Centralizer-commutator product decomposition for finite nilpotent groups

Statement
Suppose $$G$$ is a finite nilpotent group and $$H \le \operatorname{Aut}(G)$$ is such that the orders of $$G$$ are relatively prime. Define:


 * $$[G,H]$$ as the subgroup generated by all elements of the form $$g\sigma(g)^{-1}$$ where $$g \in G, \sigma \in H$$.
 * $$C_G(H)$$ as the subgroup of $$G$$ comprising those $$g \in G$$ such that $$\sigma(g) = g$$ for all $$\sigma \in H$$.

Then, we have the following:

$$G = [G,H]C_G(H)$$.

Further, if $$K \le G$$ is a $$H$$-invariant subgroup such that $$G = KC_G(H)$$, then $$[G,H] \le K$$.

Facts used

 * 1) uses::Stability group of subnormal series of finite group has no other prime factors
 * 2) uses::Centralizer-commutator product decomposition for Abelian groups: This states that the result holds when $$G$$ is Abelian (in fact, something stronger holds: $$G$$ is an internal direct product of the two subgroups).
 * 3) uses::Commutator of a group and a subgroup of its automorphism group is normal
 * 4) uses::Nilpotent implies center is normality-large
 * 5) uses::Nilpotence is quotient-closed

Proof
Given: A finite nilpotent group $$G$$, a subgroup $$H$$ of $$\operatorname{Aut}(G)$$ such that the orders of $$G$$ and $$H$$ are relatively prime.

To prove: $$[G,H]C_G(H) = G$$ and if $$KC_G(H) = G$$, then $$[G,H] \le K$$.

Proof: We first prove that if $$KC_G(H) = G$$, then $$[G,H] \le K$$:

Pick any $$g \in G, \sigma \in H$$. We want to show that $$g\sigma(g)^{-1} \in K$$.

Since $$G = KC_G(H)$$, we have $$g = ab$$, where $$a \in K, b \in C_G(H)$$. Thus, we have:

$$g\sigma(g)^{-1} = (ab)\sigma(ab)^{-1} = ab\sigma(b)^{-1}\sigma(a)^{-1}$$.

Since $$b \in C_G(H)$$, $$\sigma(b) = b$$, so we get:

$$g\sigma(g)^{-1} = a\sigma(a)^{-1}$$.

Since $$a \in K$$ and $$K$$ is $$H$$-invariant, $$\sigma(a) \in K$$, so $$a\sigma(a)^{-1} \in K$$, completing the proof.

We now prove that $$[G,H]C_G(H) = G$$.

Case that $$[G,H] \le Z(G)$$
We first consider the case that $$[G,H]$$ is contained in the center of $$G$$.


 * 1) The map $$\alpha_\sigma: g \mapsto g\sigma(g)^{-1}$$ is an endomorphism of $$G$$ for any $$\sigma \in H$$: For any $$a,b \in G$$, we have $$ab(\sigma(ab))^{-1} = ab\sigma(b)^{-1}\sigma(a)^{-1}$$. Since $$b\sigma(b)^{-1} \in [G,H] \le Z(G)$$, we can commute $$b\sigma(b)^{-1}$$ with $$\sigma(a)^{-1}$$ to get $$ab(\sigma(ab))^{-1} = (a\sigma(a)^{-1})(b\sigma(b)^{-1})$$. This proves the condition for being a homomorphism.
 * 2) $$C_G(H)$$ contains the subgroup $$[G,G]$$: Note that $$C_G(H)$$ can be described as the intersection of kernels of $$\alpha_\sigma$$ for all $$\sigma \in H$$. Each $$\alpha_\sigma$$ has image inside the center of $$G$$ by definition, hence is a map to an Abelian group. Thus, the kernel of each $$\alpha_\sigma$$ contains the commutator subgroup $$[G,G]$$. Hence, the intersection, which is $$C_G(H)$$, also contains $$[G,G]$$.
 * 3) Let $$A = G/[G,G]$$ and $$\pi:G \to A$$ be the quotient map. Then, $$H$$ acts on $$A$$ by the natural induced action, $$[A,H] = \pi[G,H]$$, and $$C_A(H) = \pi(C_G(H))$$:
 * 4) Observe that since $$[G,G]$$ is characteristic in $$G$$, it is $$H$$-invariant, so $$H$$ acts on the quotient group. Note that this descent satisfies $$\pi(\sigma(g)) = \sigma(\pi(g))$$ for all $$g \in G$$.
 * 5) $$[A,H] = \pi[G,H]$$: This follows directly from the definition.
 * 6) Now, suppose $$L = \pi^{-1}(C_A(H))$$. Clearly, if $$g \in C_G(H)$$, then $$\sigma(g) = g$$ for all $$\sigma \in H$$. Thus, $$\sigma(\pi(g)) = \pi(\sigma(g)) = \pi(g)$$, so $$\pi(g) \in C_A(H)$$. Thus, $$\pi(C_G(H)) \le C_A(H))$$, so $$C_G(H) \le L$$.
 * 7) Consider the subnormal series $$\{ e \} \le [G,G] \le L$$ for $$L$$. $$H$$ acts as the identity on $$[G,G]$$ since $$[G,G] \le C_G(H)$$, and it acts as the identity on $$L/[G,G]$$ since $$L/[G,G] = \pi(L) = C_A(H)$$. Thus, $$H$$ acts as stability automorphisms of this subnormal series, and its order is relatively prime to $$G$$, and hence to $$L$$. Fact (1) thus forces that $$H$$ acts as the identity on $$L$$, so $$L \le C_G(H)$$.
 * 8) Combining the previous two steps yields $$L = C_G(H)$$.
 * 9) We have $$A = C_A(H) \times [A,H]$$: This follows from fact (2), and the observation that since the orders of $$G$$ and $$H$$ are relatively prime, so are the orders of $$A$$ and $$H$$.
 * 10) $$G = C_G(H)[G,H]$$: From steps (3) and (4), we have $$A = \pi(C_G(H))\pi([G,H]) = \pi(C_G(H)[G,H])$$. Note that $$C_G(H)[G,H]$$ is a subgroup, since $$[G,H] \le Z(G)$$, hence is normal. Thus, $$C_G(H)[G,H]$$ is a subgroup of $$G$$ whose image via $$\pi$$ is the whole quotient $$A$$. On the other hand, step (2) says that $$\operatorname{ker}\pi = [G,G] \le C_G(H)$$, so $$C_G(H)[G,H]$$ contains the kernel $$[G,G]$$ and intersects every coset of it -- hence must be equal to the whole group $$G$$.

