I was learning the topic and faced the excercise:
Let G1, G2 and G3 be finite cyclic groups of the same order n, and let e(·,·) : G1 ×G2 → G3 be a pairing map. Show that, for given g1 ∈ G1, g2 ∈ G2 and all a,b ∈ Zn, the following identity holds:
e(g1^a,g2^b) = e(g1,g2)^a·b
The proof of equality above is based on the property:
e(g1 * g1', g2) = e(g1, g2) * e(g1', g2)
As e function is a multiplication of pair members, can the property above be represented as:
e(g1 * g1', g2) = e(g1, g2) * e(g1', g2) = (g1 * g2) * (g1' * g2) and if so, why dont we have g2^2 at the end ?
In other words where from the second g2 comes from here? (e(g1, g2) * e(g1', g2))
P.S. I assume that e is a multiplication function based on other example from learning materials I am using:
One of the most basic examples of a non-degenerate pairing involves the groups G1, G2 and G3 all being groups of integers with addition (Z,+). In this case, the following map defines a non-degenerate pairing:
e(·,·) : Z×Z → Z (a,b) → a·b
The example you give of $e: \mathbb Z \times \mathbb Z \to \mathbb Z$ can be represented as multiplication because all three groups involved are the same group.
However, in the general case, $G_1, G_2, G_3$ can all be different. So, if we have $g_1 \in G_1$ and $g_2 \in G_2$, $g_1 * g_2$ is not defined. You may write $e(g_1,g_2)$ as $g_1 * g_2$ but note that this does not correspond to any group operations of $G_1$, $G_2$ or $G_3$. This is simply a notational choice you may make. In particular, when you write $e(g_1 * g_1', g_2) = e(g_1, g_2) * e(g_1', g_2) = (g_1 \cdot g_2) * (g_1' \cdot g_2)$, you would want to distinguish the operations as I have done explicitly with different notation. The operation $*$ here is for $G_3$ and $\cdot$ is for the map $e$.