why $e(g,g)^N=1$ in bilinear pairings holds? I can't get the point of prime order bilinear pairings:$\mathbb{G}\times\mathbb{G}\rightarrow\mathbb{G}_T$,$g=$ generator of $\mathbb{G}$ , $N=p*q$, $p$ and $q$ primes and $e(g,g)^N=1$. why $e(g,g)^N=1$ holds? Why is it 1?
If $N$ is the order of the group $\mathbb{G}_T$, then for any element $x \in \mathbb{G}_T$ we have that $x^N = 1$. This follows from the Lagrange theorem. Since $e(g,g) \in \mathbb{G}_T$, the same applies to it.