Centralizer-commutator product decomposition for finite nilpotent groups

Statement

Suppose $G$ is a finite nilpotent group and $H\leq \operatorname {Aut} (G)$ 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 $K\leq G$ is a $H$-invariant subgroup such that $G=KC_{G}(H)$, then $[G,H]\leq K$.

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 $[G,H]\leq K$.

Proof: We first prove that if $KC_{G}(H)=G$, then $[G,H]\leq K$:

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 $[G,H]\leq Z(G)$

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 $b\sigma (b)^{-1}\in [G,H]\leq Z(G)$, 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, $\pi (C_{G}(H))\leq C_{A}(H))$, so $C_{G}(H)\leq L$.
   4. Consider the subnormal series $\{e\}\leq [G,G]\leq L$ for $L$. $H$ acts as the identity on $[G,G]$ since $[G,G]\leq C_{G}(H)$, 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 $L\leq C_{G}(H)$.
   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 $[G,H]\leq Z(G)$, 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 $\operatorname {ker} \pi =[G,G]\leq C_{G}(H)$, 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 $[[G,H],H]\leq [G,H]$, 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, $[P,H]\leq M=[G,H]\cap Z(G)$, and we know that $M$ is in the center of $G$, hence $M\leq Z(P)$. Thus, $[P,H]\leq Z(P)$, 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 $P\leq G$, we have $[P,H]\leq [G,H]$. Also, $C_{P}(H)\leq C_{G}(H)$. 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}