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I'm trying to understand how the identity point is represented in a group of prime order.

What I think is correct:

If the group has even order, then the identity point is in the group, because the identity point has order 2.

If the group has prime order, like elliptic curves such as secp256 then the identity point can still be represented as a group, but is the point at infinity. If the point could not be represented in the group, then it would not be a group.

I do not believe I completely understand the second point. What order is the identity point if it is in the group?

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  • $\begingroup$ How can an element be the identity of a group without being in the group? This is in response to your phrase 'If the identity ... then it is in the group.' $\endgroup$ – kodlu Jul 16 at 2:30
  • $\begingroup$ This answer might help you better What is the point at infinity on secp256k1 and how to calculate it? $\endgroup$ – kelalaka Jul 16 at 11:03
  • $\begingroup$ @kelalaka Very helpful. So for Weierstrass curves, the point at infinity exists, however it is not representable in affine co-ordinates. So we set a flag to denote whether it is the identity point when using affine. For Edwards curves, we do not have this problem though, do you know why? $\endgroup$ – WeCanBeFriends Jul 17 at 12:46
  • $\begingroup$ I think I understand that it has nothing to do with whether the group that the curve yields is prime or not. It is do with the curve itself, because I have never seen a "infinity flag" to denote whether the point is infinity for Edwards curves, or I am mistaken? $\endgroup$ – WeCanBeFriends Jul 17 at 12:48
  • $\begingroup$ Since it is any point on the curve $\endgroup$ – kelalaka Jul 17 at 18:39
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For something to be a group, it must have an identity. This is a definition.

The order of an element, $g$, in a group is defined to be the smallest positive number $n$ such that $g^n = 1$, where $1$ is the identity of the group.

By this definition, the order of the identity in any group is $1$.

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