is the following equality about group true Let $G,G_1,G_2,H$ be groups.
Is the following equality true:
If $G=G_1\cap G_2$ then : $ |H/G|=   |(H/G_1)\times (H/G_2)|$ 
A specific example is: Is  $|\mathbb{Z}/mn\mathbb{Z}| =  |(\mathbb{Z}/m \mathbb{Z})  (\mathbb{Z}/ n \mathbb{Z})|$  

There are at least two missing assumptions:
(1) Suppose that $G_1$ and $G_2$ are subgroups of $H$.
(2) Suppose that $G_1G_2$ is a subgroup of $H$.
Let $G=G_1\cap G_2$.  Observe that $G$ is a group.  
I am reinterpreting your questions as follows:
Is $|H/G|=|H/(G_1G_2)|$? (Note that you do not want the Cartesian product here as the Cartesian product is not a subset.)
Let's consider the finite case, in this case $|H/(G_1G_2)|=\frac{|H|}{|G_1G_2|}$.  Since $|G_1G_2|=\frac{|G_1||G_2|}{|G_1\cap G_2|}$, the original expression equals $\frac{|H|\cdot|G_1\cap G_2|}{|G_1||G_2|}$.  On the other hand, $|H/G|=\frac{|H|}{|G|}=\frac{|H|}{|G_1\cap G_2|}$.  Therefore, the equality that you're looking for holds only if
$$
\frac{|G_1\cap G_2|}{|G_1||G_2|}=\frac{1}{|G_1\cap G_2|}.
$$
In other words, you must have $|G_1\cap G_2|^2=|G_1||G_2|$.  This fails in many cases, for example, when $G_1$ is the trivial group and $G_2$ is not trivial.
