Coset Product of Normal Subgroup is Consistent with Subset Product Definition

Theorem
Let $\struct {G, \circ}$ 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:
 * $\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$

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

Proof
Consider the set:
 * $\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$

As $e \in N$, have:


 * $\paren {a \circ b} \circ N = \paren {a \circ e} \circ \paren {b \circ N} \subseteq \paren {a \circ N} \circ \paren {b \circ N}$

by of $\circ$.

Hence $\paren {a \circ b} \circ N \subseteq \paren {a \circ N} \circ \paren {b \circ N}$.

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:
 * $\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$

leads to the definition of coset product as:
 * $\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$