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.

Proof
By Subgroup is Normal iff Left Cosets are Right Cosets, the set of left cosets for $N$ equals 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.

, 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.

Thus $G / N$ is closed.

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

Thus $G / N$ 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 $x^{-1} N$ is the inverse of $x N$.

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

Also see
From Subgroup is Normal iff Left Cosets are Right Cosets, the left coset space of a normal subgroup is equal to its right coset space.

It follows that $G / N$ does not depend on whether left cosets are used to define it or right cosets. Thus we do not need to distinguish between the left quotient group and the right quotient group - the two are one and the same.