On existence of normal p-complement of finite groups with restrictions on the conjugacy class sizes

TL;DR

This paper proves that finite groups with conjugacy class sizes restricted to certain p-power conditions necessarily have a normal p-complement, under specific divisibility assumptions.

Contribution

It establishes a new criterion for the existence of a normal p-complement based on conjugacy class size restrictions in finite groups.

Findings

If conjugacy class sizes' p-part is 1 or p^α for all elements

Existence of a p-element with class size divisible by p

Group has a normal p-complement under these conditions

Abstract

The greatest power of a prime $p$ dividing the natural number $n$ will be denoted by $n_p$. Let $Ind_G(g)=|G:C_G(g)|$. Suppose that $G$ is a finite group and $p$ is a prime. We prove that if there exists an integer $\alpha>0$ such that $Ind_G(a)_p\in \{1,p^{\alpha}\}$ for every $a$ of $G$ and a $p$-element $x\in G$ such that $Ind_G(x)_p>1$, then $G$ includes a normal $p$-complement.

Equations

$$O^{p^{\prime }}(G/O_{p^{\prime }}(G))=S_1\times \cdots \times S_r\times H,$$

Full text

††thanks: The work is supported by Russian Science Foundation (project 18-71-10007).

\msc

20DXX,20E45

\VOLUME30 \DOIhttps://doi.org/10.46298/cm.9294 {paper}

1 Introduction

In this paper, all groups are finite. Denote the set of prime divisors of positive integer $n$ by $\pi(n)$, and by the set $\pi(|G|)$ for a group $G$ by $\pi(G)$. For a set of primes $\pi$ and a positive integer $n$ we will denote $n_{\pi}=\prod_{p\in\pi}|n|_{p}$. Let $G$ be a group and take $a\in G$. With $a^{G}$ standing for the conjugacy class in $G$ containing $a$, put $N(G)=\{|x^{G}|,x\in G\}\setminus\{1\}$. Denote by $|G||_{p}$ the number $p^{n}$ such that $N(G)$ contains a multiple of $p^{n}$ and avoids multiples of $p^{n+1}$. For $\pi\subseteq\pi(G)$ put $|G||_{\pi}=\prod_{p\in\pi}|G||_{p}$. For brevity, $|G||$ is meaning $|G||_{\pi(G)}$. Observe that $|G||_{p}$ divides $|G|_{p}$ for each $p\in\pi(G)$. However, $|G||_{p}$ can be less than $|G|_{p}$. Take a set of primes $\pi$, denote $\Theta_{\pi}=\{\tau\subseteq\pi\ |\ \tau\neq\varnothing,|\tau|\geq|\pi|-1\}$.

Definition 1.1**.**

Let $\pi$ be a set of primes. We say that a group $G$ satisfies the condition $\pi^{*}$ or $G$ is a $\pi^{*}$-group and write $G\in\pi^{*}$ if for every $a\in N(G)$ there exists $\tau_{a}\in\Theta_{\pi}$ such that $a_{\pi}=|G||_{\tau_{a}}$.

Ishikawa [Ishikawa] proved that a group $G$ with $N(G)=\{p^{\alpha}\}$ is nilpotent class at most $3$. Casolo, Dolfi and Jabara [Casolo] described the set of $\{p\}^{*}$-groups. In particular, they proved that any group of $\{p\}^{*}$ is solvable and includes a normal $p$-compliment. Camina [Camina] proved that a group $G$ with $\{p,q\}^{*}$-property is nilpotent if $N(G)=\{p^{n},q^{m},p^{n}q^{m}\}$. Beltram and Filipe [Beltran] extended Camina’s theorem in the following way. Let $G$ be a group whose set of conjugacy class sizes is $\{1,n,m,nm\}$, where $n$ and $m$ are coprime positive integers; then $G$ is nilpotent and the integers $n$ and $m$ are prime-power numbers; in particular $G\in\{n,m\}^{*}$. The author of [GorshkovC] investigate $\{p,q\}^{*}$-groups with trivial center. In the present paper we will investigate some generalisations of the property $\{p\}^{*}$.

Definition 1.2**.**

We say that a group $G$ satisfies the condition $R(p)$ or $G$ is a $R(p)$-group and write $G\in R(p)$ if there exists an integer $\alpha>0$ such that $a_{p}\in\{1,p^{\alpha}\}$ for each $a\in N(G)$.

Note that, if $G\in\pi^{*}$, then $G\in R(p)$ for each $p\in\pi$. The set of $R(p)$-group disjoins on two subsets $R(p)^{*}$ and $R(p)^{**}$:

1. $(i)$

$G\in R(p)^{*}$ if $G$ contains a $p$-element $h$ such that $Ind_{G}(h)_{p}>1$; 2. $(ii)$

$G\in R(p)^{**}$ if $Ind_{G}(h)_{p}=1$ for each $p$-element $h\in G$.

