Frobenius action on Carter subgroups

TL;DR

This paper investigates the structure of Carter subgroups and Fitting series in finite solvable groups under Frobenius actions, establishing bounds on their Fitting height and $ ext{pi}$-series relations.

Contribution

It generalizes previous results by analyzing the Fitting series and $ ext{pi}$-series of centralizers under Frobenius group actions, providing new bounds and structural insights.

Findings

Fitting series of $C_G(H)$ are intersections with those of $G$.

Fitting height of $G$ exceeds that of $C_G(H)$ by at most one.

The $ ext{pi}$-series of $C_G(H)$ are intersections with those of $G$.

Abstract

Let $G$ be a finite solvable group and $H$ be a subgroup of $Aut(G)$. Suppose that there exists an $H$-invariant Carter subgroup $F$ of $G$ such that the semidirect product $FH$ is a Frobenius group with kernel $F$. We prove that the terms of the Fitting series of $C_{G}(H)$ are obtained as the intersection of $C_{G}(H)$ with the corresponding terms of the Fitting series of $G$, and the Fitting height of $G$ may exceed the Fitting height of $C_{G}(H)$ by at most one. As a corollary it is shown that for any set of primes $\pi$, the terms of the $\pi$-series of $C_{G}(H)$ is obtained as the intersection of $C_{G}(H)$ with the corresponding terms of the $\pi$-series of $G$, and the $\pi$-length of $G$ may exceed the $\pi$-length of $C_{G}(H)$ by at most one. They generalize the main results of \cite{Khu}.

Equations

$$1=\prod _{f\in X}{z}^f=(\prod _{f\in X\setminus F_1}{z}^f)(\prod _{f\in X\cap F_1}{z}^f)\in (\prod _{f\in X\cap F_1}{z}^f)C_Q(W)=z^{|X\cap F_1|}{C}_Q(W).$$

$$Ker(C_Q(H)on{C}_V(H))=Ker(C_Q(H)onV).$$

$$C_{G/N}(H)\cong C_{\overline{G}/\overline{N}}(H)=C_{\overline{G}}(H)\overline{N}/\overline{N}\cong C_G(H)N/N$$

$$Ker(C_Q(H)on{C}_V(H))=Ker(C_Q(H)onV).$$

$$F_{k+1}(C_G(H))/F_k(C_G(H))\cong F(C_{\overline{G}}(H))=C_{F(\overline{G})}(H)=C_{\overline{F_{k+1}(G)}}(H)=$$

$$\overline{C_{F_{k+1}(G)}(H)}\cong C_{F_{k+1}(G)}(H)/C_{F_k(G)}(H)=C_{F_{k+1}(G)}(H)/F_k(C_G(H)$$

$$h(G)-1=h(\overline{G})\le h(C_{\overline{G}}(H))+1=h(C_G(H)/C_{F(G)}(H))+1.$$

Full text

*∗*Corresponding author

1. introduction

Let $G$ and $A$ be finite groups such that $A$ acts on $G$ by automorphisms. Both the structure of $A$ and the way it acts on $G$ has drastic consequences on the structure of $G.$ As a typical result in this framework one can mention the pioneering work of J.G.Thompson which says that a group $G$ having an automorphism of prime order fixing no elements of $G$ except the identity, is nilpotent.

The present work is motivated by our research on possible generalizations of a result due to Khukhro [11] showing that some important group theoretic invariants of a solvable group $G$ admitting a Frobenius group $A=FH$ of automorphisms with Frobenius kernel $F$ and complement $H$ are closely related to the corresponding invariants of the fixed point subgroup $C_{G}(H)$ if $F$ acts fixed-point-freely on $G.$ Under these conditions $F$ is a Carter subgroup of the semidirect product $GF$ and $H$ acts on the solvable group $GF$ leaving $F$ invariant and acts, not only fixed-point-freely, but also Frobeniusly on $F$. Here we prove that almost the same result is true if $G$ is a solvable group, $H$ a group acting on $G$ and leaving a Carter subgroup $F$ of $G$ invariant and acting Frobeniusly on it. Namely, the main results of this paper are the following theorem on the Fitting series and its corollary on $\pi$-series.

