Nilpotent residual and Fitting subgroup of fixed points in finite groups

TL;DR

This paper investigates how fixed points of automorphisms in finite groups influence the group's structure, establishing bounds on certain subgroups based on properties of centralizers under automorphisms.

Contribution

It provides new bounds on the nilpotent residual and Fitting subgroup of a finite group based on automorphism fixed points, extending understanding of automorphism actions.

Findings

Bound on the order of the nilpotent residual of G in terms of fixed point properties.

Bound on the index of the second Fitting subgroup based on centralizer properties.

Results depend solely on a parameter m related to fixed point sizes.

Abstract

Let $q$ be a prime and $A$ a finite $q$-group of exponent $q$ acting by automorphisms on a finite $q'$-group $G$. Assume that $A$ has order at least $q^3$. We show that if $\gamma_{\infty} (C_{G}(a))$ has order at most $m$ for any $a \in A^{\#}$, then the order of $\gamma_{\infty} (G)$ is bounded solely in terms of $m$. If the Fitting subgroup of $C_{G}(a)$ has index at most $m$ for any $a \in A^{\#}$, then the second Fitting subgroup of $G$ has index bounded solely in terms of $m$.

Equations

$$\{g\in G|g^a=g\text{for all}a\in A\},$$

$$\tilde{V}=V_1>V_2>\cdots V_n>V_{n+1}=0.$$

$$W=W_1\oplus \cdots \oplus W_t$$

Full text

1. Introduction

Suppose that a finite group $A$ acts by automorphisms on a finite group $G$. The action is coprime if the groups $A$ and $G$ have coprime orders. We denote by $C_{G}(A)$ the set

[TABLE]

the centralizer of $A$ in $G$ (the fixed-point subgroup). In what follows we denote by $A^{\#}$ the set of nontrivial elements of $A$. It has been known that centralizers of coprime automorphisms have strong influence on the structure of $G$.

Ward showed that if $A$ is an elementary abelian $q$-group of rank at least 3 and if $C_{G}(a)$ is nilpotent for any $a\in A^{\#}$, then the group $G$ is nilpotent [13]. Later Shumyatsky showed that if, under these hypotheses, $C_{G}(a)$ is nilpotent of class at most $c$ for any $a\in A^{\#}$, then the group $G$ is nilpotent with $(c,q)$-bounded nilpotency class [10]. Throughout the paper we use the expression “$(a,b,\dots)$-bounded” to abbreviate “bounded from above in terms of $a,b,\dots$ only”. In the recent article [3] the above result was extended to the case where $A$ is not necessarily abelian. Namely, it was shown that if $A$ is a finite group of prime exponent $q$ and order at least $q^{3}$ acting on a finite $q^{\prime}$-group $G$ in such a manner that $C_{G}(a)$ is nilpotent of class at most $c$ for any $a\in A^{\#}$, then $G$ is nilpotent with class bounded solely in terms of $c$ and $q$. Many other results illustrating the influence of centralizers of automorphisms on the structure of $G$ can be found in [9].

Recall that the nilpotent residual $\gamma_{\infty}(K)$ of a group $K$ is the last term of the lower central series of $K$. It can also be defined as the intersection of all normal subgroups of $K$ whose quotients are nilpotent. Also, recall that the second Fitting subgroup $F_{2}(G)$ of a finite group $G$ is defined as the inverse image of $F(G/F(G))$, that is, $F_{2}(G)/F(G)=F(G/F(G))$. Here $F(G)$ stands for the Fitting subgroup of $G$.

Recently, in [4] it was proved that if $A$ is an elementary abelian group of order at least $q^{3}$ acting by automorphisms on a finite $q^{\prime}$-group $G$ and if $|\gamma_{\infty}(C_{G}(a))|\leq m$ for any $a\in A^{\#}$, then the order of $\gamma_{\infty}(G)$ is $m$-bounded. If $F(C_{G}(a))$ has index at most $m$ in $C_{G}(a)$ for any $a\in A^{\#}$, then the index of $F_{2}(G)$ is $m$-bounded.

