Theorem 6.22: Let H; be a normal subgroup of the group G; * Then g/H; o is a group. Proof: associative Identity element: Let e be identity element of g ◆He=H∈G/ H is identity element of G/H ◆ Inverse element: For vha∈G/H,HaeG/H is inverse element of a where aleg is inverse element of a Theorem 6.22: Let [H;] be a normal subgroup of the group [G;]. Then [G/H;] is a group.  Proof: associative  Identity element: Let e be identity element of G.  He=HG/H is identity element of G/H  Inverse element: For HaG/H, Ha-1G/H is inverse element of Ha, where a-1G is inverse element of a