Theorem 1.1**.**

*Let $G$ be a finite solvable group and $H$ be a subgroup of $Aut(G)$. Suppose that there exists an $H$-invariant Carter subgroup $F$ of $G$ such that the semidirect product $FH$ is a Frobenius group with kernel $F$. Then we have

$(a)$* $F_{n}(C_{G}(H))=F_{n}(G)\cap C_{G}(H)$ for all $n\in\mathbb{N}$;*

$(b)$* $h(G)\leq h(C_{G}(H))+1$. In fact $G=FF_{m}(G)$ where $m=h(C_{G}(H))$.*

Corollary 1.2**.**

*Let $G$ be a finite solvable group and $H$ be a subgroup of $Aut(G)$. Suppose that there exists an $H$-invariant Carter subgroup $F$ of $G$ such that the semidirect product $FH$ is a Frobenius group with kernel $F$. Then we have

$(a)$* $O_{\pi}(C_{G}(H))=O_{\pi}(G)\cap C_{G}(H)$ for any set of primes $\pi$;*

$(b)$* $O_{\pi_{1},\pi_{2},\dots,\pi_{k}}(C_{G}(H))=O_{\pi_{1},\pi_{2},\dots,\pi_{k}}(G)\cap C_{G}(H)$ for any sets of primes

$\pi_{1},\pi_{2},\dots,\pi_{k}$;*

$(c)$* $\ell_{\pi}(G)\leq\ell_{\pi}(C_{G}(H))+1$.*

As we have already pointed out the following result due to Khukhro follows as a consequence. It should be noted that by [1] the condition $C_{G}(F)=1$ below directly implies the solvability of the group $G.$

Corollary 1.3**.**

[11*]** Let $G$ be a finite group admitting a Frobenius group $FH$ of automorphisms with kernel $F$ and complement $H$ such that $C_{G}(F)=1.$ Then we have

$(a)$* $F_{n}(C_{G}(H))=F_{n}(G)\cap C_{G}(H)$ for all $n\in\mathbb{N}$;*

$(b)$* $h(G)=h(C_{G}(H))$;*

$(c)$* $O_{\pi}(C_{G}(H))=O_{\pi}(G)\cap C_{G}(H)$ for any set of primes $\pi$;*

$(d)$* $O_{\pi_{1},\pi_{2},\dots,\pi_{k}}(C_{G}(H))=O_{\pi_{1},\pi_{2},\dots,\pi_{k}}(G)\cap C_{G}(H)$ for any sets of primes

$\pi_{1},\pi_{2},\dots,\pi_{k}$;*

$(e)$* $\ell_{\pi}(G)=\ell_{\pi}(C_{G}(H))$.*

This can be proven by regarding the Frobenius kernel $F$ as an $H$-invariant Carter subgroup of the semidirect product $G_{1}=GF$ and $H$ as a group of automorphisms of the group $G_{1}.$ Then the result follows immediately.

So far we have obtained several extensions of Khukhro’s result replacing $FH$ by a Frobenius-like group with kernel $F$ under some mild additional conditions (see [3],[4],[5],[6],[7],[8],[9]). Unfortunately, this time we must be satisfied with the present form due to the example given in [4]. Namely, there exists a group $G$ admitting a group $H$ of automorphisms of prime order such that $G=VQF$ where

$(i)$ $V=F(G)$ is an elementary abelian $p$-group for some prime $p$, $Q$ is a $q$-group for some prime $q$ with $Q\unlhd QF$;

$(ii)$ $FH$ is a Frobenius-like (but not Frobenius) group where $F$ is an extraspecial $r$-group for some prime $r\notin\{p,q\}$, and $H$ centralizes $Z(F)$;

$(iii)$ $C_{Q}(F)=1,C_{V}(F)\neq 1,$ but $C_{V}(F)H$ is a Frobenius group;

$(iv)$ $C=C_{V}(F)\times F$ is a Carter subgroup of $G$ and $CH$ is a Frobenius-like group;

$(v)$ $Ker(C_{Q}(H)\,{on}\,C_{V}(H))\neq Ker(C_{Q}(H)\,{on}\,V)=1$ and hence

$(vi)$ $F(C_{G}(H))\neq F(G)\cap C_{G}(H).$

All groups are finite throughout the paper. The notation and the terminology are standard as in [10] except the following: The Fitting height and the $\pi$-length of a group $G$ are denoted by $h(G)$ and $\ell_{\pi}(G)$, respectively.

2. KEY PROPOSITION

In this section we present the following result which makes the appearance of the main result of this paper auxiliary. It should be pointed out that it is of independent interest, too.

