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.