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

$$ $$ $$ $$ $$ $$ $$

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.