Proposition 2.1**.**

Let a group $H$ act on the solvable group $G$ and let $F$ be an $H$-invariant Carter subgroup of $G$ such that $FH$ is a Frobenius group with kernel $F$ and complement $H$. Let $Q$ be a $q$-subgroup of $G$ such that $Q\unlhd GH$, and $V$ be a $kGH$-module over a field of characteristic $p$ for distinct primes $p$ and $q$ on which $Q$ acts nontrivially. Suppose that one of the following holds:

(i) $[C_{V}(F),h]=C_{V}(F)$ for any nonidentity $h\in H;$

(ii) $chark$ is coprime to $|H|$.

Then we have $Ker(C_{Q}(H)\,{on}\,C_{V}(H))=Ker(C_{Q}(H)\,{on}\,V).$

Proof.

Suppose the proposition is false and choose a counterexample with minimum $dim_{k}V+\left|QFH\right|$. We split the proof into a sequence of steps. To simplify the notation we set $K=Ker(C_{Q}(H)\,on\,C_{V}(H))$.

(1) We may assume that $G=QF$. Furthermore $Q\not\leq F$, in particular $[Q,F]\neq 1$, and hence $O_{q^{\prime}}(F)\neq 1.$

Proof.

It can be easily seen by induction that $G=QF.$ We may also assume that $Q\not\leq F$, in particular $[Q,F]\neq 1$, because otherwise $C_{Q}(H)=1$ and the theorem follows. Note that if $QF$ is a $q$-group then $F$ is properly contained in $QF$ as $Q\not\leq F.$ But then $F$ is properly contained in ts normalizer in $QF$, which is not the case. ∎

(2) We may assume that $k$ is a splitting field for all subgroups of $GH$.

Proof.

We consider the $GH$-module $\bar{V}=V\otimes_{k}{\bar{k}}$ where $\bar{k}$ is the algebraic closure of $k.$ Notice that $C_{\bar{V}}(H)=C_{V}(H)\otimes_{k}{\bar{k}}$. Therefore once the proposition has been proven for the group $GH$ on $\bar{V}$, it becomes true for $GH$ on $V$ also. ∎

(3) $V$ is an irreducible $GH$-module on which $G$ acts faithfully.

Proof.

Let $W=X/Y$ be a $GH$-composition factor of $V$ on which $K$ acts nontrivially. If $C_{V}(F)=1$ then $C_{Y}(F)=1$ and so $C_{W}(H)=C_{X}(H)Y/Y$ by Theorem 1.5 in [11]. Otherwise by hypothesis the group $C_{V}(F)H$ is Frobenius and hence $chark$ is coprime to $|H|$ which implies that $C_{W}(H)=C_{X}(H)Y/Y$. Notice that if $W\neq V$ then $Ker(C_{Q}(H)$ on $C_{W}(H))=Ker(C_{Q}(H)$ on $W)$ holds by induction. Hence $K=Ker(K$ on $C_{W}(H))=Ker(K$ on $W)$ which contradicts the assumption that $K$ acts nontrivially on $W.$ Therefore we can regard $V$ as an irreducible $GH$-module.

We set next $\bar{G}=G/Ker(G$ on $V)$ and consider the action of the group $\bar{G}H$ on $V$. An induction argument gives $Ker(C_{\bar{Q}}(H)$ on $C_{V}(H))=Ker(C_{\bar{Q}}(H)$ on $V)$ and hence $Ker(\overline{C_{Q}(H)}$ on $C_{V}(H))=Ker(\overline{C_{Q}(H)}$ on $V)$ which leads to a contradiction. Thus we may assume that $G$ acts faithfully on $V.$ ∎

It should be noted that we need only to prove $K=1$ due to the faithful action of $Q$ on $V$. So we assume this to be false.

(4) $L=K\cap Z(C_{Q}(H))\neq 1$ since $1\neq K\trianglelefteq C_{Q}(H)$. Pick $1\neq c\in L$ of order $q$. Then $Q=\left\langle c^{F}\right\rangle$.

Proof.

Suppose first that $Q\neq\left\langle c^{F}\right\rangle$. An induction argument applied to the action of $\left\langle c^{F}\right\rangle FH$ on $V$ we get $c\in Ker(C_{\left\langle c^{F}\right\rangle}(H)\,on\,C_{V}(H))=Ker(C_{\left\langle c^{F}\right\rangle}(H)\,on\,V)=1.$ This contradiction gives $Q=\left\langle c^{F}\right\rangle$. ∎

By Clifford’s theorem the restriction of the $GH$-module $V$ to the normal subgroup $Q$ is a direct sum of $Q$-homogeneous components. Let $\Omega$ denote the set of all $Q$-homogeneous components of $V$.