We prove the following theorem.

Theorem 1.3**.**

If $G\in R(p)^{*}$, then $G$ has a normal $p$-complement.

It follows from the theorem that the center of a $R(p)^{*}$-group is not trivial.

Corollary 1.4**.**

If $G\in R(p)^{*}$ and $P\in Syl_{p}(G)$, then $Z(P)\leq Z(G)$.

Vasil’ev [Vasilev] proved that if $G$ is a $R(p)$-group with trivial center and $|G||_{p}=p$, then Sylow $p$-subgroups of $G$ are abelian. This assertion is true in the general case.

Corollary 1.5**.**

If $G\in R(p)$ and $Z(G)=1$, then Sylow $p$-subgroups of $G$ are abelian.

2 Preliminary results

Lemma 2.1** ([GorshkovA, Lemma 1.4]).**

For a finite group $G$, take $K\unlhd G$ and put $\overline{G}=G/K$. Take $x\in G$ and $\overline{x}=xK\in G/K$. The following claims hold:

1. (i)

$|x^{K}|$* and $|\overline{x}^{\overline{G}}|$ divide $|x^{G}|$.* 2. (ii)

For neighboring members $L$ and $M$ of a composition series of $G$, with $L<M$, take $x\in M$ and the image $\widetilde{x}=xL$ of $x$. Then $|\widetilde{x}^{S}|$ divides $|x^{G}|$, where $S=M/L$. 3. (iii)

If $y\in G$ with $xy=yx$ and $(|x|,|y|)=1$, then $C_{G}(xy)=C_{G}(x)\cap C_{G}(y)$. 4. (iv)

If $(|x|,|K|)=1$, then $C_{\overline{G}}(\overline{x})=C_{G}(x)K/K$. 5. (v)

$\overline{C_{G}(x)}\leq C_{\overline{G}}(\overline{x})$.

Lemma 2.2** ([Casolo, Lemma 2.1]).**

Let $x,y$ be elements of a group $G$ and assume at least one of the following conditions:

1. (i)

$x$* and $y$ commute and have coprime orders;* 2. (ii)

$x\in N,y\in M$* with $N,M\unlhd G$ and $N\cap M=1$.*

Then $C_{G}(xy)=C_{G}(x)\cap C_{G}(y)$.

Lemma 2.3** ([Casolo, Lemma 2.7]).**

Let $A$ be a group acting via automorphisms on a group $G$ and $N$ be a normal $A$-invariant subgroup of $G$. If $(|A|,|N|)=1$, then:

• (i)

$C_{G/N}(A)=C_{G}(A)N/N$;

• (ii)

$|C_{G}(A)|=|C_{N}(A)||C_{G/N}(A)|$.

Lemma 2.4** ([GorshkovA, Lemma 4]).**

Take $g\in G$. If each conjugacy class of $G$ contains an element $h$ such that $g\in C_{G}(h)$, then $g\in Z(G)$.

Lemma 2.5**.**

If $G\in R(p)$ and $N\unlhd G$ such that $|N|_{p}=|G|_{p}$, then $N\in R(p)$ or $|N||_{p}=1$.

Proof 2.6**.**

Since $N$ is a normal subgroup and $N$ includes every Sylow $p$-subgroup of $G$, we have $N$ includes every Sylow $p$-subgroup of $C_{G}(x)$ for any $x\in G$. Therefore, $Ind_{G}(x)_{p}=Ind_{N}(x)_{p}$ and the lemma is proved.

Lemma 2.7**.**

If $G\in R(p)$, $N\unlhd G$ is a $p^{\prime}$-group, then $G/N\in R(p)$ or $|G/N||_{p}=1$.

Proof 2.8**.**

Let \overline{{\color[rgb]{1,1,1}a}}:G\rightarrow G/N be a natural homomorphism. We have $Ind_{G}(h)$ is a multiple of $Ind_{\overline{G}}(\overline{h})$ for any $h\in G$. Therefore, $|G||_{p}\geq|\overline{G}||_{p}$. Assume that there exists $\overline{x}\in\overline{G}$ such that $1<Ind_{\overline{G}}(\overline{x})_{p}<|G||_{p}$. Let $H$ be a Sylow $p$-subgroup of $C_{\overline{G}}(\overline{x})$. Therefore, $|G|_{p}>|H|>|G|_{p}/|G||_{p}$. Put $T<G$ is a $p$-group such that $\overline{T}=H$. From Lemma 2.3 follows that $C_{G}(T)$ contains $y$ such that $yN=\overline{x}$. Since $C_{G}(y)\geq T$, we obtain $Ind_{G}(y)_{p}\leq|G|_{p}/|T|<|G||_{p}$. Therefore, $Ind_{G}(y)_{p}=1$ and consequently $Ind_{\overline{G}}(yN)_{p}=1=Ind_{\overline{G}}(x)_{p}$; a contradiction.

