Coset Product is Well-Defined/Proof 2

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

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

Let $a, b \in G$.

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

is well-defined.

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

Consider $\left({a \circ N}\right) \circ \left({b \circ N}\right)$ as a subset product:
 * $\left({a \circ N}\right) \circ \left({b \circ N}\right) = \left\{{a \circ n_1 \circ b \circ n_2: n_1, n_2 \in N}\right\}$

This is justified by Coset Product 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 $a \circ n_1 \circ b \circ n_2 \in \left({a \circ N}\right) \circ \left({b \circ N}\right)$, it follows that:

That is:
 * $\left({a \circ N}\right) \circ \left({b \circ N}\right) \subseteq \left({a \circ b}\right) \circ N$

Then:

So:
 * $\left({a \circ N}\right) \circ \left({b \circ N}\right) \subseteq \left({a \circ b}\right) \circ N$

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

The result follows by definition of set equality.