# 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.
• So the reason is because: $e(g,g)^N=e(g^q,g^p)=e(1,1)=1$ Right? If i have e$(g^q,x) =e(1,x)$ this is equal to 1 also? – curious Sep 27 '12 at 13:14
• No, if the order of $g$ is the composite $N$, then $g^q$ and $g^p$ will be different than $1$. The reason that $e(g,g)^N = 1$ holds is that $e(g,g)$ is the generator of $\mathbb{G}_T$ and has order $N$. – Conrado Sep 27 '12 at 16:22
• $e(g^{a1},g^{a2})^N=1$ holds also? – curious Sep 27 '12 at 16:52
• Yes, since $e(g^{a_1},g^{a_2})^N = e(g,g)^{N a_1 a_2} = (e(g,g)^N)^{a_1 a_2} = 1^{a_1 a_2} = 1$. – Conrado Sep 27 '12 at 19:03
The answer can be more simplest ! By definition a pairing e, is a non degenerate bilinear form with image in the group of N-root of Unity. Then $\forall P, Q \in E(\bar{K}): e(P,Q)^N = 1$
$E(\bar{K})$ is the torsion group of the elliptic curve E over the field K.