p-Singular characters and normal Sylow p-subgroups

1. Introduction

All groups and characters considered in this paper are finite and complex, respectively. The letter G always denotes a finite group. A character degree is p-singular if it is divisible by a prime p. The degrees of the complex irreducible characters of G often encode much structural information about G. We list two important results of this kind.

Thompson Theorem (see [18, Theorem 1]).Let G be a group and p a given prime. If every nonlinear irreducible character degree of G is p-singular, then G has a normal p-complement.

Itô–Michler Theorem (see [8, Theorem] and [12, Theorem 5.4]).The degree of every ordinary irreducible character of a finite group G is coprime to a prime p if and only if the Sylow p-subgroups of G are abelian and normal.

The interaction between the structure of finite groups and the character degrees has long been of interest. Some studies associate invariants with the character degrees of a finite group, and then study the effect of these invariants on the structure of the group, such as solvability, supersolvability, nilpotency, p-solvability and normality of p-complement, see for example [3, 4, 7, 9, 10, 11, 13, 14, 15, 16, 17, 19]. Many authors have studied the structure of finite groups using the ratios related to the character degree sum. Let $\mathrm{\text{T}}(G)$ be the sum of the degrees of all irreducible characters of G and $\mathrm{\text{R}}(G)=\frac{|G|}{\mathrm{\text{T}}(G)}$. Nekrasov and Berkovich [14] classified the finite nonabelian groups G with $\mathrm{\text{R}}(G)<p$, where p is the smallest prime divisor of $|G|$ such that a Sylow p-subgroup is not central. Tong-Viet [19, Theorem A] showed that if R$(G)<\frac{15}{4}$, then G is solvable and the bound cannot be improved. Maróti and Hung [11, Theorem 1.1] proved that if $\mathrm{\text{R}}(G)\le \frac{p}{\sqrt{3}}$ for a prime p, then G is p-solvable. They also showed that G is supersolvable if $\mathrm{\text{R}}(G)<2$.

Let $\mathrm{\text{Irr}}(G)$ be the set of all irreducible complex characters of G. In the past few decades, scholars have explored whether there exist subsets of $\mathrm{\text{Irr}}(G)$ that accurately capture the normality of Sylow subgroups. Naturally, it makes sense to consider the character degrees which are linear or p-singular.

Let G be a group and p a prime. We write

Note that $|G|={\sum }_{\chi \in \mathrm{\text{Irr}}(G)}\chi {(1)}^2$. Inspired by the study of $\mathrm{\text{R}}(G)$, we consider the following invariant:

Hence the Itô–Michler Theorem is equivalent to the following: ${\mathrm{\text{R}}}_p(G)=1$ if and only if the Sylow p-subgroups of G are abelian and normal. Pan et al. [15] proved that if ${\mathrm{\text{R}}}_2(G)<\frac{3}{2}$, then G has a normal Sylow 2-subgroup. An obvious question is the following: is there an analogue of the result for odd prime? Let p be an odd prime. Define

Inspired by [4], we obtain the following results.

Theorem 1.1. Let G be a group and p an odd prime. If ${\mathrm{\text{R}}}_p(G)<f(p)$, then G is p-solvable.

As usual, we use $O^{p^{\prime }}(G)$ to denote the smallest normal subgroup M of G such that $G/M$ is a $p^{\prime }$-group.

Theorem 1.2. Let G be a group and p an odd prime. If ${\mathrm{\text{R}}}_p(G)<f(p)$, then $O^{p^{\prime }}(G)$is solvable.

Remark 1.3. Note that $O^{3^{\prime }}({\mathrm{\text{A}}}_5)={\mathrm{\text{A}}}_5$ and $O^{p^{\prime }}({\mathrm{\text{PSL}}}_2(p))={\mathrm{\text{PSL}}}_2(p)$ for prime $p\ge 5$. Since ${\mathrm{\text{R}}}_3({\mathrm{\text{A}}}_5)=\frac{19}{7}$ and ${\mathrm{\text{R}}}_p({\mathrm{\text{PSL}}}_2(p))=\frac{1+p^2}{1+p}$ for prime $p\ge 5$, the bounds in Theorems 1.1 and 1.2 are best possible.

Theorem 1.4. Let G be a group and p a prime. If ${\mathrm{\text{R}}}_p(G)<\frac{p+1}{2}$, then G has a normal Sylow p-subgroup.

Remark 1.5. The bound in Theorem 1.4 may not be the best possible for all primes. If the cyclic group $Z_p$ admits a faithful nontrivial action on an abelian group A of order $p+1$, then the bound is best possible (in this case, we can construct the semidirect product $G=Z_p⋉A$ with ${\mathrm{\text{R}}}_p(G)=\frac{p+1}{2}$). In particular, if p is a Mersenne prime, then the bound is best possible. On the other hand, for example, $Z_5$ can only act trivially on an abelian group of order 6, so the best bound may not be 3 for $p=5$. For prime $p\ge 5$, ${\mathrm{\text{R}}}_p({\mathrm{\text{PSL}}}_2(p))=\frac{1+p^2}{1+p}$. We conjecture that if $p\ge 5$ is not a Mersenne prime, then the optimal bound is $\frac{1+p^2}{1+p}$.