In the present article we extend the results obtained in [4] to the case where $A$ is not necessarily abelian.

We obtain the following results.

Theorem 1.1**.**

Let $q$ be a prime and $m$ a positive integer. Let $A$ be a finite $q$-group of exponent $q$ acting by automorphisms on a finite $q^{\prime}$-group $G$. Assume that $A$ has order at least $q^{3}$ and $|\gamma_{\infty}(C_{G}(a))|\leq m$ for any $a\in A^{\#}$. Then $|\gamma_{\infty}(G)|$ is $m$-bounded.

Theorem 1.2**.**

Let $q$ be a prime and $m$ a positive integer. Let $A$ be a finite $q$-group of exponent $q$ acting by automorphisms on a finite $q^{\prime}$-group $G$. Assume that $A$ has order at least $q^{3}$ and $F(C_{G}(a))$ has index at most $m$ in $C_{G}(a)$ for any $a\in A^{\#}$. Then the index of $F_{2}(G)$ is $m$-bounded.

In the next section we give some helpful lemmas that will be used in the proofs of the above results. Section 3 deals with the proof of Theorem 1.2. In Section 4 we prove Theorem 1.1.

2. Preliminaries

If $A$ is a group of automorphisms of a group $G$, the subgroup generated by elements of the form $g^{-1}g^{\alpha}$ with $g\in G$ and $\alpha\in A$ is denoted by $[G,A]$. The subgroup $[G,A]$ is an $A$-invariant normal subgroup in $G$. Our first lemma is a collection of well-known facts on coprime actions (see for example [7]). Throughout the paper we will use it without explicit references.

Lemma 2.1**.**

Let $A$ be a group of automorphisms of a finite group $G$ such that $(|G|,|A|)=1$. Then

1. i)

$G=[G,A]C_{G}(A)$. 2. ii)

$[G,A,A]=[G,A]$. 3. iii)

$A$* leaves invariant some Sylow $p$-subgroup of $G$ for each prime $p\in\pi(G)$.* 4. iv)

$C_{G/N}(A)=C_{G}(A)N/N$* for any $A$-invariant normal subgroup $N$ of $G$.* 5. v)

If $A$ is a noncyclic elementary abelian group and $A_{1},\dots,A_{s}$ are the maximal subgroups in $A$, then $G=\langle C_{G}(A_{1}),\ldots,C_{G}(A_{s})\rangle$. Furthermore, if $G$ is nilpotent, then $G=\prod_{i}C_{G}(A_{i})$.

Let $\varphi$ be an automorphism of prime order $q$ of a finite group $G$ and assume that $|C_{G}(\varphi)|=m$. In [6] Fong proved using the classification of finite simple groups that $G$ has a soluble subgroup of $(q,m)$-bounded index. Later Hartley and Meixner in [8] proved that if $G$ is soluble, then $G$ contains a nilpotent subgroup of $(q,m)$-bounded index. And lastly, it was shown by Khukhro that if $G$ is nilpotent, then $G$ has a subgroup of $(q,m)$-bounded index which is nilpotent of $q$-bounded class (see, for example, [9, Theorem 5.3.1]). Combining this results we get the following theorem.

Theorem 2.2**.**

[9, Corollary 5.4.1]** If $\varphi$ is an automorphism of prime order $q$ of a finite group $G$ and $|C_{G}(\varphi)|=m$, then $G$ contains a nilpotent subgroup of $(q,m)$-bounded index whose nilpotency class is $q$-bounded.

The following lemma is an application of the three subgroup lemma.

Lemma 2.3**.**

Let $A$ be a group of automorphisms of a finite group $G$ and let $N$ be a normal subgroup of $G$ contained in $C_{G}(A)$. Then $[[G,A],N]=1$. In particular, if $G=[G,A]$, then $N\leq Z(G)$.

Proof.

Indeed, by the hypotheses, $[N,G,A]=[A,N,G]=1$. Thus, $[G,A,N]=1$ and the lemma follows. ∎

We finish this section with some results about elementary abelian groups acting as automorphisms.

Lemma 2.4**.**

[10, Lemma 2.6]** Let $q$ be a prime and $A$ be an elementary abelian group of order $q^{2}$ acting on a finite $q^{\prime}$-group $G$. Let $C=\cap_{a\in A^{\#}}F(C_{G}(a))$. Then $C\leq F(G)$.

Lemma 2.5**.**

[2, Lemma 3.1]** Let $q$ be a prime and $m$ a positive integer. Let $A$ be an elementary abelian group of order $q^{2}$ acting on a finite $q^{\prime}$-group $G$ in such a manner that $|C_{G}(a)|\leq m$ for any $a\in A^{\#}$. Then the order of $G$ is $m$-bounded.

We denote by $R(H)$ the soluble radical of a finite group $H$, that is, the largest normal soluble subgroup of $H$.

Theorem 2.6**.**

[4, Theorem 2.5]** Let $q$ be a prime and $m$ a positive integer such that $m<q$. Let $A$ be an elementary abelian group of order $q^{2}$ acting on a finite $q^{\prime}$-group $G$ in such a way that the index of $R(C_{G}(a))$ in $C_{G}(a)$ is at most $m$ for any $a\in A^{\#}$. Then $[G,A]$ is soluble.

3. Proof of Theorem 1.2

Let $F(G)$ denote the Fitting subgroup of a group $G$. Write $F_{0}(G)=1$ and let $F_{i+1}(G)$ be the inverse image of $F(G/F_{i}(G))$ for any $i=0,1,\ldots$. If $G$ is soluble, the least number $h$ such that $F_{h}(G)=G$ is called the Fitting height $h(G)$ of $G$.

Let us now assume the hypothesis of Theorem 1.2. Thus, $A$ is a finite group of prime exponent $q$ and order at least $q^{3}$ acting on a finite $q^{\prime}$-group $G$ in such a manner that $F(C_{G}(a))$ has index at most $m$ in $C_{G}(a)$ for any $a\in A^{\#}$. We wish to show that $F_{2}(G)$ has $m$-bounded index in $G$. If $A$ contains an abelian subgroup of order $p^{3}$, the result is immediate from [4]. Therefore without loss of generality we assume that all subgroups of order $p^{3}$ in $A$ are non-abelian. Clearly, $A$ must contain subgroups of order $p^{3}$. Thus, replacing if necessary $A$ by one of its subgroups of order $p^{3}$ we may assume that $A$ is an extra-special group of order $p^{3}$. As usual, we denote by $A^{\prime}$ the commutator subgroup of $A$. Since $A$ is extra-special, $A^{\prime}=Z(A)$ and $|Z(A)|=p$.

First, using Theorem 2.2 we prove that $F_{2}(G)$ has $(q,m)$-bounded index in $G$. In what follows we denote by $B$ the subgroup $Z(A)$. Note that $B$ is contained in all subgroups of order $p^{2}$ of $A$. Moreover, since $B=Z(A)$ we have that all centralizers $C_{G}(a)$ are $B$-invariants and $C_{G}(B)$ is $A$-invariant.

Lemma 3.1**.**

The index of $F_{2}(G)$ is $(q,m)$-bounded.

Proof.

Let $A_{1},\ldots,A_{q+1}$ be the subgroups of order $q^{2}$ of $A$. Note that $C_{G}(B)$ admits the action of the quotient-group $\overline{A}=A/B$ which is an elementary abelian group of order $q^{2}$. Thus, $C_{G}(B)=\langle C_{C_{G}(B)}(\overline{a})\ |\ \overline{a}\in\overline{A}^{\#}\rangle$. An alternative way of expressing this is to say that $C_{G}(B)$ is generated by the subgroups $C_{G}(A_{i})$ (see for example [3, Lemma 2.3]) .

