0
$\begingroup$

on pg. 378 section 2 (Overview) it says "We let G be a multiplicative group of prime order p , and g be a generator of G. We let e : G x G --> $G_T$" be a bilinear map.

If somebody could please break each piece of this into smaller parts I would really appreciate it. Here is my attempt:

G is a multiplicative group in order p. Since G is cyclic this means any member of G multiplied by any integer mod p yields identity (1).

g being the generator, means that you always start with g. So you can raise g to a power or you can multiply g by a random integer. But you always have to get 1 mod p for it to be a member of G.

Before looking at binlinear map, I thought I should first read what a linear map is. According to wikipedia, a linear map always yields the same subspace of the input subspaces. So bilinear I think you can end up with something that is not linear (like an elliptic curve?).

My understanding of all this is quite fuzzy and I would appreciate it if someone could explain these to me in simple English or easy-to-understand drawings.

$\endgroup$

1 Answer 1

1
$\begingroup$

$G$ is not the group of integers mod $p$.

  • $G$ is a group, that is a set with a composition that obeys the group axioms.
  • it is multiplicative, that is we agree to use $\cdot$ (and not $+$, say) as symbol for the composition
  • it s of order $p$, that is its underlying set has $p$ elements
  • $g$ is a generator, which im plies that $G$ is cyclic
  • $e\colon G\times G\to G_T$ is bilinear, that is for each $a\in G$, the map $G\to G_T$, $x\mapsto e(a,x)$ is linear and $x\mapsto e(x,a)$ is also linear
$\endgroup$
2
  • $\begingroup$ What is bilinear then? Did I get that right? Is there an example of a simple bilinear function ? $\endgroup$ Commented Oct 14, 2013 at 19:25
  • $\begingroup$ Knowing a bit more of the context would be handy. Could it be that $G$ is in fact a subset of some ring, so that the distinction of additive vs. miultiplicative group is not just a notational one? $\endgroup$ Commented Oct 14, 2013 at 21:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .