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If A stabilizes some normal series, then A = 1. Proof: Let 1 = H0 ··· H1 Hr = G be a series, and suppose that φ ∈ A stabilizes this series. Then H1 that if Hi if Hi CG (φ), then Hi+1 CG (φ). We will show CG (φ), proving the result, since if G = CG (φ), φ = 1. 2) Hi+1 /Hi = CHi+1 /Hi (φ) = CHi+1 (φ)Hi /Hi = CHi+1 (φ)/Hi , since Hi CG (φ) ∩ Hi+1 . Thus Hi+1 = CHi+1 (φ) CG (φ), as required. 5 (Thompson’s Critical Subgroup Theorem) Let G be a finite p-group. Then there is a characteristic subgroup C of G such that: (i) every non-trivial p -automorphism of G induces a non-trivial p -automorphism on C; (ii) CG (C) = Z (C); (iii) [G, C] Z (C) (or equivalently [G, C, C] = 1); and (iv) C has class at most 2, and C/ Z (C) is elementary abelian.
If X is cyclic, then G contains a cyclic subgroup of index 2, and preimage of X we easily see that G has maximal class, so we assume that X = x, z is isomorphic with n−3 C2n−2 × C2 . As with the previous theorem, we get x2 four-group, generated by x 2n−3 ∈ Z (G), and so Z (G) is a Klein = y and z. Now consider H = G/ y . This is again of maximal class, and the image of X in H is isomorphic with C2 × C2n−3 . Since no group of maximal class has a non-cyclic abelian subgroup of order 8, this forces n = 4.
Then G is an abelian group of odd order. 2. This implies that xφ = x−1 , and so this map is an automorphism. It is also an anti-automorphism, and so G possesses an automorphism that is also an anti-automorphism; thus G is abelian. 8 (B. Neumann, 1956) Let φ be a fixed-point-free automorphism of G, and suppose that φ has order 3. Then G is nilpotent of class at most 2. 2, we see that xxφ xφ = 1 for all x ∈ G. Let y = x−1 , so that 2 yφ yφy = 1 for all y ∈ G. Therefore 2 [x, xφ ] = yy φ xxφ = xφ xxφ = 1.