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$n$-torsion points have the structure $\operatorname{ker}([n])\cong \mathbb{Z}_n\times \mathbb{Z}_n$ , so $\operatorname{ker}([2]) \cong \mathbb{Z}_2×\mathbb{Z}_2$ , gives us three $2$-torsion points (2-torsion points).

But, $\operatorname{ker}([4]) \cong \mathbb{Z}_4\times \mathbb{Z}_4$, this means we have $5$ subgroups of order $4$ . In each subgroup, there is a generator $\pi$. My question is, these five $\pi$'s are all different order $2$ points right? Does not this means we have five $4$-torsion points ??

What's wrong here?

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  • $\begingroup$ With $f$ the multiplication by $2$ then $f(\Bbb{Z/4Z\times Z/4Z})=\Bbb{2Z/4Z\times 2Z/4Z\cong Z/2Z\times Z/2Z}$ $\endgroup$
    – reuns
    May 30, 2020 at 5:36
  • $\begingroup$ In the $4$-torsion, I count six cyclic subgroups of order $4$. Each order $2$ point appears in two of them. $\endgroup$ May 30, 2020 at 5:45
  • $\begingroup$ @AnginaSeng Thanks for your reply. I can count these six subgroups too, 3 pairs, each 2 of them share the same 2-torsion point, right? But, the "ker([4]) ≅ Z4×Z4" statement confuses me all the time, ker([4]) ≅ Z4×Z4 means that it can be think as 5 cyclic groups of order 4 right? So the are 10 4-torsion points (each subgroup, 1P and 3P are all 4-torsion points)? Why? I don't know where goes wrong? $\endgroup$
    – rzxh
    May 30, 2020 at 7:13
  • $\begingroup$ @reunsThanks for your reply. I can understand this calculation in some sense I think, but I can not draw the pictuce describing the structure of those points like the one in my question. So just like my reply to Angina Seng, I can not understand where is my problem. Would you please explain it with detail? thanks $\endgroup$
    – rzxh
    May 30, 2020 at 7:43

1 Answer 1

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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 $2\Bbb Z/4\times 2\Bbb Z/4\cong \Bbb Z/2\times \Bbb Z/2$.

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