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 :) 

You can produce a bijection of the left coset space $L_G=\{gH:g\in G\}$ onto the right coset space $L_R=\{Hg:g\in H\}$ by $gH\mapsto Hg^{-1}$. Regarding the fact that $|H|\;\mid \;|G|$, use the fact that each coset has the same cardinality: Take $zH$ and $xH$ different cosets. Then $zH\to xH:u\mapsto xz^{-1}u$ is a bijection. In particular, $|H|=|gH|$ for any coset, so $|G|=m|H|$ for $m$ the number of cosets in the coset space, usually denoted by $m=|G:H|$. It is also crucial that given two cosets $zH,xH$, $zH\cap xH\neq\varnothing\implies zH=xH$. That is, the cosets form a partition of the original group. As julien noted, $x\sim y\iff y^{-1}x\in H$ is an equivalence relation on $G$. This can be seen as $x\sim y \iff \exists h\in H:hx=y$ (the cosets are the $G$ orbits of $H$ under the left translation or right translation, which is an action, which always induces a partition into orbits)