(5) $K$ acts trivially on the sum of components in any regular $H$-orbit in $\Omega$. Therefore there exists $W\in\Omega$ such that $Stab_{H}(W)\neq 1$.

Proof.

Let $U$ be an element in $\Omega$ such that $\{U^{y}:y\in H\}$ is a regular $H$-orbit in $\Omega$ and let $X$ be the sum of components. Then $K$ acts trivially on $C_{X}(H)=\{\Sigma_{y\in H}v^{y}:v\in U\}$ and hence trivially on $X$. ∎

(6) $F$ acts transitively on $\Omega$ and $H$ fixes an element, say $W$, of $\Omega$ and $H$ acts regularly and $K$ acts trivially on the set $\Omega\setminus\{W\}.$

Proof.

By (5) there exists $W\in\Omega$ such that $Stab_{H}(W)\neq 1$. Let $\Omega_{1}$ be the $F$-orbit on $\Omega$ containing $W$. Clearly, we have $Stab_{H}(W)\leq H_{1}=Stab_{H}(\Omega_{1}).$ So $H_{1}\neq 1$.

The group $H$ acts transitively on $\left\{\Omega_{i}:i=1,2,\ldots,s\right\},$ the collection of $F$-orbits on $\Omega$. Let now $V_{i}=\bigoplus_{W\in\Omega_{i}}W$ for $i=1,2,\ldots,s.$ Suppose that $s>1.$ Then $H_{1}$ is a proper subgroup of $H.$ Applying induction to the action of $GH_{1}$ on $V_{1}$ we obtain

$Ker(C_{Q}(H_{1})$ on $C_{V_{1}}(H_{1}))=Ker(C_{Q}(H_{1})$ on $V_{1})$.

It follows that $Ker(C_{Q}(H)$ on $C_{V_{1}}(H_{1}))=Ker(C_{Q}(H)$ on $V_{1})$ holds as $C_{Q}(H)\leq C_{Q}(H_{1}).$ On the other hand we have $C_{V}(H)=\left\{u^{x_{1}}+u^{x_{2}}+\cdots+u^{x_{s}}:u\in C_{V_{1}}(H_{1})\right\}$ where $x_{1},\ldots,x_{s}$ is a complete set of right coset representatives of $H_{1}$ in $H$. By definition $K$ acts trivially on $C_{V}(H)$ and normalizes each $V_{i}$. Then $K$ is trivial on $C_{V_{1}}(H_{1})$ and hence on $V_{1}.$ As $K$ is normalized by $H$ we see that $K$ acts trivially on each $V_{i}$ and hence on $V.$ This contradiction shows that $F$ acts transitively on $W$ and hence $H_{1}=H.$

Let now $S=Stab_{FH}(W)$ and $F_{1}=F\cap S$. Then $\left|F:F_{1}\right|=\left|\Omega\right|=\left|FH:S\right|.$ Notice next that $(\left|S:F_{1}\right|,\left|F_{1}\right|)=1$ as $(\left|F\right|,\left|H\right|)=1$. Let $S_{1}$ be a complement of $F_{1}$ in $S.$ Then we have $\left|F:F_{1}\right|=\left|F\right|\left|H\right|/\left|F_{1}\right|\left|S_{1}\right|$ which implies that $\left|H\right|=\left|S_{1}\right|.$ Therefore we may assume that $S=F_{1}H,$ that is $W$ is $H$-invariant.

Finally let $x\in F$ and $1\neq h\in H$ such that $(W^{x})^{h}=W^{x}$ holds. Then $[h,x^{-1}]\in F_{1}$ and so $F_{1}x=F_{1}x^{h}=(F_{1}x)^{h}$ implying the existence of an element $g\in F_{1}x\cap C_{F}(h)$ by Theorem 3.27 in [10]. Now the Frobenius action of $H$ on $F$ gives that $x\in F_{1}.$ This means that $W$ is the only element in $\Omega$ which is stabilized by some nonidentity element of $H$ and hence all the orbits of $H$ on $\Omega$ except $\{W\}$ are regular. ∎

*(7) $C_{Z(Q)}(F)=1$. *

Proof.

Due to the scalar action of $Z(Q/C_{Q}(W))$ on $W$, we have $[Z(Q),H]\leq C_{Q}(W)$ and hence $[C_{Z(Q)}(F),H]\leq C_{Q}(W)$. As $[C_{Z(Q)}(F),H]$ is centralized by $F$, $[C_{Z(Q)}(F),H]\leq\bigcap_{f\in F}C_{Q}(W^{f})=C_{Q}(V)=1.$ It follows that $C_{Z(Q)}(F)\leq N_{G}(F)\cap C_{Q}(H)\leq C_{F}(H)=1.$ ∎

(8) Final Contradiction.

