Re: Multiplicative Homomorphic Mapping


"Gerry Myerson" <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:gerry-E32158.14204822082005@xxxxxxxxxxxxxxxxxxxxx
> g generates G, h generates H, a is in G, so a = g^x for some (very
> hard to calculate) x, you want to know h^x; I don't see how you're
> going to get it without getting x first, or anyway doing as much
> work as it would take to get x. Do you have any reason to think
> otherwise?
>
> --
> Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)

I am not sure if I understand the difficulty of this problem well enough.
Given g^x for unknown x, I don't see why finding h^x has to be as
difficult as finding x. At the very least, given an algorithm doing so,
it appears that the DL problem cannot be solved.

I am not sure if the following could be possible counterexample:

Suppose G is a supersingular elliptic over F_p with a non-degenerate
pairing to F_p^l, let P in G be a subgroup of order q (call it S_1), then
the pairing e(P,P) is a generator of a subgroup in F_p^l (call it S_2)
of the same order.

Given xP in S_1, we can find e(P,P)^x = e(xP,P).


This is a very special instance and what I want to find is other more
general
instances.


.

Relevant Pages

  • Re: generators be bound
    ... >> definition a generator of that subgroup. ... According to the folk here that's not a generator. ... Generator of a proper subgroup. ... "the multiplicative group of maximal order" is ...
    (sci.crypt)
  • Re: normalizer of Q8 in SL(2,q)
    ... Edwin Clark wrote: ... This commutes with the first generator, ... automorphism group by the inner automorphisms is isomorphic to S_3. ... Hence N/Q_8 is isomorphic to a subgroup of S_3. ...
    (sci.math)
  • Re: Parameters for Diffie-Hellman-Merkle
    ... Richard Heathfield wrote: ... >Paul Crowley wrote: ... then it's a generator of the order Q subgroup. ...
    (sci.crypt)
  • Re: generators be bound
    ... > Tom St Denis wrote: ... if you have a subgroup of a cyclic ... According to the folk here that's not a generator. ... be a sub-group of it as well. ...
    (sci.crypt)
  • Re: Easy question in algebra
    ... >I suppose this is an easy question in algebra: ... Is the reason for that, that g^q =1 and 1 is not a generator of G? ... In this subgroup, since q is prime and since the order of an element ...
    (sci.math)