Product operation for cosets In Artin's book, he gives a lemma that for a normal subgroup $N \subset G$ and cosets $aN$, $bN$, that the product set $(ab)N$ is also a coset; the proof simply uses the definition of normality:
$$(aN)(bN) = a(Nb)N = a(bN)N = (ab)(NN) = (ab)N.$$
My first question is how to put this proof in the context of demonstrating that this product operation is well-defined, which requires proving that if $aN = a'N$ and $bN = b'N$, then $abN = a'b’N$. I am not sure if this is in fact a proof of well-definedness and know that we can define a set product, say $AB$ of elements $ab$ where $a \in A, b \in B$. But we can't "use" this product unless it is well-defined, in which case it is by definition a coset, which leads me to beleive that Artin is in fact saying that this proof allows the operation to "work."
Can someone help me understand this?

Edit: I misinterpreted the question posed, so what's below is largely unhelpful. I will keep my answer up regardless for anyone who would like to know how to prove that coset multiplication is well defined for context.
To prove that coset multiplication is well defined I will prove the following statement adapted from the book Abstract Algebra by John Beachy and William D. Blair:

Let $N$ be a normal subgroup of $G$, and let $a,b,c,d\in G$. If $aN = cN$ and $bN = dN$, then $abN = cdN$. Which implies that multiplication of left cosets is well defined.

Proof: Suppose that $aN = cN$ and $bN = dN$ then it is necessarily the case that $a^{-1}c\in N$ and $b^{-1}d\in N$. However, we know $N$ is normal and so $d^{-1}(a^{-1}c)d\in N$. Since $b^{-1}d\in N$ we have $(ab)^{-1}cd = (b^{-1}d)(d^{-1}a^{-1}cd)\in N$ which immediately implies that $abN = cdN$. As such, left coset multiplication is indeed well defined given that $N$ is a normal subgroup.
