Quotient Groups
There is another group associated with an invariant subgroup, called the quotient group.
Consider the invariant subgroup $H \triangleleft G$, with $\lvert H \rvert = m$ and $\lvert G \rvert = n$. Lagrangre’s theorem tells us that there are $p = n / m$ distinct cosets that can be formed.
This set of cosets is $Q = \{ He, Hg_{2}, \dots, Hg_{p} \}$, where $\{ e, g_{2}, \dots, g_{p} \} \in G$ are the elements which produce distinct cosets (not to be confused with all the elements in $G$).
The cosets can be used in an operation called coset multiplication.
Definition 1.12 – Coset Multiplication
The multiplication of cosets, $(H g_{a}) \cdot (Hg_{b})$, is defined as the product of all the elements in $Hg_{a}$ with all the elements in $H g_{b}$. Any repeated elements are eliminated from the resulting set.
The set of distinct cosets $Q$ with coset multiplication forms the quotient group.
Definition 1.13 – Quotient Group
Given an invariant subgroup $H \triangleleft G$, the quotient group $G / H$ is
- the set of distinct cosets $Q = \{ He, Hg_{2}, \dots, Hg_{p}\}$, where $\{ e, g_{2}, \dots, g_{p} \}$ are the $p$ elements of $G$ which create distinct cosets,
- with the group product being coset multiplication.
Proof that the Quotient Group is a Group
Closure
$$ (Hg_{a})(Hg_{b}) = (H(g_{a}H)g_{b}) = (H(Hg_{a})g_{b}) = (HH)(g_{a}g_{b}) = H (g_{a} g_{b}) \in Q $$Associativity
The associativity of the quotient group is inherited from the associativity of group $G$.
Identity
$$ H(Hg_{a}) = (HH)g_{a} = Hg_{a} \implies e = G $$Inverse
$$ (Hg_{a}^{-1})(Hg_{a}) = H(g_{a}^{-1} H)g_{a} = H(Hg_{a}^{-1})g_{a} = HH(g_{a}^{-1}g_{a}) = H \implies (Hg_{a})^{-1} = Hg_{a}^{-1} $$Example
Consider the group of integers $\{0, 1, 2, 3\}$ with the modulo-4 addition; also known as group $\mathbb{Z}_{4}$. The multiplication table is
$$ \begin{array}{c|cccc} \mathbb{Z}_{4} & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 2 & 3 & 0 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 0 & 1 & 2 \end{array} $$The subgroup $H = \{ 0, 2 \}$ is invariant. Note that this subgroup is isomorphic to group $Z_{2}$.
$$ \begin{array}{c|cc} H & 0 & 2 \\ \hline 0 & 0 & 2 \\ 2 & 2 & 0 \end{array} \quad \to \quad \begin{array}{c|cc} Z_{2} & e & a \\ \hline e & e & a \\ a & a & e \end{array} $$We can define the set of distinct cosets.
$$ H\cdot 0 = \{ 0, 2 \} \quad H\cdot 1 = \{ 1, 3 \} \quad H \cdot 2 = \{2, 0 \} \quad H\cdot 3 = \{3, 1 \} $$$$ \therefore \quad Q = \left\{ \{ 0, 2 \}, \{ 1, 3 \} \right\} = \mathbb{Z}_{4} / Z_{2} $$We can build the multiplication table for $\mathbb{Z}_{4} / Z_{2}$ by performing the coset product of the element cosets.
$$ \begin{align} \{ 0,2 \} \cdot \{ 0, 2 \} &= \{ 0, 2, 2, 0 \} = \{0, 2 \} \\ \{ 0,2 \} \cdot \{ 1, 3 \} &= \{ 1, 3, 3, 1 \} = \{1, 3 \} \\ \{ 1,3 \} \cdot \{ 0, 2 \} &= \{ 1, 3, 3, 1 \} = \{1, 3 \} \\ \{ 1,3 \} \cdot \{ 1, 3 \} &= \{ 0, 2, 2, 0 \} = \{0, 2 \} \end{align} $$$$ \begin{array}{c|cc} \mathbb{Z}_{4} / Z_{2} & \{0,2\} & \{1,3\} \\ \hline \{0,2\} & \{0, 2\} & \{1,3\} \\ \{1,3\} & \{1,3\} & \{0,2\} \end{array} $$