# Examples of Quotient Groups

Example 1: If $H$ is a normal subgroup of a finite group $G$, then prove that
$o\left( {G|H} \right) = \frac{{o\left( G \right)}}{{o\left( H \right)}}$

Solution: $o\left( {G|H} \right) =$ number of distinct right (or left) cosets of $H$ in $G$, as $G|H$ is the collection of all right (or left) cosets of $H$ in $G$
$= \frac{{{\text{number}}\,{\text{of}}\,{\text{distinct}}\,{\text{elements}}\,{\text{in}}\,{\text{G}}}}{{{\text{number}}\,{\text{of}}\,{\text{distinct}}\,{\text{elements}}\,{\text{in}}\,{\text{H}}}}$
$= \frac{{o\left( G \right)}}{{o\left( H \right)}}$      by Lagrange’s Theorem

Example 2: Show that every quotient group of a cyclic group is cyclic, but not conversely.

Solution:
Let $H$ be a subgroup of a cyclic group $G$. Then $H$ is also cyclic because every cyclic group is abelian. Therefore $H$ is a normal subgroup of $G$.

Let $a$ be a generator of $G$ and ${a^n}$ be any element of $G$, where $n$ is an integer. Then $H{a^n}$ is any element of $G|H$.

Also, it can be easily proved that ${\left( {Ha} \right)^n} = H{a^n}$ for every integer $n$. Therefore, $G|H$ is cyclic and its generator is $Ha$.

Its converse is not true; for example if ${P_3}$ and ${A_3}$ are the symmetric and alternating groups of the three symbols $a,b,c$ then the quotient group ${P_3}|{A_3}$ is cyclic, whereas ${P_3}$ is not.

Example 3: Show that every quotient group of an abelian group is abelian but its converse is not true.

Solution:
Let $a,b \in G$ be arbitrary, then $Ha,Hb$ are any two elements of the quotient group $G|H$. Then we have
$\begin{gathered} \left( {Ha} \right)\left( {Hb} \right) = Hab = Hba \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {Hb} \right)\left( {Ha} \right) \\ \end{gathered}$

Therefore, $G|H$ is an abelian.

Its converse is not true; for example if ${P_3}$ and ${A_3}$ are the symmetric and alternating groups of the three symbols $a,b,c$ then the quotient group ${P_3}|{A_3}$ being of order 2 is abelian whereas ${P_3}$ is not.