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

## Statement

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

## Facts used

1. 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. Nilpotence is subgroup-closed
3. Lagrange's theorem
4. 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:
• Since $G$ is nilpotent, fact (2) tells us that $G_1$ is nilpotent.
• 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 $G_1 = [G_1,A]C_{G_1}(A) = G_2C_{G_1}(A)$.
4. $G = G_2C_G(A)$: 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 $G = G_2C_G(A)$.
5. $G_2$ is $A$-invariant: Since $[G_2,A] = G_3 \le G_2$, we in particular have that $G_2$ is $A$invariant.
6. $G_1 \le G_2$: By the preceding two steps, $G_2$ is $A$-invariant and $G = G_2C_G(A)$, so by fact (1), we have $G_1 \le G_2$.
7. $G_1 = G_2$: This follows from the preceding step, combined with step (1).