Tag Info

New answers tagged


If you define $u$ as element of $G_1$, then you can't just use it in $G_2$ (in the first equation), like you did in your equation. Besides, you missed to state, if $u$ and $g$ are generators of those groups. The following holds: $$e(u^a,g^{\frac{b}{a}}) = e(u,g)^b = e(u^b,g)$$ Anyway, you didn't do that in the actual question, where I don't really see the ...


Not only is $\mathbb G_1 = \mathbb G_2 = \mathbb G_T = \mathbb Z_p$ with $e(x,y) = x \cdot y \pmod{p}$ an example of a "bilinear group" (a triple of groups with a bilinear map) but every bilinear group of prime order is isomorphic to one of this form. So anything you want to know in principle about such groups, you can calculate on this example. I've used it ...

Top 50 recent answers are included