# The Theory of p-Groups [Lecture notes] by David A. Craven By David A. Craven

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Clearly if φ is fixed-point-free, then φτg is fixed-point-free, where τg is conjugation by g. 2 Therefore xg commutes with xφ and xφ for all g and x, and so x commutes with xg . We now note that G can have no elements of order 3: if it were to, then since x and xφ commute, x, xφ is a φ-invariant elementary abelian of order 9, and contains a fixed point under φ, which is demonstrably impossible. We finish by quoting a standard result. If G is a group such that [g, x, x] = 1 for all g, x ∈ G and G contains no 3-torsion, then G is nilpotent of class at most 2.

Thus CG/N (A) CG (A)N/N. It suffices to show the reverse inclusion; that is, we need to find an element x ∈ Cg (A) lying in every A-invariant coset gN of N . Proceed by induction on the number of factors of |A|, noting that the case where |A| is a prime is clear: for then, gN is split up into orbits of size 1 or p, and the fact that |N | and p are coprime proves that there is an orbit of size 1. The proof of this will be omitted. 2 has an important corollary, for which we first need a defintion.

We define [x, φ] = x−1 (xφ). Notice that [x, φ] = 1 if and only if x ∈ CG (φ), which is analogous to the statement [x, y] = 1 if and only if x ∈ CG (y), which holds for any group. Consider the group G ¯ for the image of X in this semidirect product. We need write X ¯ [x, φ] = [¯ x, φ]. 38 φ again, and The proof of this is in the exercises. Suppose that this is true: what good does it do us? Well, all of the commutator relations that we have will work equally well for commutators involving automorphisms.