It is said that if two elliptic curves $E_1, E_2$ defined over a finite field $K$ are isomorphic, then $E_1(K), E_2(K)$ groups are also isomorphic. But the converse is not true. Now,

  1. By definition, when $E_1, E_2$ are isomorphic over $K$ then $E_1$ can be transformed to $E_2$ by admissible change of variables. If I am not wrong, this implies that a point $P$ on $E_1$ is transformed to a point $P'$ on $E_2$.

  2. If $E_1(K), E_2(K)$ groups are isomorphic implies that there exist a bijective map between the two groups where the elements are points. So, the points are mapped again.

Since we are interested in the curve groups so, does it matter whether one take isomorphic curve or isomorphic group? Similarly, I don't understand what are the advantages of 1. over 2.?

  • $\begingroup$ I'm not sure if I would say it's an advantage (with respect to what metric?) but an isomorphism between $E_1$,$E_2$ simply maps more points. An isomorphism between $E_1(K)$,$E_2(K)$ only maps the $K$-rational points that solve your curve equation. -- Just out of curiosity: What material did you get this from? $\endgroup$ – Elias Oct 28 '16 at 13:22
  • $\begingroup$ @Elias, I am reading the book "Guide to elliptic curve cryptography" and the topic is "Isomorphism classes", section 3.1.5, page#84. $\endgroup$ – user110219 Oct 28 '16 at 15:10
  • $\begingroup$ The difference between 1 and 2 is that on 1 the isomorphism map can be given by polynomials with coefficients in K, while on 2 the polynomials may have coefficients in an extension field of K. $\endgroup$ – user27950 Oct 29 '16 at 8:02

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