Coset Product is Well-Defined

Theorem
Let $N \triangleleft G$ where $G$ is a group (that is, let $N$ be a normal subgroup of $G$.

Let $a, b \in G$.

Let the coset product of $a N$ and $b N$ be defined as:


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

Alternatively, the product of the cosets $a N$ and $b N$ can be defined as:


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

that is, using the definition from subset product.

The coset product is well-defined.

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

Proof 1
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.

Proof 2
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.