Now, for any subgroup $A_{i}$ set $C_{i}=\cap_{a\in A_{i}^{\#}}F(C_{G}(a))$. The subgroup $C_{i}$ has $(q,m)$-bounded index in $C_{G}(A_{i})$ and by Lemma 2.4 we have that $C_{i}\leq F(G)$. Let $\overline{G}=G/F(G)$. Then the image of each subgroup $C_{G}(A_{i})$ in $\overline{G}$ has $(q,m)$-bounded order and so using Lemma 2.5 we obtain that $C_{\overline{G}}(B)$ has $(q,m)$-bounded order. Therefore by Theorem 2.2 $F(\overline{G})$ has $(q,m)$-bounded index and the proof is complete. ∎

The above lemma says that if $q\leq m$, then $F_{2}(G)$ has $m$-bounded index. We will therefore assume that $q>m$. In this case, $B$ acts trivially on $C_{G}(a)/F(C_{G}(a))$ for any $a\in A^{\#}$ and so $[C_{G}(a),B]\leq F(C_{G}(a))$ for any $a\in A^{\#}$.

Now our main goal is to show that $[G,B]$ is nilpotent.

Lemma 3.2**.**

Let $A_{i}$ be a subgroup of $A$ of order $q^{2}$ containing $B$ and suppose that $G=[G,B]$. Then $G$ is generated by the subgroups $[C_{G}(a),B]$ where $a\in A_{i}\setminus B$.

Proof.

Since $A_{i}$ is an elementary abelian group of order $p^{2}$ we have that $G=\langle C_{G}(a)\ |\ a\in A_{i}^{\#}\rangle$. If $G$ is abelian, the result is obvious since the subgroups $C_{G}(a)$ are $B$-invariants. If $G$ is nilpotent, the result can be seen by considering the action of $A$ on the abelian group $G/\Phi(G)$. Finally, the general case follows from the nilpotent case and Lemma 2.4 of [11]. ∎

Lemma 3.3**.**

The subgroup $[G,B]$ is nilpotent.

Proof.

We argue by contradiction. By Lemma 2.6, the subgroup $[G,B]$ is soluble. Suppose $G$ is a counterexample of minimal possible order. Let $V$ be a minimal $A$-invariant normal subgroup of $G$. Then $V$ is an elementary abelian $p$-group and $G/V$ is an $r$-group for some primes $r\neq p$. Write $G=VH$ where $H$ is an $A$-invariant Sylow $r$-subgroup such that $H=[H,B]$. By the minimality we also have $V=[V,H]$, so that $C_{V}(H)=1$.

Let $A_{1},\ldots,A_{q+1}$ be the subgroups of order $q^{2}$ containing $B$. For each $a\in A^{\#}$ we denote by $V_{a}$ and $H_{a}$ the centralizers $C_{V}(a)$ and $C_{H}(a)$ respectively. It is clear that $V_{a}$ is normal in $C_{G}(a)$. Let $E_{a}=[H_{a},B]$. Note that $V_{a}E_{a}\leq F(C_{G}(a))$ since $[C_{G}(a),B]\leq F(C_{G}(a))$. By Lemma 3.2 we have that $H=\langle E_{x}\ |\ x\in A_{i}\setminus B\rangle$ for any $A_{i}$. Then $[C_{V}(A_{i}),H]=1$ for any $A_{i}$. Since $C_{G}(B)$ is generated by the subgroups $C_{G}(A_{i})$ we obtain that $[C_{V}(B),H]=1$. So by the minimality of $G$ we have that $C_{V}(B)=1$. Note that $E_{a}$ centralizes $C_{V}(a)$ for any $a\in A^{\#}$ but there exists $a\in A^{\#}$ such that $E_{a}$ acts nontrivially on $V$. Our aim is a contradiction following from these assumptions.

Now, we regard $V$ as an $\mathbb{F}_{p}HA$-module and extend the ground field to a finite field $k$ that is a splitting field for $HA$. We now obtain a $kHA$-module $\widetilde{V}=V\otimes_{\mathbb{F}_{p}}k$. Many of the above-mentioned properties of $V$ are inherited by $\widetilde{V}$. In particular, $C_{\widetilde{V}}(H)=0$ and $E_{a}$ centralizes $C_{\widetilde{V}}(a)$ for any $a\in A^{\#}$.

