$H\triangleleft G$ How to show that the congruence relation modulo $H$ is compatible with $G$'s structure?
Let $H$ be a normal subgroup of $G$.
Let $Q$ be the set of equivalence classes of $G$ under the congruence relation modulo $H$, that is the relation $\mathscr R$ such that :
$$x\mathscr R y \iff x\in yH \iff x\in Hy.$$
[ Let $q : G \rightarrow Q$ be the function such that , for all $g\in _G$ , $q(g) = [g]$, that is, the surjective function that maps every element of $g$ onto its equivalence class under $\mathscr R$. ]
In order to apply a theorem that allows to "turn" $Q$ into a group, it is required to first be sure that $\mathscr R$ is compatible with $G$'s structure.
Note : $\underset {with \space G's \space structure}{\mathscr R \space compatible}$$\underset {df}{\iff} \bigg[(x\mathscr R x'\land (y\mathscr R y')] \implies (xy) \mathscr R (x'y')\bigg] $
( with $xy$ and $x'y'$, products in $G$).
I'd like to prove that $\mathscr R$ is compatible with G's structure.
Suppose that $x\mathscr R x'$ and that $ y\mathscr R y'$, with $x, x', y, y' \in G$.
(1) We have :
$x\mathscr R x'\implies x\in x'H \implies x= x'\overline {h} , \space \space \overline {h}\in H$
$y\mathscr R y'\implies y\in y'H \implies y= y'\overline {\overline {h}} , \space \space\overline {\overline {h}}\in H$.
(2) Therefore :
$xy= x'\space \overline {h}\space y'\space \overline {\overline {h}} $.
(3) But, since $H$ is normal, $gH=Hg$ ( $\forall g_{\in G}$) , which means, in particular, that $Hy' = y'H$ ,i.e. $\overline {h}\space y'= y' \overline{\overline {\overline {h}}}$ , for some $\overline{\overline {\overline {h}}}\in H$.
(4) So, $xy= x'\space y'\space \overline{\overline {\overline {h}}}\space \overline {\overline {h}}= (x'\space y')\space (\overline{\overline {\overline {h}}}\space \overline {\overline {h}})$ ( associativity in $G$ allowing all groupings).
(5) Because $H$ is closed under the group operation , $\overline{\overline {\overline {h}}}\space \overline {\overline {h}}\in H$.
(6) From this follows that $xy$ is of the form $ (x'y') h , \space h\in H$, i.e. that : $(xy)\Large {\mathscr R}$ $ (x'y')$ ( as desired) .
Is this proof ok? In particular, step (3)?
Note : sorry for the strange symbols I used to denote elements of $H$, I realize retrospectively that using subscripts would have been better.