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Yes, as @DrLecter said in the comments, that equation holds from the bilinear property. Here is a step-by-step proof. Let $e : \mathbb G_1 \times \mathbb G_2 \to \mathbb G_T$ be a bilinear pairing. The bilinear property states that: \begin{align}e(g_1 ^ a, g_2 ^b) = e(g_1,g_2)^{ab}\end{align} Since you don't seem to distinguish between $\mathbb G_1$ and ...

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