Proof.

By (4), $Q=\left\langle c^{F}\right\rangle$. It follows by (6) that for any $x\in F\setminus F_{1},$ $c^{x}$ centralizes $W$. Thus we have $Q=\left\langle c^{F_{1}}\right\rangle C_{Q}(W)$.

Let $m$ be the nilpotency class of $Q/C_{Q}(W)$ which is of course equal to the nilpotency class of $Q/C_{Q}(W^{f})$ for any $f\in F.$ As $Q=Q/\cap_{f\in F}C_{Q}(W^{f})$ is isomorphic to a subgroup of $\oplus_{f\in F}Q/C_{Q}(W^{f})$ we see that $m$ is equal to the nilpotency class of $Q$. Then there exists $z=[c^{y_{1}},\ldots,c^{y_{m}}]$ where $y_{i}\in F_{1},i=1,\ldots,m$ such that $z\notin C_{Q}(W)$. Clearly $z\in Z(Q).$ Notice that for all $x\in F\setminus F_{1},$ we have $z^{x}=[c^{y_{1}x},\ldots,c^{y_{m}x}]\in C_{Q}(W).$ Furthermore we have $[Z(Q),F_{1}]\leq C_{Q}(W)$ due to the scalar action of $Z(Q/C_{Q}(W))$ on $W$. Let $X=O_{q^{\prime}}(F).$ By (7) $C_{Z(Q)}(X)=1$. Then,

[TABLE]

implying that $q$ divides $\left|X\cap F_{1}\right|$, which is impossible. This completes the proof. ∎

The following is an important consequence of the above proposition and appears as another version of Proposition 4.1 in [2] showing that the condition that $C_{V}(F)=1$ can be replaced by the condition that $C_{Q}(F)=1$ without assuming any coprimeness condition.

Corollary 2.2**.**

Let $FH$ be a Frobenius group with kernel $F$ and complement $H$ acting on a $q$-group $Q$ for some prime $q$. Let $V$ be a $kQFH$-module over a field of characteristic is coprime to $|QH|$. If $C_{Q}(F)=1$ then we have

[TABLE]

Proof.

As $C_{Q}(F)=1$, we can regard $F$ as an $H$-invariant Carter subgroup ofthe semidirect product $QF$. Then we appeal to the above proposition by letting $G=QF.$ ∎

3. PROOFS OF THE MAIN RESULTS

The following proposition will be needed in proving Theorem 1.1 and Corollary 1.2.

Proposition 3.1**.**

Let $G$ be a group and $H$ be a subgroup of $Aut(G)$. Suppose that there exists an $H$-invariant Carter subgroup $F$ of $G$ such that the semidirect product $FH$ is a Frobenius group with kernel $F$. For any $H$-invariant solvable normal subgroup $N$ of $G$ we have $C_{G/N}(H)=C_{G}(H)N/N.$

Proof.

We proceed by induction on the order of $G.$ As $N$ is solvable there exists a prime $p$ such that $O_{p}(N)\neq 1.$ Set $C_{O_{p}(N)}(F)=1=G/O_{p}(N).$ We first observe that $C_{\overline{G}}(H)=\overline{C_{G}(H)}:$ This follows from Theorem 1.5 in [11] in case $C_{O_{p}(N)}(F)=1.$ Otherwise $1\neq C_{O_{p}(N)}(F)\leq F$ which implies that $p$ divides $|F|.$ Then $p$ is coprime to $|H|$ and the claim follows.

As $\overline{F}$ is an $H$-invariant Carter subgroup of $\overline{G}$, an induction argument gives that $C_{\overline{G}/\overline{N}}(H)=C_{\overline{G}}(H)\overline{N}/\overline{N}.$ Notice that

[TABLE]

which proves the claim. ∎

Proof of Theorem 1.1  (a) To prove the result for $n=1$ we use induction on the order of $G.$ Set $\overline{G}=G/F(G)$. Then $\overline{F}$ is an $H$-invariant Carter subgroup of $\overline{G}$ such that the semidirect product $\overline{F}H$ is a Frobenius group with kernel $\overline{F}$. Then the result holds for $\overline{G}$ by induction, that is $F(C_{\overline{G}}(H))=F(\overline{G})$. In particular $\overline{F(C_{G}(H))}\leq F(C_{\overline{G}}(H))=F(\overline{G})$ by Proposition 3.1, implying that $F(C_{G}(H))\leq F_{2}(G).$

