Proof of The Product of Cosets of a Normal Subgroup Is Well Defined and Makes the Cosets a Group

Throughout, products in $G$ are regrouped freely by associativity, part of Groups and Abelian Groups (i), and juxtaposition $gh$ stands for $g\cdot h$, as in Semigroups (i); routine facts about the relation of lying in a common left coset are used without comment (Toolkit for AL1).

We first describe the cosets of $N$ as equivalence classes. Let

$$R=\{(g,h)\in G\times G: g^{-1}h\in N\}.$$

Since $N$ is a subgroup of $G$ (Normal Subgroups (i)), Left Cosets Partition a Group and All Have the Same Cardinality (iv) shows that $R$ is an equivalence relation on $G$, that the equivalence class $[g]_R$ of each $g\in G$ is the coset $gN$, and that the quotient $G/R$ is the set $G/N$ of left cosets.

Consequently the canonical projection $\pi_R$, which by Equivalence Classes Partition the Set: Cover, Disjointness and Representatives; the Quotient Is a Set and the Canonical Projection Is a Surjection (vi) is a surjective function from $G$ to $G/R=G/N$ with value $[g]_R=gN$ at each $g\in G$, is exactly the function $g\mapsto gN$ of (iii). We write $\pi$ for it.

Next, $R$ is compatible with the group operation. Suppose that $g\,R\,g'$ and $h\,R\,h'$, so that $g^{-1}g'\in N$ and $h^{-1}h'\in N$. By Cancellation Laws and Basic Inverse Identities in a Group (iv),

$$(gh)^{-1}(g'h')=\bigl(h^{-1}(g^{-1}g')h\bigr)\bigl(h^{-1}h'\bigr).$$

The first factor lies in $N$ because $N$ is normal: Normal Subgroups (i), applied to the element $h^{-1}$, whose inverse is $h$ by Cancellation Laws and Basic Inverse Identities in a Group (iii), and to $g^{-1}g'\in N$, gives $h^{-1}(g^{-1}g')h\in N$. The second factor lies in $N$ by hypothesis, and so does their product, $N$ being a subgroup (Subgroups (i)). Hence $(gh)\,R\,(g'h')$.

The group operation is a binary operation on $G$ (Groups and Abelian Groups (i)), and we have just seen that $R$ is compatible with it. Therefore Compatible Operations, Maps and Relations Pass to the Quotient (i) gives exactly one binary operation $\ast$ on $G/R=G/N$ such that $[g]_R\ast[h]_R=[gh]_R$ for all $g,h\in G$. Since $[g]_R=gN$ for every $g$, this is precisely the assertion (i).

We now verify the group laws. Every element of $G/N$ is a coset $gN$ for some $g\in G$ (Cosets and the Index of a Subgroup (iii)), so it suffices to compute with cosets. For $g,h,k\in G$, (i) and the group laws of $G$ give

$$\bigl((gN)\ast(hN)\bigr)\ast(kN)=((gh)k)N=(g(hk))N=(gN)\ast\bigl((hN)\ast(kN)\bigr),$$ $$(eN)\ast(gN)=gN=(gN)\ast(eN),\qquad (gN)\ast(g^{-1}N)=eN=(g^{-1}N)\ast(gN).$$

Thus $(G/N,\ast)$ is a monoid with neutral element $eN$ in which every element is a unit (Monoids (i), Units of a Monoid (i)), and hence a group by Groups and Abelian Groups (i). As the identity and inverses of a group are unique (Uniqueness of the Identity Element and of Inverses in a Group (i) and Uniqueness of the Identity Element and of Inverses in a Group (ii)), its identity is $eN$ and the inverse of $gN$ is $g^{-1}N$, as (ii) asserts. It remains to see that $eN=N$. By the description of $R$ above, $h\in[e]_R$ if and only if $e^{-1}h\in N$; and $e^{-1}h=eh=h$ because $e^{-1}=e$ (Cancellation Laws and Basic Inverse Identities in a Group (ii)). So $eN=[e]_R=N$.

Finally, for $g,h\in G$, (i) gives

$$\pi(gh)=(gh)N=(gN)\ast(hN)=\pi(g)\ast\pi(h),$$

so $\pi$ is a group homomorphism (Group Homomorphisms and Isomorphisms (i)), and we saw above that it is surjective. Since $eN$ is the identity of $G/N$, the kernel of $\pi$ (Kernel and Image of a Group Homomorphism (i)) is $\{g\in G: gN=eN\}$. By Left Cosets Partition a Group and All Have the Same Cardinality (ii), $gN=eN$ holds if and only if $e^{-1}g\in N$, that is, since $e^{-1}g=g$, if and only if $g\in N$. So the kernel of $\pi$ is $N$, which completes the proof of (iii).