Consider an unrefinable series of $kHA$-submodules

[TABLE]

Let $W$ be one of the factors of this series. It is a nontrivial irreducible $kHA$-module. If $c\in H$ acts trivially on every such $W$, then $c$ acts trivially on $\widetilde{V}$, as the order of $H$ is coprime to the characteristic $p$ of the field $k$. Therefore without loss of generality we assume that $H$ does not centralizes $W$.

By Clifford’s theorem [7, Theorem 3.4.1], $W$ is the direct sum

[TABLE]

of Wedderburn components $W_{i}$ with respect to $H$, which are $kH$-modules transitively permuted by $A$. Furthermore, on each of the $W_{i}$ the center of $H$ is represented by scalar multiplication.

We denote by $\Omega$ the set of Wedderburn components $\{W_{1},\ldots,W_{t}\}$. Since $H$ does not centralizes $W$ we can choose a Wedderburn component $W_{1}$ on which $H$ acts non-trivially. Thus, in order to get a contradiction it is sufficient to prove that $H$ acts trivially on such $W_{1}$.

Lemma 3.4**.**

Let $a\in A$. The subgroup $H$ acts trivially on the sum of components in any regular $\langle a\rangle$-orbit in $\Omega$.

Proof.

Let $e\in E_{a}$ and consider the regular $\langle a\rangle$-orbit $\{W_{j},W_{j}^{a},\ldots,W_{j}^{a^{q-1}}\}$. Note that $a\not\in B$ since $C_{V}(B)=1$. Let $A_{1}=\langle a,B\rangle$. Of course $|A_{1}|=q^{2}$.

The element $e$ leaves invariant all components $W_{j}^{a^{i}}$ since $e\in H$. On the other hand, $e$ fixes every element of the form $w_{j}+w_{j}^{a}+\cdots+w_{j}^{a^{q-1}}$ which belongs to $C_{W}(a)$. Hence $e$ acts trivially on all components $W_{j}^{a^{i}}$.

Now, note that for any $x\in A_{1}\setminus B$ the $\langle x\rangle$-orbit $\{W_{j},W_{j}^{x},\ldots,W_{j}^{x^{q-1}}\}$ is also regular since $C_{V}(B)=1$ (each component $W_{j}$ must be $B$-invariant). Hence, any element of $E_{x}$ acts trivially on $W_{j}$. Then $H$ acts trivially on all components $W_{j}^{a^{i}}$ since $H$ is generated by the subgroups $E_{x}$ such that $x\in A_{1}\setminus B$. ∎

By the above lemma we conclude that $W_{1}^{a}=W_{1}$ for any $a\in A^{\#}$. Recall that $A$ is not abelian and $B=A^{\prime}=Z(A)$. Let $A_{1}$ be a subgroup of order $q^{2}$ containing $B$. The centralizers $C_{G}(x)$ such that $x\in A_{1}\setminus B$ are permuted by $A$ since $C_{G}(x)^{a}=C_{G}(x^{a})$ and $x^{a}=x[x,a]$. Then $C_{W_{1}}(a)\neq 0$ for any $a\in A_{1}\setminus B$ since $W_{1}=\langle C_{W_{1}}(x)\ |\ x\in A_{1}\setminus B\rangle$ and the subgroups $C_{W_{1}}(x)$ are permuted by $A$.

By the minimality we have that $[N,B]=1$ for any proper normal subgroup of $H$. Then $H$ is a special $r$-group, that is, $H$ is an elementary abelian group or has nilpotency class 2 and $H^{\prime}=Z(H)=\Phi(H)$ is elementary abelian.

