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 :) 
 A: Denote $g\sim g'$ if $g^{-1}g'\in H$. Check that this defines an equivalence relation on $G$ and that equivalence classes are of the form $gH$ for some $g\in G$. Taking a set of representatives $\{g_i\;;\; i\in I\}$ for these equivalence classes yields a partition
$$
G=\bigsqcup_{i\in I} g_iH.
$$
Now note that $H$ is in bijection with each $g_iH$. This proves the claim for left cosets. The cardinality of $I$ is the index of $H$ in $G$, denoted by $[G:H]$. If the order of $G$ is finite, then the order of $H$  and $[G:H]$ are finite, and we get $$|G|=[G:H]|H|.$$ 
So the order of $H$ divides the order of $G$, which is known as Lagrange's theorem.
Do the same with $g\sim' g'$ if $g'g^{-1}\in H$ to get the right cosets. Note that the isomorphism $g\longmapsto g^{-1}$ exchanges left cosets and right cosets. In particular, $\{g_i^{-1}\;;\;i\in I\}$ is now a system of representatives for the right cosets. 
$$
G=\bigsqcup_{i\in I} Hg_i^{-1}.
$$
It follows that we can define $[G:H]$ without ambiguity, as the cardinality of the set of left cosets, or the cardinality of the set of right cosets.
A: 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.
A: 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)
