# 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.