In order to prove that $H$ acts trivially on $W_{1}$ it suffices to prove that each subgroup $E_{a}$ acts trivially on $W_{1}$. Let $a\in A^{\#}$ such that $E_{a}$ does not act trivially on $W_{1}$. If $E_{a}$ is not abelian, then $E_{a}^{{}^{\prime}}\leq Z(H)$ and acts trivially on $C_{W_{1}}(a)$. Therefore, $E_{a}^{{}^{\prime}}$ acts trivially on $W_{1}$ since $Z(H)$ acts by scalar multiplication on $W_{1}$. In fact, if $E_{a}$ is not abelian, then $E_{x}$ is not abelian for any $x\in\langle a,B\rangle\setminus B$ since such subgroups are permuted by $A$. Moreover, since $C_{W_{1}}(x)\neq 0$ for any $x\in\langle a,B\rangle\setminus B$ we have that $Z=\langle E_{x}^{{}^{\prime}}\ |\ x\in\langle a,B\rangle\setminus B\rangle$ acts trivially on $W_{1}$. Since $Z$ is a normal $A$-invariant subgroup of $H$, we can factor out $Z$ and assume that $E_{a}$ is abelian. But in this case $B$ acts fixed point freely on $W_{1}E_{a}$ and then $W_{1}E_{a}$ is a nilpotent group by Thompson’s theorem [12] which means that $E_{a}$ acts trivially on $W_{1}$ that is a contraction and the proof is complete. ∎

We can now easily complete the proof of Theorem 1.2. By the above lemma $B$ acts trivially on the quotient $G/F(G)$. Therefore $G=F(G)C_{G}(B)$. This shows that $F(C_{G}(B))\leq F_{2}(G)$. Since the index of $F(C_{G}(B))$ in $C_{G}(B)$ is at most $m$, the result follows.

4. Proof of Theorem 1.1

We say that a finite group $G$ is metanilpotent if $\gamma_{\infty}(G)\leq F(G)$. The following elementary lemma will be useful (for the proof see for example [1, Lemma 2.4]).

Lemma 4.1**.**

Let $G$ be a metanilpotent finite group. Let $P$ be a Sylow $p$-subgroup of $\gamma_{\infty}(G)$ and $H$ be a Hall $p^{\prime}$-subgroup of G. Then $P=[P,H]$.

Let us now assume the hypothesis of Theorem 1.1. Thus, $A$ is a finite group of prime exponent $q$ and order at least $q^{3}$ acting on a finite $q^{\prime}$-group $G$ in such a manner that $\gamma_{\infty}(C_{G}(a))$ has order at most $m$ for any $a\in A^{\#}$. We wish to show that $\gamma_{\infty}(G)$ has $m$-bounded order. It is clear that $A$ contains a subgroup of order $q^{3}$. As in the previous section, we may assume that $A$ is an extra-special group of order $p^{3}$ and we denote by $B$ the subgroup $A^{\prime}=Z(A)$. Since $\gamma_{\infty}(C_{G}(a))$ has order at most $m$, we obtain that $F(C_{G}(a))$ has index at most $m!$ (see for example [9, 2.4.5]). By [5, Theorem 1.1] $\gamma_{\infty}(G)$ has $(q,m)$-bounded order. Without loss of generality we will assume that $m!<q$. In particular, $[G,B]$ is nilpotent by Lemma 3.3.

Lemma 4.2**.**

If $G$ is soluble, then $\gamma_{\infty}(G)=\gamma_{\infty}(C_{G}(B))$.

Proof.

We will use induction on the Fitting height $h$ of $G$. Suppose first that $G$ is metanilpotent. Let $P$ be a Sylow $p$-subgroup of $\gamma_{\infty}(G)$ and $H$ a Hall $A$-invariant $p^{\prime}$-subgroup of $G$. By Lemma 4.1 we have that $P=[P,H]$. It is sufficient to show that $P\leq\gamma_{\infty}(C_{G}(B))$. Therefore, without loss of generality, we assume that $G=PH$. With this in mind, observe that $\gamma_{\infty}(C_{G}(a))=[C_{P}(a),C_{H}(a)]$ for any $a\in A^{\#}$.