The prime graph $GK(G)$ of a finite group $G$ is defined as follows. The vertex set of $GK(G)$ is the set $\pi(G)$. Two distinct primes $p,q\in\pi(G)$ considered as vertices of the graph are adjacent by the edge if and only if there is an element of order $pq$ in $G$. Denote by $s(G)$ the number of connected components of $GK(G)$ and by $\pi_{i}(G),i=1,...,s(G)$, its $i$-th connected component. If $G$ has even order, then put $2\in\pi_{1}(G)$.

Lemma 2.9**.**

[Williams, Theorem A]* If a finite group $G$ has disconnected prime graph, then one of the following conditions holds:*

1. (a)

$s(G)=2$* and $G$ is a Frobenius or 2-Frobenius group;* 2. (b)

there is a nonabelian simple group $S$ such that $S\leq G=G/F(G)\leq Aut(S)$, where $F(G)$ is the maximal normal nilpotent subgroup of $G$; moreover, $F(G)$ and $G/S$ are $\pi_{1}(G)$-subgroups, $s(S)\geq s(G)$, and for every $i$ with $2\leq i\leq s(G)$ there is $j$ with $2\leq j\leq s(S)$ such that $\pi_{i}(G)=\pi_{j}(S)$.

Lemma 2.10** ([Gorenstein, Lemma 5.3.4]).**

Let $A\times B$ be a group of automorphisms of the $p$-group $P$ with $A$ a $p^{\prime}$-group and $B$ a $p$-group. If $A$ acts trivially on $C_{P}(B)$, then $A=1$.

Lemma 2.11** ([Gorenstein, Lemma 5.2.3]).**

Let $A$ be a $p^{\prime}$-group of automorphisms of the abelian group $P$. Then we have $P=C_{P}(A)\times[P,A]$

Lemma 2.12**.**

[Camina, Lemma 1]* If, for some prime $p$, every $p^{\prime}$-element of a group $G$ has index prime to $p$, then the Sylow $p$-subgroup of $G$ is a direct factor of $G$.*

Lemma 2.13**.**

[VVasilev, Lemma 3.6]* For distinct primes $s$ and $r$, consider a semidirect product $H$ of a normal $r$-subgroup $T$ and a cyclic subgroup $C=\langle g\rangle$ of order $s$ with $[T,g]\neq 1$. Suppose that $H$ acts faithfully on a vector space $V$ of positive characteristic $t$ not equal to $r$. If the minimal polynomial of $g$ on $V$ does not equal $x^{s}-1$, then*

1. (i)

$C_{T}(g)\neq 1$; 2. (ii)

$T$* is nonabelian;* 3. (iii)

$r=2$* and $s$ is a Fermat prime.*

Lemma 2.14**.**

[GorshkovB, Lemma 11]* If $S\leq A\leq Aut(S)$, where $S$ is a nonabelian simple group, then $|A|=|A||$.*

Lemma 2.15**.**

[Navarro, Theorem B]* Let $G$ be a finite group and $p$ a prime. Suppose that for every $p$-element $x$ the number $|x^{G}|$ is a $p^{\prime}$-number. Then,*

[TABLE]

where $H$ has an abelian Sylow $p$-subgroup, $r\geq 0$ , and $S_{i}$ is a nonabelian simple group with either

1. (i)

$p=3$* and $S_{i}\simeq Ru$ or $J_{4}$ or $S_{i}\simeq\ ^{2}F_{4}(q)^{\prime}$, $9\nmid(q+1)$; or* 2. (ii)

$p=5$* and $S_{i}\simeq Th$ for all $i$.*

3 Proof

Let $G$ be a counterexample for assertion of the theorem of minimal order.

Lemma 3.1**.**

$O_{p^{\prime}}(G)=1$**

Proof 3.2**.**

From Lemma 2.7 it follows that $G/O_{p^{\prime}}(G)\in R(p)$ or $|G/O_{p^{\prime}}(G)||_{p}=1$. We can think that $G/O_{p^{\prime}}(G)$ does not include a normal $p$-complement, else $G$ contains a normal $p$-complement. Therefore, $G/O_{p^{\prime}}$ a counterexample for assertion of theorem; a contradiction with minimality $G$. If $|G/O_{p^{\prime}}(G)||_{p}=1$, then Lemma 2.12 implies that $G/O_{p^{\prime}}(G)$ is a $p$-group. Therefore, $O_{p^{\prime}}(G)$ is a normal $p$-compliment of $G$; a contradiction.

