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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 ??

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  • $\begingroup$ In $\mathbf Z/4\mathbf Z \times \mathbf Z/4\mathbf Z$, can you list the elements of order $2$? My count is not $5$. $\endgroup$
    – user69015
    May 30 '20 at 8:25
  • $\begingroup$ @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? $\endgroup$
    – rzxh
    May 30 '20 at 8:59
  • $\begingroup$ 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 $2$. $\endgroup$
    – user69015
    May 30 '20 at 9:30
  • $\begingroup$ What is the source of this image? $\endgroup$
    – kelalaka
    May 30 '20 at 9:37
  • $\begingroup$ @kelalaka From "Supersingular isogeny key exchange for beginners" (Craig Costello 2019) $\endgroup$
    – rzxh
    May 30 '20 at 9:43

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