Let G be a finite group in which a Sylow2 subgroup is cyclic

Let G be a finite group in which a Sylow-2 subgroup is cyclic. Prove that there exists a normal subgroup N of odd order such that the index [G:N] is a power of 2.

Solution

A) Every 2-Sylow subgroup of G is cyclic (they are isomorphic to each other).                                              Let P₂ be a 2-Sylow subgroup of G, with generator h0, so h₀ has order 2ⁿ . First we will show G has a subgroup of index 2, which is necessarily normal since all subgroups of index 2 are normal.                                                                                         Consider the action of G on G by left multiplication. This action is a group homomorphism ` : G Sym(G).                                                                       Let’s look in particular at `(h0), the left multiplication by h0 on G. Decompose G into m right H-cosets Hg1, . . . , Hgm. Each Hgi = {gi , h₀gi , h² 0 gi , . . . ,    h^{2 n1} 0 gi} is an orbit for left multiplication by h0 on G, so the left multiplication by h₀ on G is a permutation that consists of m different cycles of length 2n . If |G| = 2m where m is odd then G contains a normal subgroup of size m and all elements of order 2 in G are conjugate to each other. B) A 2-Sylow subgroup of G has size 2, which must be cyclic.   there is a normal subgroup of size m. Since the 2-Sylow subgroups of G have size 2, any two elements of order 2 generate conjugate subgroups by Sylow II, and therefore the elements themselves are conjugate.