Lemma 3.3**.**

Orders of minimal normal subgroups of $G$ are multiples of $p$.

Proof 3.4**.**

It follows from Lemma 3.1.

Lemma 3.5**.**

Each minimal normal subgroup of $G$ is a $p$-group.

Proof 3.6**.**

Let $H$ be the socle of $G$. Then $H$ has expression in form $S\times X$, where $S=S_{1}\times S_{2}\times...\times S_{n}$, for nonabelian simple groups $S_{1},...,S_{n}$ and a $p$-group $X$. It follows from Lemma 3.3 that $p$ divides the order of $S_{i}$ for all $1\leq i\leq n$. Assume that $G$ contains a $p$-element $x$ such that $S_{1}^{x}\neq S_{1}$. Let $D=\langle aa^{x}a^{x^{2}}...a^{|x|-1}|a\in S_{1}\rangle$. We have $D=C_{S_{1}\times S^{x}_{1}...}(x)$ and $D\simeq S_{1}$. Since $|S_{1}|$ is a multiple of $p$, we see that $Ind_{G}(x)_{p}>1$. By Lemma 2.14, we obtain $D$ contains a $p^{\prime}$-element $y$ such that $Ind_{D}(y)_{p}=|D|_{p}$. Thus, $Ind_{G}(xy)_{p}>Ind_{G}(x)_{p}=|G||_{p}$; a contradiction. It follows that $S_{i}^{y}=S_{i}$, for any $1\leq i\leq n$ and for any $p$-element $y$. Take $h_{1}\in S_{1}$ and $h_{2}\in S_{2}$. We have a $p$-element $y\in C_{G}(h_{1}h_{2})$ iff $y\in C_{G}(h_{1})\cap C_{G}(h_{2})$. Assume that $n>1$. Since $|S_{i}|=|S_{i}||$, we see that $S_{i}$ contains an element $h_{i}$ such that $|h_{i}^{S_{i}}|_{p}=|S_{i}|_{p}$, in particular $|h_{i}^{G}|_{p}>1$. Let $A$ be a Sylow $p$-subgroup of $C_{G}(h_{1}h_{2})$ and $B$ be a Sylow $p$-subgroup of $C_{G}(h_{1})$. Then $A<C_{G}(h_{1})\cap C_{G}(h_{2})$ and $|G|_{p}>|B|>|A|$; a contradiction. Therefore, $n=1$. From Lemma 2.14, it follows that $S$ contains an element $h$ such that a $p$-element $y\in C_{G}(h)$ iff $y\in C_{G}(S)$. If $C_{G}(S)$ contains $y$ such that $|y^{C_{G}(S)}|_{p}>1$, then $|(yh)^{G}|_{p}>|h^{G}|_{p}$; a contradiction. From Lemma 2.12 it follows that $C_{G}(S)=O_{p}(G)$. Moreover $O=O_{p}(G)$ is abelian. We have $|h^{G}|_{p}=|G|_{p}/|O|_{p}$. Take $a\in S$ such that $|a^{S}|_{p}<|S|_{p}$. Hence $|a^{G}|_{p}<|h^{G}|_{p}$. This implies that $|a^{S}|_{p}=1$. From Lemma 2.15 it follows that a Sylow $p$-subgroup of $S$ is abelian or $S$ is isomorphic to one of groups $J_{4},Ru,\ ^{2}F_{4}(q)^{\prime},Th$. Also it signifies that $S\in R(p)$. Assume that there exists a $p$-element $x\in G\setminus H$ such that $Ind_{G}(x)_{p}>1$ and $x$ acts on $S$ as an outer automorphism. From Lemma 2.12 and the equation $Ind_{C_{G}(x)}(y)_{p}=1$ for each $p^{\prime}$-element $y\in C_{G}(x)$ it follows that $C_{G}(x)=L\times T$, where $L$ is a Sylow $p$-subgroup of $C_{G}(x)$. Therefore, $C_{S}(x)=\widetilde{P}\times\widetilde{L}$, where $\widetilde{P}$ is a Sylow $p$-subgroup of $C_{S}(x)$. Assume that $S\simeq Alt_{n}$, where $n\geq 5$. Since $\pi(Out(S))=\{2\}$, we obtain $p=2$. If $n\neq 6$, then $C_{S_{1}}(x)\simeq Alt_{n-2}$; a contradiction. If $n=6$, then $C_{S}(x)\simeq Alt_{4}$ or $C_{S_{1}}(x)\simeq Sym_{3}$; a contradiction. From [Conway] it follows that $S$ is not isomorphic to a sporadic simple group or the Tits group. Therefore, $S$ is a group of Lie type. Assume that a Sylow $p$-subgroup of $S$ is nonabelian. Since $S$ is not isomorphic to one of a sporadic groups, it follows