<p>(Disclaimer: this is not my field.)</p>

<p>Let g be the chosen generator and n its (prime) order.  In the ECDSA algorithm, these are publicly known.  It is true that g generates a cyclic (abelian) group isomorphic to Z/nZ.</p>

<p>Now a private key consists of an integer k, and the corresponding public key is the group element h = kg. (I use additive notation since we are in an abelian group, so kg = g+g+...+g (k times)).  If we were actually working in Z/nZ, it would be trivial to recover k from h: just divide h by g (mod n) using the Euclidean algorithm.  Then we would have no security at all.</p>

<p>The point, as I understand it, is that in the elliptic curve group, there is no known efficient way to "divide".  And while there certainly exist isomorphisms from <code>&lt;g&gt;</code> to Z/nZ (map g to any element of Z/nZ you like), the inverse of such an isomorphism is not trivial to compute.</p>

CC BY-SA 4.0
2019-05-26T15:06:33.417Z
Interesting. A follow up question now would be why is this isomorphism hard to find? I.e is there a provable reason or just social consensus that people were not able to find it?
Jernej
CC BY-SA 4.0
2019-05-26T15:08:43.677Z
Now you're beyond the limits of my knowledge.  You might find more experts at http://crypto.stackexchange.com.
Nate Eldredge
CC BY-SA 4.0
2019-05-26T17:17:20.977Z
@Jernej The reason is simply that people have tried and failed to find an algorithm for the discrete logarithm (given group elements A and B, find x such that xA=B). This is essentially true for all cryptography: people find a hard problem to base things on, and hope that problem gets attention from people trying to break it.
Pieter Wuille
CC BY-SA 4.0
2019-05-26T18:50:24.067Z
@PieterWuille So essentially saying that finding the said group isomorphism is equivalent to the discrete log problem?
Jernej
CC BY-SA 4.0
2019-05-26T18:51:53.177Z
Yes, the ismorphism from G to Z/nZ is exactly f(h) = DL(h wrt g).
Pieter Wuille
CC BY-SA 4.0
2019-05-26T15:01:29.280Z
2019-05-26T15:01:29.280Z
Nate Eldredge
<p>As I understand things so far, the ECDSA scheme used by Bitcoin uses a group G generated by a point on the elliptic curve y^2 = x^2 + 7.</p>

<p>Now G is a cyclic group of finite order n and is, hence, isomorphic to the additive group Z/nZ.</p>

<p>My question is: why not work within Z/nZ directly then? I am assuming there are certain security benefits of using G but I am lacking the general overview to understand why.</p>

<cryptography><ecdsa>
ECDSA - why not using the cyclic additive group
Jernej
# ECDSA - why not using the cyclic additive group

(Asked by: username_0 on 2019-05-26)

As I understand things so far, the ECDSA scheme used by Bitcoin uses a group G generated by a point on the elliptic curve y^2 = x^2 + 7.

Now G is a cyclic group of finite order n and is, hence, isomorphic to the additive group Z/nZ.

My question is: why not work within Z/nZ directly then? I am assuming there are certain security benefits of using G but I am lacking the general overview to understand why.

**Tags:** cryptography, ecdsa


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## 1 answers

### Answer by username_1 on 2019-05-26. Score: 2

(Disclaimer: this is not my field.)

Let g be the chosen generator and n its (prime) order. In the ECDSA algorithm, these are publicly known. It is true that g generates a cyclic (abelian) group isomorphic to Z/nZ.

Now a private key consists of an integer k, and the corresponding public key is the group element h = kg. (I use additive notation since we are in an abelian group, so kg = g+g+...+g (k times)). If we were actually working in Z/nZ, it would be trivial to recover k from h: just divide h by g (mod n) using the Euclidean algorithm. Then we would have no security at all.

The point, as I understand it, is that in the elliptic curve group, there is no known efficient way to "divide". And while there certainly exist isomorphisms from  to Z/nZ (map g to any element of Z/nZ you like), the inverse of such an isomorphism is not trivial to compute.

> **Comments:**
> **username_0**: Interesting. A follow up question now would be why is this isomorphism hard to find? I.e is there a provable reason or just social consensus that people were not able to find it?  
> **username_1**: Now you're beyond the limits of my knowledge.  You might find more experts at http://crypto.stackexchange.com.  
> **username_2**: @username_0 The reason is simply that people have tried and failed to find an algorithm for the discrete logarithm (given group elements A and B, find x such that xA=B). This is essentially true for all cryptography: people find a hard problem to base things on, and hope that problem gets attention from people trying to break it.  
> **username_0**: @PieterWuille So essentially saying that finding the said group isomorphism is equivalent to the discrete log problem?  
> **username_2**: Yes, the ismorphism from G to Z/nZ is exactly f(h) = DL(h wrt g).  

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