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?
 A: The equations are just statements about equalities of sets. They use the following facts:

*

*For subsets $A$ and $B$ of $G$, $AB=\{ab\mid a\in A,b\in B\}$.

*For any subgroup $H$ of $G$, $HH=H$.

*For any normal subgroup $N$ of $G$, and every $x\in G$, $xN=Nx$ as sets.

*“Multiplication” of sets is associative: for any subsets $A$, $B$, and $C$ of $G$, $(AB)C = A(BC)$.

Using these we are not making a definition that the product of the set $aN$ and the the set $bN$ is going to be designated, by fiat, as $abN$. Rather, we are saying that the set $(aN)(bN)$ and the set $(ab)N$ are equal as sets. That means there is no issue of “well-definedness of the product”: we are making an assertion about equality of sets.
Because $(aN)(bN)=abN$ as sets; and $(a’N)(b’N)=a’b’N$, as sets, if $aN=a’N$ as sets and $bN=b’N$ as sets, then
$$abN = (aN)(bN) = (a’N)(b’N)=a’b’N.$$
We are not defining “products of cosets”, we are noting that the product of two cosets of a normal subgroup is again a coset of that normal subgroup; and along the way that the specific coset can be found by multiplying representatives. But we are just applying the definition of “multiplication of sets”, and the properties associated with it and with cosets of a normal subgroup.
This is different from starting from the set of cosets and saying “we define a multiplication of cosets by saying that $(aN)*(bN) = (ab)N$.” Such an assertion would require showing that it is well defined. Here, instead, we are using the product of sets, which has no definition issues.
A: 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.