that $S\simeq\ ^{2}F_{4}(q)^{\prime},9\nmid q+1$ and $p=3$. Therefore, $x$ acts on $S$ as a field automorphism. Thus, $q=2^{3(2m+1)}$; this contradicts with that $9\nmid q+1$. Thus, Sylow $p$-subgroups of $S$ are abelian. Assume that $p=2$. From description of simple groups with abelian Sylow $2$-subgroup [Walter] it follows that $S$ is isomorphic to one of a groups $L_{2}(q)$ where $q=2^{f}$ or $q\equiv 3,5(mod\ 8)$, $J_{1}$ or ${}^{2}G_{2}(q)$ where $q=3^{2m+1}$ and $m\geq 1$. Put $P\in Syl_{2}(S)$. From [Kondrat'ev, Theorems 1 and 7] it follows that $C_{S}(P)=Z(P)$ or $S$ is isomorphic to $L_{2}(q)$ where $q$ odd. Let $S$ be isomorphic to $L_{2}(q)$ for some odd $q$. If $x$ acts on $S$ as a field automorphism, then $C_{S}(x)\simeq L_{2}(d)$, where $d$ divides $q$, in particular $C_{S}(x)$ is not a direct product of a Sylow $p$-subgroup and a $p$-complement; a contradiction. Assume that $x$ acts on $S$ as diagonal automorphism or a diagonal-field automorphism. Therefore, $S$ contains a $2$-element $z$ such that $C_{G}(z)\cap x^{G}=\varnothing$. Consequently $1<|z^{G}|_{p}<|h^{G}|_{p}$; a contradiction. Hence, $S$ does not isomorphic $L_{2}(q)$ for odd $q$. It follows that $C_{S}(P)=Z(P)$. Since $|S|=|S||$ and $S\in R(p)$, we see that $\{p\}$ is a connected component of $GK(S)$. The group of outer automorphisms of $J_{1}$ is trivial, therefore $S$ does not isomorphic $J_{1}$. If $S\simeq^{2}G_{2}(q)$, then from [VasilevVdovin] it follows that $2$ is not a connected components of $GK(S)$. If $S\simeq L_{2}(q)$ for even $q$, then $Out(S)$ is isomorphic to a group of field automorphisms. By analogy as before we can assume that $C_{S}(x)$ is not a direct product of a Sylow $p$-subgroup and a $p$-complement; a contradiction. Thus, $p>2$. From description of finite simple groups with an abelian Sylow $p$-subgroup [Shen] it follows that $p$ does not divide the orders of graph and diagonal automorphism groups. Lemma 2.5 and fact that subgroup of field automorphisms is a normal subgroup of $Out(S)$, implies $G\simeq(S\times X).F$, where $F$ is a some cyclic $p$-group. In particular, we get that $X.F$ is a $p$-group. We can assume that $F=\langle xX\rangle$. As noted above $C_{X}(x)<X$; a contradiction with fact that $|a^{G}|_{p}=1$ for each $a\in X$. Let $x\in G$ be a $p$-element such that $Ind_{G}(x)_{p}>1$. We have $x\in SC_{G}(S)$. Since $S$ and $C_{G}(S)$ are normal subgroups of $G$ with trivial intersection, we obtain $x$ has unique expression in form $x=x_{S}x_{C}$ where $x_{S}\in S,x_{C}\in C_{G}(S)$. Moreover from Lemma 2.2 it follows that $C_{G}(x)=C_{G}(x_{S})\cap C_{G}(x_{C})$. From Lemma 2.14 it follows that $S$ contains $h$ such that a $p$-element $y\in C_{G}(h)$ iff $y\in C_{G}(S)$. Therefore, for each $A\in Syl_{p}(C_{G}(h))$ there exists $B\in Syl_{p}(C_{G}(x_{S}))$ such that $A<B$. Since $Ind_{S}(x_{S})_{p}<Ind_{S}(h)_{p}$ and $x_{C}\in C_{G}(h)$, we obtain $Ind_{G}(x)_{p}<In_{G}(hx_{C})$; a contradiction.

Let $O=O_{p}(G)$. Lemma 3.5 implies that $O$ includes the socle of $G$. Therefore, $C_{G}(O)=Z(O)$, and for each $h\in G\setminus O$ we have $Ind_{G}(h)_{p}=|G||_{p}$.

