Bilinear pairing arithmetic 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  $
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 $\mathbb G = &lt;g&gt;$ (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}