# Centralizer-commutator product decomposition for finite nilpotent groups

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## 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. Stability group of subnormal series of finite group has no other prime factors
2. 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. Commutator of a group and a subgroup of its automorphism group is normal
4. Nilpotent implies center is normality-large
5. 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 $C_A(H) = \pi(C_G(H))$:
1. 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 $g \in G$.
2. $[A,H] = \pi[G,H]$: This follows directly from the definition.
3. 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 $C_G(H) \le L$.
4. 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 $L \le C_G(H)$.
5. Combining the previous two steps yields $L = C_G(H)$.
4. 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 $H$.
5. $G = C_G(H)[G,H]$: 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.

## References

### Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 180, Theorem 3.5, Section 5.3 (p'-automorphisms of p-groups), (proved only for groups of prime power order, but the same proof technique)More info