Lemma 3.7**.**

$|G|_{p}>|O|$**

Proof 3.8**.**

Assume that $|G|_{p}=|O|$. Let $x\in O$ such that $Ind_{O}(x)>1$ and $h\in G$ be a $p^{\prime}$-element. We have $|C_{O}(x)|>|Z(O)|$. Therefore, $|C_{O}(h)|>|Z(O)|$, consequently $C_{O}(h)$ contains an element $y$ such that $Ind_{O}(y)>1$. Hence $C_{O}(y)=C_{O}(h)$. From Lemma 2.10 it follows that $C_{O}(h)=O$; a contradiction.

From Lemma 3.7 if follows that $|G|_{p}>|O|$. Let $h\in G$ be a $p^{\prime}$-element and $x\in C_{G}(h)\setminus O$ be a $p$-element. Using Lemma 2.10 we can show that $Ind_{G}(x)_{p}=1$, so $x\in C_{G}(O)=Z(O)$, which is a contradiction. In particular $p$ is a connected component of $GK(G/O)$.

Lemma 3.9**.**

The group $O$ is abelian.

Proof 3.10**.**

Assume that there exists $x\in O\setminus Z(O)$. Put $h\in G$ is a $p^{\prime}$-element. We have $|C_{G}(h)|_{p}=|C_{G}(x)|_{p}$, in particular $C_{G}(h)$ contains a $p$-element $y$ such that $Ind_{G}(y)_{p}>1$. From Lemma 2.10 it follows that $O<C_{G}(h)$; a contradiction.

Lemma 3.11**.**

$Ind_{G}(x)_{p}=1$* for each $x\in O$.*

Proof 3.12**.**

Assume that there exists $x\in O$ such that $Ind_{G}(x)_{p}>1$. Lemma 3.9 implies that $O$ is abelian. Therefore, $C_{G}(x)\geq O$. Put $h\in G$ is a $p^{\prime}$-element. We know that $p$ is a connected component of $GK(G/O)$, hence for each Sylow $p$-subgroup $P$ of $C_{G}(h)$ we have $P\leq O$. Since $C_{G}(O)=Z(O)$, we see that $P\neq O$. Consequently $|G|_{p}/|C_{O}(x)|\geq|G|_{p}/|C_{G}(x)|_{p}=|G||_{p}$. Therefore, $|P|\geq|O|$; a contradiction.

Lemma 3.13**.**

The group $G$ is non solvable.

Proof 3.14**.**

Assume that $G$ is solvable. From Lemma 2.9 it follows that $G/O$ is a Frobenius or $2$-Frobenius group. Since kernel of $G/O$ is a $p^{\prime}$-group and Lemma 2.5, we obtain $G/O$ is a Frobenius group with $p^{\prime}$-kernel $\overline{K}$ and complement $\overline{F}$, else $G$ is not minimal. Put $K<G$ be a minimal subgroup such that $KO/O~{}=~{}\overline{K}$. From Frattini argument we have $N(K)O/O\simeq\overline{K}$. Let $F\leq N_{G}(K)$ be a minimal subgroup such that $FO/O=\overline{F}$. Since $G/O$ is a Frobenius group with the complement $\overline{F}$, we see that $N_{G}(F)<OF$; in particular $\pi(N_{G}(F))=\{p\}$. Let $H<K$ be maximal with respect to inclusion subgroup of $K$ such that $C_{O}(H)>Z(G)$. We show that $H$ is not trivial. For each $h\in K$ and $y\in F\setminus O$ we have $|h^{G}|_{p}=|y^{G}|_{p}$ and $|C_{G}(y)|_{p}>|Z(G)|_{p}$. Therefore, $C_{O}(h)>Z(G)$, in particular $H>1$. Let $x\in C_{O}(H)\setminus Z(G)$. Lemma 3.11 implies that $x\in Z(P)$ for some Sylow $p$-subgroup $P$ of $G$. Hence $C_{G}(x)$ includes a subgroup $V$ which is conjugated with $F$. We have $H<C_{G}(x)$. Since $H$ is a maximal $p^{\prime}$-subgroup with a non trivial centralizer in $O$, we see that $H$ is a Hall $p^{\prime}$-subgroup of $C_{G}(x)$. In particular $C_{G}(x)=O.\overline{H}.\overline{F}$, where $\overline{H}=HO/O$. Let $y\in C_{O}(H)\setminus Z(G)$. We can show that $C_{G}(y)=O.\overline{H}\leftthreetimes\overline{R}$. Since $\overline{R}<N_{G/O}(\overline{H})$ and $\overline{F}<N_{G/O}(\overline{H})$, we have $\overline{H}\overline{R}\overline{F}$ is a Frobenius group with the kernel $\overline{H}$ and the complement $\overline{F}$. Therefore, $\overline{H}\overline{R}\overline{F}=\overline{H}\overline{R}=\overline{H}\overline{F}$ and $C_{G}(y)=C_{G}(x)$. Thus, $C_{G}(x)=C_{G}(C_{O}(H))$. Let $N=N_{K}(H)$. Since $K$ is nilpotent, we get that $H=K$ or $N>H$. We have $N<N_{G}(C_{G}(H))$. Therefore, $N<N_{G}(C_{G}(C_{O}(H)))$. If $N>H$, then $N\not\leq C_{G}(C_{O}(H))$. Thus, $N_{G}(F)$ includes a subgroup $X$ such that $(X/H)\simeq N/H$; a contradiction with fact that $N_{G}(F)$ is a $p$-group. It follows that $H=K$. We have $F<C_{G}(C_{O}(H))$ and $O$ is abelian. Therefore, $C_{O}(H)\leq Z(G)$; a contradiction.

