Examples of Quotient Groups

Example 1: If H be 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 is G.
Let a be a generator of G and {a^n} be any element of G, where n is some integer. Then H{a^n} is any element of G|H.
Also, it can be proved easily 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} be the symmetric and alternating groups on 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} be the symmetric and alternating groups on 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.

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