Coset Product is Well-Defined/Proof 2

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $N$ be a normal subgroup of $G$.

Let $a, b \in G$.

Then the coset product:
 * $\left({a N}\right) \left({b N}\right) = \left\{{x y: x \in a N, y \in b N}\right\}$

is well-defined.

That is, the congruence modulo a subgroup is compatible with the group product.

Proof
We need to demonstrate that:
 * $\left\{{x y: x \in a N, y \in b N}\right\} = \left\{{\left({a b}\right) n: n \in N}\right\}$

Let:
 * $P = \left\{{x y: x \in a N, y \in b N}\right\}$
 * $Q = \left\{{\left({a b}\right) n: n \in N}\right\}$

First we show that $P \subseteq Q$.

By definition, $z \in P \implies \exists n_1, n_2 \in N: z = a n_1 b n_2$.

Now $n_1 b \in N b = b N$ as $N$ is normal.

Thus $\exists n_3 \in N: n_1 b = b n_3$.

Thus $z = a n_1 b n_2 = a b n_3 n_2 \in Q$.

Thus $P \subseteq Q$.

Now let $z \in Q$.

Thus $\exists n \in N: z = a b n$.

But $a \in a N, b n \in b N$ so $z = \left({a}\right) \left({b N}\right) \in P$.

Thus $Q \subseteq P$.

The result follows.