From Lemmas 3.13 and 2.9 it follows that $G/O\simeq F.S$, where $F$ is a nilpotent $\pi_{1}(G/O)$-group, and $S$ is a nonabelian simple group. Let $g\in G$ be a $p^{\prime}$-element. Put $H_{g}\leq C_{G}(g)$ a subgroup generated by all $(\pi(|g|)\cup\{p\})^{\prime}$-elements. We have $H_{g}\leq C_{G}(C_{O}(g))$. Therefore, $H_{g}$ acts regularly on $O/C_{O}(g)$. Hence Sylow subgroups of $H_{g}$ are cyclic or quaternion groups.

Lemma 3.15**.**

$F=1$.

Proof 3.16**.**

Assume that $|\pi(F)|>1$. Let $h\in F$ be a $t$-element for some $t\in\pi(F)$. We have $C_{F}(h)$ includes some Hall $t^{\prime}$-subgroup $T$ of $F$. Since Sylow subgroups of $H_{h}$ are cyclic or quaternion, we get $S$ acts trivially on $T$. Therefore, $F$ is a $t$-group for some prime $t\neq p$. Assume that there exists $h\in Z(F)$ such that $|\pi(C_{F.S}(h))|>1$. Let $g\in C_{F.S}(h)$ be a $t^{\prime}$-element. Assume that $Z(F)<C_{F.S}(g)$. Since $Z(F)$ is a normal subgroup of $F.S$, we have $\langle g^{G}\rangle<C_{F.S}(Z(F))$. The group $S$ is a simple group, therefore, $\langle g^{G}\rangle F/F=S$. In particular $F.S$ contains an element of order $pt$; a contradiction. Hence $[Z(F),g]>1$. We have $H_{h}$ acts regularly on the $O/C_{O}(h)$ and $\langle g^{C_{F.S}(h)}\rangle\leq H_{h}$. This assertion contradicts with Lemma 2.13. Therefore, $\pi(C_{F.S}(h))=\{t\}$ for each $h\in Z(F)$. Assume that there exists $a\in F.S$ a $t$-element such that $|\pi(C_{F.S}(a))|>1$. Since $C_{F.S}(a)\cap Z(F)>1$, we get a contradiction with Lemma 2.13. Therefore, $t$ is a connected component of $GK(F.S)$. Since $t\in\pi_{1}(GK(F.S))$, we get $t=2$. From a description of the prime graph of finite simple groups [VasilevVdovin] it follows that $S\simeq Alt_{5}$. From Brauer $2$-character tables [Conway] it follows that $F.S$ contains an element $x$ such that $|x|\in\{6,10\}$; a contradiction.

From Lemmas 3.13 and 3.15 it follows that $G/O$ is a simple group and $p$ is a connected component of $GK(G/O)$.

