Group theory - left/right $H$-cosets and quotient sets $G/H$ and $G \setminus H$. Let $G$ be a group and  $H$ be a subgroup of $G$. The left $H$-cosets are the sets $gH, g \in G$. The set of left $H$-cosets is the quotient set $G/H$. The right $H$-cosets are the sets $Hg, g\in G$. The set of right $H$-cosets is the quotient $H \setminus G.$ The set of $G$ is the union of left(respectively, right) $H$-cosets, each of which has $\left|H\right|$ elements. We deduce that the order of $H$ divides the order of $G$, and that the number of left $H$-cosets equals the number of right $H$-cosets.
What I wrote can be find in Groups and symmetries - Yvette Kosmann-Schwarzbach. 
What should I understand from that statement ? 
is ok if I say that : $$G=gH \cup Hg?$$
$$\left|G/H\right|=\left|H \setminus G\right|?$$
What means that the number of left $H$-cosets equals the number of right $H$-cosets? 
And why we deduce that the order of $H$ divides the order of $G$? 
Thanks :) 

It means that $G=\cup_{g\in G} gH=\cup_{g\in G} Hg$, that the map $x\rightarrow gx$ is a bijection between $H$ and $gH$ and therefore $|gH|=|H|$. It can be deduced from this that $|G|=|G/H||H|$. A similar statement for right cosets.