We will prove that $P=[C_{P}(B),C_{H}(B)]$. Let $A_{1},\ldots,A_{q+1}$ be the subgroups of order $q^{2}$ of $A$. Note that $A_{i}$ acts trivially on $\gamma_{\infty}(C_{G}(a))$ for any $a\in A_{i}^{\#}$ since $q>m$. Hence $\gamma_{\infty}(C_{G}(a))\leq C_{P}(A_{i})$ for any $a\in A_{i}^{\#}$. In particular, $\gamma_{\infty}(C_{G}(B))\leq C_{P}(A)$ since $\gamma_{\infty}(C_{G}(B))$ is $A$-invariant.

First, assume that $P$ is abelian. Given $a,b\in A_{i}$ we have that $[[C_{P}(a),C_{H}(a)],C_{H}(b)]\leq[C_{P}(A_{i}),C_{H}(b)]\leq[C_{P}(b),C_{H}(b)]$. On the other hand, $H=\prod_{b\in A_{i}^{\#}}C_{H}(b)$. Therefore, the subgroup $N=\prod_{a\in A^{\#}}[C_{P}(a),C_{H}(a)]$ is a normal subgroup. Since $N$ is $A$-invariant, we obtain that $A$ acts on $G/N$ in such a way that $C_{G}(a)$ is nilpotent for any $a\in A^{\#}$. Thus $G/N$ is nilpotent by [5, Lemma 2.5]. Therefore, $P=\prod_{a\in A^{\#}}[C_{P}(a),C_{H}(a)]$. In particular, $P=C_{P}(B)$.

Assume that $P$ is not abelian. Consider the action of $B$ on $P/\Phi(P)$. By the above, $P/\Phi(P)=C_{P}(B)\Phi(P)/\Phi(P)$. We see that $P=C_{P}(B)\Phi(P)$, which implies that $P=C_{P}(B)$.

Since $P=C_{P}(B)$ is a normal subgroup of $G$, we have that $[B,P,H]=[P,H,B]=1$. Then $[[H,B],P]=1$. Therefore, $P=[C_{P}(B),C_{H}(B)]$ since $H=C_{H}(B)[H,B]$.

If $G$ is soluble and its Fitting height is $h>2$, then we consider the quotient group $G/\gamma_{\infty}(F_{2}(G))$ which has Fitting height $h-1$ and $\gamma_{\infty}(F_{2}(G))\leq\gamma_{\infty}(G)$. Hence, by induction we have that $\gamma_{\infty}(G)=\gamma_{\infty}(C_{G}(B))$.

∎

Recall that under our assumptions $[G,B]$ is nilpotent and $C_{G}(B)$ has a normal nilpotent subgroup of index at most $m!$. Let $R$ be the soluble radical of $G$. Since $G=[G,B]C_{G}(B)$, the index of $R$ in $G$ is at most $m!$. Lemma 4.2 shows that the order of $\gamma_{\infty}(R)$ is at most $m$. We pass to the quotient $G/\gamma_{\infty}(R)$ and without loss of generality assume that $R$ is nilpotent. If $G=R$, we have nothing to prove. Therefore assume that $R<G$ and use induction on the index of $R$ in $G$. Since $[G,B]\leq R$, it follows that each subgroup of $G$ containing $R$ is $A$-invariant. If $T$ is any proper normal subgroup of $G$ containing $R$, by induction the order of $\gamma_{\infty}(T)$ is $m$-bounded and the theorem follows. Hence, we can assume that $G/R$ is a nonabelian simple group. We know that $G/R$ is isomorphic to a quotient of $C_{G}(B)$ and so, being simple, $G/R$ has order at most $m$.

As usual, given a set of primes $\pi$, we write $O_{\pi}(U)$ to denote the maximal normal $\pi$-subgroup of a finite group $U$. Let $\pi=\pi(m!)$ be the set of primes at most $m$. Let $N=O_{\pi^{\prime}}(G)$. Our assumptions imply that $G/N$ is a $\pi$-group and $N\leq F(G)$. Thus, by the Schur-Zassenhaus theorem [7, Theorem 6.2.1] the group $G$ has an $A$-invariant $\pi$-subgroup $K$ such that $G=NK$. Let $K_{0}=O_{\pi}(G)$.

