I'm proving a chain of 10 equivalent definitions of normal subgroup. I cannot complete one step: (7) implies (8), where
(7) $\forall x, y, a, b \in G ( x \in aN\land y \in bN\to xy \in (ab)N)$.
(8) $N =\bigcup_{a\in N}\textrm{Cl}(a)$.
My start: Since $a\in\textrm{Cl}(a)$, it follows that $N \subseteq \bigcup_{a\in N}\textrm{Cl}(a)$. For the converse, given $x\in \bigcup_{a\in N}\textrm{Cl}(a)$, there is $a\in N$ such than $x\in\textrm{Cl}(a)$, hence $x=cac^{-1}$, thus $xc=ca\in cN$, and applying (7) $xcca\in c^2N$ or $(xc)^2\in c^2N$
... and some how should I conclude $x\in N$.
Note:
$\textrm{Cl}(a):=\{b\in G: \exists c\in G(b=cac^{-1})\}.$
We only know, I suppose, that $N$ is just a subgroup in this context.