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I was implementing point arithmetic operation, and was exploring the properties of point arithmetic, and I am unable to conclude whether $$ k^{-1}(kP) = P $$ where P is a point over elliptic curve $ F_p $ and $ k^{-1}.k = 1 $ mod p

Please help me to understand its correct or not ?
And if it is correct how it can be proved ?

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No, this is not correct.

Let $E:y^2=x^3+1$ defined over $\mathbb{F}_5$. Let $P=(0,1)$. Then $2P=(0,4)$, but $3(2P)=6P=\mathcal{O}=[0,1,0]$, i.e., zero, or point at infinity, even though $2\cdot 3 \equiv 6 \equiv 1 \bmod 5$.

The reason for this is $|E(\mathbb{F}_5)|=6$ in this case, so $6P=\mathcal{O}$ for any point $P\in E(\mathbb{F}_5)$.

Perhaps what you are interested in is the following. Let $d=|E(\mathbb{F}_p)|$ and let $P\in E(\mathbb{F}_p)$. Then, $dP=\mathcal{O}$. Now, let $k\geq 1$ be relatively prime to $d$, i.e., $\gcd(k,d)=1$, and let $h$ be a natural number such that $kh\equiv 1 \bmod d$, i.e., there is some $n\geq 1$ such that $hk=1+nd$. Then: $$h\cdot (kP)=(hk)P = (1+nd)P = P+n(dP)=P+\mathcal{O}=P.$$

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  • $\begingroup$ Thanks for help, I was mixing up order n of point with field prime p. $\endgroup$ Apr 2, 2013 at 6:10

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