Suppose that $K_{0}=1$. Then $G$ is a semidirect product of $N$ by $K=C_{K}(A)$. For an automorphism $a\in A^{\#}$ observe that $[C_{N}(a),K]\leq\gamma_{\infty}(C_{G}(a))$ since $C_{N}(a)$ and $K$ have coprime order. On the one hand, being a subgroup of $\gamma_{\infty}(C_{G}(a))$, the subgroup $[C_{N}(a),K]$ must be a $\pi$-group. On the one hand, being a subgroup of $N$, the subgroup $[C_{N}(a),K]$ must be a $\pi^{\prime}$-group. We conclude that $[C_{N}(a),K]=1$ for each $a\in A^{\#}$. Since $N$ is a product of all such centralizers $C_{N}(a)$, it follows that $[N,K]=1$. Since $K_{0}=1$ and $K$ is a $\pi$-group, we deduce that $K=1$ and so $G=N$ is a nilpotent group.

In general $K_{0}$ does not have to be trivial. However considering the quotient $G/K_{0}$ and taking into account the above paragraph we deduce that $G=N\times K$. In particular, $\gamma_{\infty}(G)=\gamma_{\infty}(K)$ and so without loss of generality we can assume that $G$ is a $\pi$-group. It follows that the number of prime divisors of $|R|$ is $m$-bounded and we can use induction on this number. It will be convenient to prove our theorem first under the additional assumption that $G=G^{\prime}$.

Suppose that $R$ is an $p$-group for some prime $p\in\pi$. Note that if $s$ is a prime different from $p$ and $H$ is an $A$-invariant Sylow $s$-subgroup of $G$, then in view of Lemma 4.2 we have $\gamma_{\infty}(RH)\leq\gamma_{\infty}(C_{G}(B))$ because $RH$ is soluble. We will require the following observation about finite simple groups (for the proof see for example [5, Lemma 3.2]).

Lemma 4.3**.**

Let $D$ be a nonabelian finite simple group and $p$ a prime. There exists a prime $s$ different from $p$ such that $D$ is generated by two Sylow $s$-subgroup.

In view of Lemma 4.3 and the fact that $G/R$ is simple we deduce that $G/R$ is generated by the image of two Sylow $s$-subgroup $H_{1}$ and $H_{2}$ where $s$ is a prime different from $p$. Both subgroups $RH_{1}$ and $RH_{2}$ are soluble and $B$-invariant since $[G,B]\leq R$. Therefore both $[R,H_{1}]$ and $[R,H_{2}]$ are contained in $\gamma_{\infty}(C_{G}(B))$.

Let $H=\langle H_{1},H_{2}\rangle$. Thus $G=RH$. Since $G=G^{\prime}$, it is clear that $G=[R,H]H$ and $[R,G]=[R,H]$. We have $[R,H]=[R,H_{1}][R,H_{2}]$ and therefore the order of $[R,H]$ is $m$-bounded. Passing to the quotient $G/[R,G]$ we can assume that $R=Z(G)$. So we are in the situation where $G/Z(G)$ has order at most $m$. By a theorem of Schur the order of $G^{\prime}$ is $m$-bounded as well (see for example [9, 2.4.1]). Taking into account that $G=G^{\prime}$ we conclude that the order of $G$ is $m$-bounded.

Suppose now that $\pi(R)=\{p_{1},\dots,p_{t}\}$, where $t\geq 2$. For each $i=1,\dots,t$ consider the quotient $G/O_{p_{i}^{\prime}}(G)$. The above paragraph shows that the order of $G/O_{p_{i}^{\prime}}(G)$ is $m$-bounded. Since also $t$ is $m$-bounded, the result follows.

Thus, in the case where $G=G^{\prime}$ the theorem is proved. Let us now deal with the case where $G\neq G^{\prime}$. Let $G^{(l)}$ be the last term of the derived series of $G$. The previous paragraph shows that $|G^{(l)}|$ is $m$-bounded. Consequently, $|\gamma_{\infty}(G)|$ is $m$-bounded since $G/G^{(l)}$ is soluble and $G^{(l)}\leq\gamma_{\infty}(G)$. The proof is now complete.