Commutator of finite nilpotent group with coprime automorphism group equals second commutator

Statement
Let $$G$$ be a fact about::finite nilpotent group and $$A$$ be a subgroup of $$\operatorname{Aut}(G)$$ whose order is relatively prime to the order of $$G$$. Then, we have:

$$[[G,A],A] = [G,A]$$,

where:

$$[G,A] = \langle g\sigma(g)^{-1} \mid g \in G, \sigma \in A \}$$.

Related facts

 * Commutator of finite group with cyclic coprime automorphism group equals second commutator
 * Commutator of finite group with coprime automorphism group equals second commutator
 * Centralizer-commutator product decomposition for abelian groups
 * Centralizer-commutator product decomposition for finite nilpotent groups
 * Centralizer-commutator product decomposition for finite groups and cyclic automorphism group
 * Centralizer-commutator product decomposition for finite groups

Facts used

 * 1) uses::Centralizer-commutator product decomposition for finite nilpotent groups: This states that under the same conditions ($$G$$ a finite nilpotent group and $$A$$ a subgroup of $$\operatorname{Aut}(G)$$ of order coprime to the order of $$G$$), we have $$G = [G,A]C_G(A)$$, and further, if $$K$$ is an $$A$$-invariant subgroup of $$G$$ such that $$G = KC_G(A)$$, then $$[G,A] \le K$$.
 * 2) uses::Nilpotence is subgroup-closed
 * 3) uses::Lagrange's theorem
 * 4) uses::Order of quotient group divides order of group

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

To prove: $$[[G,A],A] = [G,A]$$.

Proof: Let $$G_1 = [G,A], G_2 = [G_1,A], G_3 = [G_2,A]$$.


 * 1) $$G_3 \le G_2 \le G_1 \le G$$: Since $$G_1 \le G$$, we have $$[G_1,A] \le [G,A]$$, so $$G_2 \le G_1$$. Thus, $$[G_2,A]\le [G_1,A]$$, so $$G_3 \le G_2$$. Thus, $$G_3 \le G_2 \le G_1 \le G$$.
 * 2) $$G = G_1C_G(A)$$: This follows directly from fact (1).
 * 3) $$G_1 = G_2C_{G_1}(A)$$: This follows by applying fact (1), replacing $$G$$ by $$G_1$$, and observing that:
 * 4) * Since $$G$$ is nilpotent, fact (2) tells us that $$G_1$$ is nilpotent.
 * 5) * Since $$G$$ has order relatively prime to that of $$A$$, the order of $$G_1$$ is also relatively prime to $$A$$ (using fact (3)). Further, since $$G_1$$ is $$A$$-invariant, there is a homomorphism $$A \to \operatorname{Aut}(G_1)$$, with image $$A_1$$ of order dividing the order of $$A$$, and so that the action of $$A$$ on $$G_1$$ factors through this homomorphism. The order of $$A_1$$ divides the order of $$A$$ by fact (4), so we get the conditions for fact (1), yielding $$G_1 = [G_1,A_1]C_{G_1}(A_1)$$. Finally, since the action of $$A$$ factors via the homomorphism, we get $$G_1 = [G_1,A]C_{G_1}(A) = G_2C_{G_1}(A)$$.
 * 6) $$G = G_2C_G(A)$$: By the two preceding steps, $$G = G_2C_{G_1}(A)C_G(A)$$. But $$C_{G_1}(A) \le C_G(A)$$, so we get $$G = G_2C_G(A)$$.
 * 7) $$G_2$$ is $$A$$-invariant: Since $$[G_2,A] = G_3 \le G_2$$, we in particular have that $$G_2$$ is $$A$$invariant.
 * 8) $$G_1 \le G_2$$: By the preceding two steps, $$G_2$$ is $$A$$-invariant and $$G = G_2C_G(A)$$, so by fact (1), we have $$G_1 \le G_2$$.
 * 9) $$G_1 = G_2$$: This follows from the preceding step, combined with step (1).