Coset Product is Well-Defined/Proof 2

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

Consider $\paren {a \circ N} \circ \paren {b \circ N}$ as a subset product:
 * $\paren {a \circ N} \circ \paren {b \circ N} = \set {a \circ n_1 \circ b \circ n_2: n_1, n_2 \in N}$

This is justified by Coset Product of Normal Subgroup is Consistent with Subset Product Definition.

Since $N$ is normal, each conjugate $b^{-1} \circ N \circ b$ is contained in $N$.

So for each $n_1 \in N$ there is some $n_3 \in N$ such that $b^{-1} \circ n_1 \circ b = n_3$.

So, if $a \circ n_1 \circ b \circ n_2 \in \paren {a \circ N} \circ \paren {b \circ N}$, it follows that:

That is:
 * $\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$

Let $n \in N$ be arbitrary.

Then:

So:
 * $\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$

and
 * $\paren {a \circ b} \circ N \subseteq \paren {a \circ N} \circ \paren {b \circ N}$

The result follows by definition of set equality.