Notice that if $F(C_{G}(H))$ is not contained in $F(G)$ then there exists a prime $q$ such that $Q_{0}=O_{q}(C_{G}(H))$ is not contained in $F(G)$. Let $Q\in Syl_{q}(F_{2}(G))$ such that $Q_{0}\leq Q.$ So there exists a prime $p\neq q$ such that $[V,Q_{0}]\neq 1$ where $V=O_{p}(G)/\Phi(O_{p}(G))$. We have $C_{V}(H)=C_{O_{p}(F(G))}(H)\Phi(O_{p}(G))/\Phi(O_{p}(G))$ by Proposition 3.1. Now by Proposition 2.1 applied to the action of $GH$ on $V$ we get

[TABLE]

and hence $Q_{0}$ centralizes $V,$ which is a contradiction. So we have the result for $n=1$.

Suppose that the result is true for a fixed but arbitrary $k$, that is, $F_{k}(C_{G}(H))=F_{k}(G)\cap C_{G}(H).$ Set $\overline{G}=G/F_{k}(G).$ Now using Proposition 3.1 and the induction assumption we get

[TABLE]

[TABLE]

and the result follows for $k+1.$ This completes the proof of part (a).

(b) We proceed by induction on the order of $G$. Set $\overline{G}=G/F(G)$. Then $\overline{F}$ is an $H$-invariant Carter subgroup of $\overline{G}$ such that the semidirect product $\overline{F}H$ is a Frobenius group with kernel $\overline{F}$. Therefore the theorem is true for $\overline{G}$ by induction, that is,

[TABLE]

By (a) we have $C_{F(G)}(H)=F(C_{G}(H)).$ Then $h(G)-1=h(\overline{G})\leq h(C_{G}(H))$ and so $h(G)\leq h(C_{G}(H))+1$, as desired. Let $m=h(C_{G}(H)).$ Then $G/F_{m}(G)$ is nilpotent and hence is covered by the homomorphic image of $F$ as the Carter subgroup of a nilpotent group is not proper. This shows that $G=F_{m}(G)F.$ $\Box$

Proof of Corollary 1.2 (a)  Clearly, we have $O_{\pi}(G)\cap C_{G}(H)\leq O_{\pi}(C_{G}(H)).$ Suppose that $O_{\pi}(C_{G}(H))\neq 1$ . Then there exists a prime $p\in\pi$ such that $O_{p}(C_{G}(H))\neq 1.$ As $O_{p}(C_{G}(H))\leq F(C_{G}(H))\leq F(G)$ by Theorem 1.1, we see that $O_{p}(G)\neq 1.$ Set $\overline{G}=G/O_{p}(G).$ Since $|\overline{G}|\leq|G|$ we get by induction that $O_{\pi}(C_{\overline{G}}(H))\leq O_{\pi}(\overline{G})\leq\overline{O_{\pi}(C_{G}(H))}.$ As $C_{\overline{G}}(H)=\overline{C_{G}(H)}$ by Proposition 3.1 and $O_{\pi}(C_{\overline{G}}(H))=O_{\pi}(\overline{C_{G}(H)})$ we obtain $O_{\pi}(C_{G}(H))\leq O_{\pi}(G)\cap C_{G}(H)$, and the claim follows. Now part (b) follows immediately from (a) by induction.

To prove part (c) we proceed by induction on the order of $G$. Let $\ell_{\pi}(C_{G}(H))=m$ and $M=O_{{\pi}^{\prime},\pi,{\pi}^{\prime},\ldots,\pi,{\pi}^{\prime}}(G)$ where the number of $\pi$’ s is equal to $m.$ By (b) we have $M\cap C_{G}(H)=O_{{\pi}^{\prime},\pi,{\pi}^{\prime},\ldots,\pi,{\pi}^{\prime}}(C_{G}(H))=C_{G}(H)$ and hence $C_{G}(H)\leq M$ due to coprimeness. That is $C_{G/M}(H)=1$. By Lemma 1.3 in [11] it follows that $[G,F]\leq M$ and so $G/M\leq N_{G/M}(FM/M)=FM/M.$ That is $G/M$ is nilpotent and hence is a $\pi$-group. This shows that $\ell_{\pi}(G)\leq m+1$ as claimed. $\Box$