Quotient Group is Group

Theorem
Let $G$ be a group.

Let $N$ be a normal subgroup of G.

Then the quotient group $G / N$ is indeed a group.

Corollary
It follows that if $G$ is finite, then $\left|{G / N}\right| = \dfrac {\left|{G}\right|} {\left|{N}\right|}$.

Proof
First note that as $N$ is normal, which means that the set of left cosets is equal to the set of right cosets.

It follows that $G / N$ does not depend on whether left cosets are used to define it or right cosets.

WLOG we will work with the left cosets.

By definition of quotient group, the elements of $G / N$ are the cosets of $N$ in $G$, where the group product is defined as:
 * $\left({a N}\right) \left({b N}\right) = \left({a b}\right) N$

The operation has been shown in Coset Product is Well-Defined to be a well-defined operation.

Now we need to demonstrate that $G / N$ is a group.

G0: Closure
This follows from Coset Product is Well-Defined.

As $a b \in G$, it follows that $\left({a b}\right) N$ is a left coset.

G1: Associativity
The associativity of coset product follows directly from Subset Product of Associative is Associative:

G2: Identity
The left coset $e N = N$ serves as the identity:

Similarly $\left({x N}\right) N = x N$.

G3: Inverses
We have $\left({x N}\right)^{-1} = x^{-1} N$:

Similarly $\left({x^{-1} N}\right) \left({x N}\right) = N$.

Thus all the group axioms are fulfilled, and $G / N$ has been shown to be a group.

Proof of Corollary
From the definition of Subgroup Index, $\left|{G / N}\right| = \left[{G : N}\right]$.

From Lagrange's Theorem, we have $\left[{G : N}\right] = \dfrac {\left|{G}\right|} {\left|{N}\right|}$.