# Quotient Groups

Definition: If $G$ is a group and $N$ is a normal subgroup of group $G$, then the set $G|N$ of all cosets of $N$ in $G$ is a group with respect to the multiplication of cosets. It is called the quotient group or factor group of $G$ by $N$. The identity element of the quotient group $G|N$ by $N$.

Theorem: The set of all cosets of a normal subgroup is a group with respect to multiplication of complexes as the composition.

Proof:
Let$N$ be a normal subgroup of a group $G$. Since $N$ is normal in $G$, therefore each right coset will be equal to the corresponding left coset.

Thus there is no distinction between right and left cosets and we shall simply call them cosets. Let $G|N$ be the collection of all cosets of $N$ in $G$, i.e. let
$G|N = \left\{ {Na:a \in G} \right\}$

Closure Property: Let $a,b \in G$, then
$\begin{gathered} \left( {Na} \right)\left( {Nb} \right) = N\left( {aN} \right)b \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = N\left( {Na} \right)b \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = NNab = Nab \\ \end{gathered}$

Since $ab \in G$, therefore $Nab$ is also a coset of $N$ in $G$. So $Nab \in G|N$. Thus $G|N$ is closed with respect to coset multiplication.

Associatively: Let $a,b,c \in G$. Then $Na,Nb,Nc \in G|N$. We have
$\begin{gathered} Na\left[ {\left( {Nb} \right)\left( {Nc} \right)} \right] = Na\left( {Nbc} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Na\left( {bc} \right) = N\left( {ab} \right)c \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {Nab} \right)Nc \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {\left( {Na} \right)\left( {Nb} \right)} \right]Nc \\ \end{gathered}$

Thus the product of $G|N$ satisfies the associative law.

Existence of Identity: We have $N = Ne \in G|N$. Also if $Na$ is any element of $G|N$, then
$\begin{gathered} N\left( {Na} \right) = \left( {Ne} \right)\left( {Na} \right) = Nea = Na \\ \left( {Na} \right)N = \left( {Na} \right)\left( {Ne} \right) = Nae = Na \\ \end{gathered}$

Therefore the coset $N$ is the identity element.

Existence of Inverse: Let $Na \in G|N$, then $N{a^{ – 1}} \in G|N$. We have
$\begin{gathered} \left( {Na} \right)\left( {N{a^{ – 1}}} \right) = Na{a^{ – 1}} = Ne = N \\ \left( {N{a^{ – 1}}} \right)\left( {Na} \right) = N{a^{ – 1}}a = Ne = N \\ \end{gathered}$

Therefore the coset $N{a^{ – 1}}$ is the inverse of $Na$. Thus each element of $G|N$ possesses an inverse.

Hence $G|N$ is a group with respect to the product of cosets.