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Is this $e(g^x,g^yH^z) = e(g^x,g^y)e(g^x,H^z)$ expression is true?

where $ g$ is the generator and $ H \in G $

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  • $\begingroup$ Yes, just apply bilinearity. $\endgroup$ – DrLecter May 8 '15 at 6:26
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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 $\mathbb G_2$, we will assume that the pairing is symmetric, so $\mathbb G_1$ and $\mathbb G_2$ are the same group and will be denoted as $\mathbb G$, with generator $g$.

Now, you said that $H \in \mathbb G$. Since $\mathbb G = <g>$ (i.e., $g$ generates the cyclic group $\mathbb G$), then we can assume that $H = g^h$, for some unknown $h$ (we don't need to know $h$, it is just needed for proving that the equation holds).

Using elementary algebra, we have that:

\begin{align}e(g^x,g^yH^z) = e(g^x,g^y\cdot (g^h)^z) = e(g^x,g^{y + hz})\end{align}

Next, we apply the bilinear property:

\begin{align}e(g^x,g^{y + hz}) = e(g,g)^{x(y + hz)}\end{align}

Again, applying basic properties of the exponentation:

\begin{align}e(g,g)^{x(y + hz)} = e(g,g)^{xy} \cdot e(g,g)^{xhz}\end{align}

Finally, you just "undo" some of the transformations, using the bilinear property:

\begin{align}e(g,g)^{xy} \cdot e(g,g)^{xhz} = e(g^x,g^y) \cdot e(g^x,g^{hz}) = e(g^x,g^y) \cdot e(g^x,H^z)\end{align}

Using this technique, you can prove any of the derived identities of bilinear pairings. The following is a general identity that involves the product of pairings:

\begin{align}e(g^a, g^b) \cdot e(g^c, g^d) = e(g,g)^{ab+cd}\end{align}

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  • $\begingroup$ +1, but you do not need to make it that complicated. Just ignore the exponents and write $g^x$, $g^y$ and $H^z$ as group elements $a$, $b$ and $c$. If you substitute, you get $e(a,b\cdot c)=e(a,b)\cdot e(a,c)$ , which is a direct application of the definition of bilinearity. $\endgroup$ – DrLecter May 8 '15 at 13:04

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