Coset Product of Normal Subgroup is Consistent with Subset Product Definition

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

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

Let $a, b \in G$.

Let $a \circ N$ and $b \circ N$ be the left cosets of $a$ and $b$ by $N$.

Then the coset product:
 * $\left({a \circ N}\right) \circ \left({b \circ N}\right) = \left({a \circ b}\right) \circ N$

is consistent with the definition of the coset product as the subset product of $a \circ N$ and $b \circ N$:
 * $\left({a \circ N}\right) \circ \left({b \circ N}\right) = \left\{{x \circ y: x \in a \circ N, y \in b \circ N}\right\}$

Proof
Consider the set:
 * $\left({a \circ N}\right) \circ \left({b \circ N}\right) = \left\{{x \circ y: x \in a \circ N, y \in b \circ N}\right\}$

As $e \in N$, have:


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

by associativity of $\circ$.

Hence $\left({a \circ b}\right) \circ N \subseteq \left({a \circ N}\right) \circ \left({b \circ N}\right)$.

Now let $x \in a \circ N$ and $y \in b \circ N$.

Then by the definition of subset product:
 * $\exists n_1 \in N: x = a \circ n_1$
 * $\exists n_2 \in N: y = b \circ n_2$

It follows that:

So the definition by subset product:
 * $\left({a \circ N}\right) \circ \left({b \circ N}\right) = \left\{{x \circ y: x \in a \circ N, y \in b \circ N}\right\}$

leads to the definition of coset product as:
 * $\left({a \circ N}\right) \circ \left({b \circ N}\right) = \left({a \circ b}\right) \circ N$