Case that $$[G,H]$$ is not contained in the center $$Z(G)$$

 * 1) $$[G,H]$$ is normal in $$G$$: This follows from fact (3).
 * 2) $$[G,H]$$ intersects the center $$Z(G)$$ nontrivially: This follows from fact (4), and the fact that $$[G,H]$$ is normal.
 * 3) Let $$M = [G,H] \cap Z(G)$$. Then, $$M$$ is nontrivial, $$H$$-invariant, and normal: $$Z(G)$$ is characteristic, and hence $$H$$-invariant, while we have $$[[G,H],H] \le [G,H]$$, so $$[G,H]$$ is $$H$$-invariant. Thus, the intersection is $$H$$invariant.  Further, $$M$$ is normal since it is the intersection of the normal subgroups $$[G,H]$$ and $$Z(G)$$ (or alternatively, is a subgroup of $$Z(G)$$).
 * 4) Let $$N = G/M$$, and $$\rho:G \to N$$ be the quotient map. Then, the action of $$H$$ on $$G$$ descends to an action on $$N$$: This follows from step (3), where it is observed that $$M$$ is normal and $$H$$-invariant.
 * 5) For the induced action of $$H$$ on $$N$$, we have $$N = [N,H]C_N(H)$$: Since $$N$$ is the quotient of $$G$$ by a nontrivial subgroup, it has strictly smaller order. Since $$G$$ is nilpotent, so is $$N$$ (fact (5)). Thus, the induction hypothesis applies to $$N$$.
 * 6) Let $$P = \rho^{-1}(C_N(H))$$. Then, $$G = [G,H]P$$: Clearly, we have $$\rho([G,H]P) = \rho([G,H])\rho(P) = \rho([G,H])C_N(H) = [N,H]C_N(H) = N$$. Thus, $$\rho([G,H]P) = N$$, while $$[G,H]$$ also contains the kernel of $$\rho$$. Hence, $$[G,H]P$$ is a subgroup (it is one because $$[G,H]$$ is normal, step (1)) containing the kernel of $$\rho$$ and intersecting every coset of this kernel, forcing $$[G,H]P = G$$.
 * 7) $$P = [P,H]C_P(H)$$: Note that since $$\rho(P) = C_N(H)$$, we have $$\rho([P,H]) = [\rho(P),H] = [C_N(H),H]$$ which is trivial. Thus, $$[P,H] \le M = [G,H] \cap Z(G)$$, and we know that $$M$$ is in the center of $$G$$, hence $$M \le Z(P)$$. Thus, $$[P,H] \le Z(P)$$, and the previous case kicks in for $$P$$.
 * 8) $$G = [G,H][P,H]C_P(H)$$: This follows by piecing together the previous two steps.
 * 9) $$G = [G,H]C_G(H)$$: Since $$P \le G$$, we have $$[P,H] \le [G,H]$$. Also, $$C_P(H) \le C_G(H)$$. Substituting these gives the required result.