Let be a finite group, , and a normal subgroup. Use the conjugacy part of Sylow’s Theorem to prove that is a Sylow -subgroup of . Deduce that is a Sylow -subgroup of .
Let be a Sylow -subgroup of . is a -subgroup of , so that for some . Then since is normal, hence , thus . Now is a subgroup of , and has maximal -power order in , since . Since is a -subgroup of , is a maximal -power subgroup of , thus a Sylow -subgroup.
Note that by the Second Isomorphism Theorem, so that is a -subgroup of . By the Third Isomorphism Theorem, ; since is a maximal -power order subgroup of , does not divide , so does not divide . Hence is a Sylow -subgroup of .