pi$.

• The group generated by $\pi=(1,0)$ (or $-\pi=(3,0)$) has $2\pi=(2,0)$.
• The group generated by $\pi=(0,1)$ (or $-\pi=(0,3)$) has $2\pi=(0,2)$.
• The group generated by $\pi=(1,1)$ (or $-\pi=(3,3)$) has $2\pi=(2,2)$.
• The group generated by $\pi=(2,1)$ (or $-\pi=(2,3)$) has $2\pi=(0,2)$.
• The group generated by $\pi=(1,3)$ (or $-\pi=(3,1)$) has $2\pi=(2,2)$.
• The group generated by $\pi=(1,2)$ (or $-\pi=(3,2)$) has $2\pi=(2,0)$.

So the six diffrent $2\pi$-values lead to three$=6/2$ different values, which are the three generators of $2\Bbb Z/4\times 2\Bbb Z/4\cong \Bbb Z/2\times \Bbb Z/2$.