Prove that if where is odd and has a cyclic Sylow 2-subgroup then has a normal subgroup of order .
We proceed by induction on .
For the base case, suppose where is odd. Note that every Sylow 2-subgroup is cyclic. By this previous exercise, has a subgroup of index 2, hence order , which is necessarily normal.
For the inductive step, suppose that for some and all odd , any group of order which has a cyclic Sylow 2-subgroup has a normal subgroup of order . Let be a group of order and with a cyclic Sylow 2-subgroup. Let be a generator of the Sylow 2-subgroup; i.e. . Let be the left regular representation of ; by this previous exercise, is an odd permutation. By this previous exercise, has a subgroup of index 2, which is necessarily normal in . Note that all Sylow 2-subgroups of are cyclic, and that every Sylow 2-subgroup of is contained in a Sylow 2-subgroup of . Thus has a cyclic Sylow 2-subgroup, in particular since has a unique subgroup of order . By the induction hypothesis, has a normal subgroup of order . Note that , and that since is maximal, .
We claim that is the unique subgroup of of order . To prove this claim, suppose to the contrary that is a subgroup of order . Now is a subgroup and strictly contains and . Moreover, we have , so that by Lagrange, is odd. This is a contradiction, though, because is the largest odd integer dividing . Thus is the unique subgroup of order . Hence is characteristic in . By this previous exercise, then, is normal in .
By induction, the conclusion holds for all groups of order having a cyclic Sylow 2-subgroup.