Coset Product is Well-Defined/Proof 1

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:
 * $\left({a N}\right) \left({b N}\right) = \left({a b}\right) N$

is well-defined.

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

Proof
Let $N \triangleleft G$ where $G$ is a group.

Let $a, a', b, b' \in G: a N = a' N, b N = b' N$.

To show that the coset product is well-defined, we need to demonstrate that $\left({a b}\right) N = \left({a' b'}\right) N$.

So:

By Equal Cosets iff Product with Inverse in Coset‎:
 * $\left({a b}\right)^{-1} \left({a' b'}\right) \in N \implies \left({a b}\right) N = \left({a' b'}\right) N$

and the job is finished.