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I am working through some content in L.C. Washington's Elliptic Curves, Number Theory, and Cryptography and I am unsure about what the group structure of a certain group looks like.

Some background: let $a/b \in \mathbb{Q}$ such that $\gcd(a,b) = 1$. Now write $a/b = p^r \cdot a_{1}/b_{1}$ where $p$ is a prime and $p \nmid a_{1}b_{1}$. Define the $p$-adic valuation to be $$v_{p}(a/b)=r.$$ Also, for an elliptic curve $E$ defined to be $y^2 = x^3 + Ax + B$, let $r \geq 1$ be an integer and let $E_{r}$ denote $$E_{r} = \{ (x,y) \in E(\mathbb{Q}) \mid v_{p}(x) \leq -2r, v_{p}(y) \leq -3r \} \cup \{\infty\}.$$

The problem that I am working on is the following: The map $$\lambda_{r}: E_{r}/E_{5r} \rightarrow \mathbb{Z}/p^{4r}\mathbb Z$$ defined by the mapping $(x,y) \mapsto p^{-r}x/y (\mod p^{4r})$ and $\infty \mapsto 0$ is an injective homomorphism (where $\mathbb{Z}_{^{4r}}$ is a group under addition).

My question is, what does the group structure of $E_{r}/E_{5r}$ look like? Also, in the text they say that "$\lambda_{r}$ should be regarded as the logarithm for the group $E_{r}/E_{5r}$ since it changes the law of composition in the group to addition in $\mathbb{Z}/p^{4r}\mathbb Z$." Can someone explain why I should regard this map as the logarithm?

Edit: My motivation behind this question is attempting to understanding Nigel Smart's attack on anomalous elliptic curves and understanding his algorithm on cracking the elliptic curve discrete logarithm problem in linear time on such curves.

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  • $\begingroup$ Since $\mathbb Z/p^{4r}\mathbb Z$ is a cyclic group of order $p^{4r}$ and $\lambda_r$ is injective, the group $E_r/E_{5r}$ is a cyclic group of order dividing $p^{4r}$. As for you second question, the answer is the one you already mentioned: the usual logarithm is a group homomorphism $\mathbb R^*_{>0}\to \mathbb R$, thus it "changes multiplication into addtion", since $\log(xy)=\log(x)+\log(y)$. Analogously, $\lambda_r$ "changes composition of points on $E$ into addition modulo $p^{4r}$". $\endgroup$ – Ferra Dec 16 '15 at 9:49

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