# How many 2-torsion points in an elliptic curve?

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

• In $\mathbf Z/4\mathbf Z \times \mathbf Z/4\mathbf Z$, can you list the elements of order $2$? My count is not $5$.
– user69015
May 30 '20 at 8:25
• @corpsfini Because ker([4]) ≅ Z4×Z4 , which leads to 5 subgroup of order 4. So these 5 cyclic groups have generator Pi, and [2]Pi is an order 2 point. So there is 5. Can you give me the way how you count this?
– rzxh
May 30 '20 at 8:59
• If you take the $5$ generators of the subgroups of order $4$, called $P_i$, then for some, we have $[2]P_i = [2]P_j$ with $i \neq j$. The values $[2]P_i$ will give only three distincts values, each are an element of order $2$.
– user69015
May 30 '20 at 9:30
• What is the source of this image? May 30 '20 at 9:37
• @kelalaka From "Supersingular isogeny key exchange for beginners" (Craig Costello 2019)
– rzxh
May 30 '20 at 9:43