Lemma 3.17**.**

$G\simeq O$**

Proof 3.18**.**

Assume that $S$ is an alternating simple group of degree $n$. An alternating group has disconnected prime graph iff one of the numbers $n,n-1,n-2$ is prime, and this number is a connected component of the prime graph. In particular, if $n>6$, then $3\in\pi_{1}(S)$. If $n>6$, then $S$ contains an element $g$ of order $3$ such that $C_{S}(g)\simeq\langle g\rangle\times Alt_{n-3}$. Therefore, in this case $H_{g}$ includes a Frobenius group; a contradiction with assertion that $H_{g}$ acts regular on $O/C_{O}(g)$. Let $n\in\{5,6\}$. Therefore, $S$ contains an element $g$ such that for each $p$-element $h\in S$ we have $\langle h,g\rangle=S$. We have $C_{O}(g)>Z(G)$. Put $x\in C_{O}(g)\setminus Z(G)$. It follows from Lemma 3.9 that $|x^{G}|_{p}=1$. Hence, $C_{G}(x)$ includes a Sylow $p$-subgroup $P$ of $G$. In particular $\langle g,PO/O\rangle=S$. That signifies that $C_{G}(x)=G$; a contradiction. Assume that $S$ is a group of Lie type. If Lie rank of $S$ is more then $2$, then $S$ contains an element $g$ such that $H_{g}$ includes a Frobenius group. Therefore, we can assume that Lie rank of $S$ is $1$ or $2$. Assume that $S\simeq L_{2}(q)$. We have that $S$ is generated by a pair $a,b$ where $|a|=(q+1)/2,|b|=(q-1)/2$. Since $p=(q-1)/2$ or $p=(q+1)/2$, we can assume that $C_{O}(g)\leq Z(G)$; a contradiction. Groups $L_{3}(q)$ and $U_{3}(q)$ contain an element $g$ such that $C_{S}(g)$ includes $L_{2}(q)$. Therefore, $S$ is not isomorphic to one of a $L_{3}(q)$ or $U_{3}(q)$. Similarly, it can be shown that $S$ is not isomorphic to $B_{2}(q),^{2}B_{2}(q),G_{2}(q),^{2}G_{2}(q)$ and sporadic groups.

Lemma 3.17 completes proof of the theorem.

4 Proof of Corollaries

Proof of Corollary 1.

Proof 4.1**.**

If $G$ is a $p$-group, then the corollary is satisfied. Let $h\in G$ be a $p^{\prime}$-element. We have $|C_{G}(h)|_{p}>|Z(P)|$, where $P\in Syl_{p}(G)$. Let $H\in Syl_{p}(C_{G}(h))$. It follows from Lemma 2.7 that $G/O_{p^{\prime}}(G)\in R^{*}(p)$. From Theorem 1 we get that $G/O_{p^{\prime}}(G)$ is a $p$-group. Since $|H|>|Z(P)|$, we get that $H$ contains $x$ such that $Ind_{G/O_{p^{\prime}}(G)}(xO_{p^{\prime}}(G))=p^{e}$, for some $e>0$. Therefore, $Ind_{G/O_{p^{\prime}}(G)}(xO_{p^{\prime}}(G))_{p}=Ind_{G}(x)_{p}=p^{e}$. Hence $C_{G}(h)$ includes some Sylow $p$-subgroup of $C_{G}(x)$. From Lemma 2.4 it follows that $Z(P)\leq Z(G)$.

Proof of Corollary 2.

Proof 4.2**.**

If $G\in R^{*}(p)$, then from Corollary 1 it follows that $Z(G)>1$; a contradiction. Therefore, $Ind_{G}(x)_{p}=1$ for each $p$-element $x$ of $G$. From Lemma 2.15 it follows that $O^{p^{\prime}}(G/O_{p^{\prime}}(G))=S_{1}\times\cdot\cdot\cdot\times S_{r}\times H$. From Lemmas 2.5 and 2.7 it follows that $O^{p^{\prime}}(G/O_{p^{\prime}}(G))\in R(p)$. Since $S_{1}$ is a subnormal subgroup of $G/O_{p^{\prime}}(G)$, we get $|x^{G/O_{p^{\prime}}(G)}|$ is a multiple of $|x^{S_{1}}|$ for each $x\in S_{1}$. Assume that $p=3$ and $S_{1}\simeq Ru$. We have $S_{1}$ contains $x$ of order $12$ such that $|C_{S_{1}}(x)|=24$. Therefore, $1<|x^{S_{1}}|_{3}<|G||_{3}$; a contradiction. If $S_{1}\simeq J_{4}$, then $S_{1}$ contains $x$ of order $21$ such that $|C_{S_{1}}(x)|=42$; a contradiction with definition of $R^{*}(p)$-groups. Assume that $S_{1}\simeq^{2}F_{4}(q)^{\prime}$. According to the results of [Shinoda], there is just one conjugacy class of elements of order $3$ in $S_{1}$. Therefore, $N_{S_{1}}(\langle x\rangle)\simeq C_{S_{1}}(x):2=3.U_{3}(q):2$ where $x\in S_{1}$ is an element of order $3$ [Malle]. Thus, $S_{1}$ contains an element $y$ of order $6$ such that $1<|y^{S_{1}}|_{3}<|G||_{3}$; a contradiction. Assume that $p=5$ and $S_{1}\simeq Th$. In this case $S_{1}$ contains an element $x$ of order $8$ such that $|C_{S_{1}}(x)|=96$; a contradiction. The assertion of Corollary 2 follows from Lemma 2.15.