We end this section with some convenient notation. We denote by $\mathrm{\text{Lin}}(G)$ the set of all linear characters of G. If H is a subgroup of G and $\lambda \in \mathrm{\text{Irr}}(H)$, then we write $\mathrm{\text{Irr}}(G|\lambda )$ and $\mathrm{\text{Lin}}(G|\lambda )$ for the set of all irreducible and linear constituents of the induced character ${\lambda }^G$, respectively. The intersection of $\mathrm{\text{Irr}}(G|\lambda )$ and ${\mathrm{\text{Irr}}}_p(G)$ is denoted by ${\mathrm{\text{Irr}}}_p(G|\lambda )$. Given a positive integer d, we write $n_d(G)$ to denote the number of irreducible characters of G that have degree d. If N is a normal subgroup of G and $𝜗\in \mathrm{\text{Irr}}(N)$, then the inertia group of $𝜗$ in G is denoted by $I_G(𝜗)$, and $\mathrm{\text{Irr}}(G|N)$ denotes the set of all irreducible characters of G whose kernels do not contain N, and furthermore the intersection of $\mathrm{\text{Irr}}(G|N)$ and ${\mathrm{\text{Irr}}}_p(G)$ is denoted by ${\mathrm{\text{Irr}}}_p(G|N)$. For a nonempty subset $\mathrm{\Omega }$ of ${\mathrm{\text{Irr}}}_p(G)$, we write

Moreover, if ${\mathrm{\text{Irr}}}_p(G|𝜗)$ and ${\mathrm{\text{Irr}}}_p(G|N)$ are nonempty sets, then we replace ${\mathrm{\text{R}}}_p({\mathrm{\text{Irr}}}_p(G|𝜗))$ and ${\mathrm{\text{R}}}_p({\mathrm{\text{Irr}}}_p(G|N))$ with ${\mathrm{\text{R}}}_p(G|𝜗)$ and ${\mathrm{\text{R}}}_p(G|N)$, respectively.

All unexplained notation and terminology is standard and appears for example in [6].

2. Preliminaries

In this section, for the sake of convenience, we prove a lemma and state some known results used in the remainder of this paper.

Lemma 2.1 (see [3, Proposition 3.2]). Let G be a group with a nonabelian minimal normal subgroup N. Assume that there exists some $\psi \in \mathrm{\text{Irr}}(N)$such that $\psi$is extendible to $I_G(\psi )$. Then $n_1(G)\le n_d(G)|G:I_G(\psi )|$, where $d=\psi (1)|G:I_G(\psi )|$.

Lemma 2.2 (see [4, Theorem 3.3]). Let G be a group with a nonabelian minimal normal subgroup N, and p a prime. If $p||N|$, then there exists $\mu \in \mathrm{\text{Irr}}(N)$such that $p|\mu (1)$and $\mu$is extendible to $I_G(\mu )$.

Lemma 2.3. Let G be a group and p a prime. Suppose that ${\mathrm{\text{Irr}}}_p^1(G)$is the subset of ${\mathrm{\text{Irr}}}_p(G)$which contains all linear characters of G. If ${\mathrm{\text{R}}}_p(G)\le p$, then

In particular, if $N⊴G$with $N\le G^{\prime }$and ${\mathrm{\text{R}}}_p(G)\le p$, then ${\mathrm{\text{R}}}_p(G/N)\le {\mathrm{\text{R}}}_p(G)$.

Proof. By hypothesis,

and It follows that The proof is complete. □

Lemma 2.4 (see [4, Proposition 2.4]). Let G be a group with minimal normal subgroup $N\cong N_1\times \cdots \times N_n$, where $n\ge 2$and all of the $N_i$’s are isomorphic to a nonabelian simple group S. Let K be the kernel of the action of G on $\{N_1,\dots ,N_n\}$. If $|G/K|$is divisible by a prime p, then there exists $\mu \in \mathrm{\text{Irr}}(N)$such that $\mu (1)\ge 12$, $\mu$is extendible to a character of $I_G(\mu )$, and $p||G:I_G(\mu )|$.

3. Proof of Theorems 1.1 and 1.2

Proof of Theorem 1.1. Suppose that the theorem is false and let G be a counterexample of minimal order. In particular, G is not p-solvable and ${\mathrm{\text{R}}}_p(G)<f(p)$. We may assume that $G^{\prime }\ne 1$. Let N be a minimal normal subgroup of G with $N\le G^{\prime }$. By Lemma 2.3, ${\mathrm{\text{R}}}_p(G/N)\le {\mathrm{\text{R}}}_p(G)<f(p)$. The choice of G ensures that $G/N$ is p-solvable. Therefore N is not p-solvable and so $p||N|$. Without loss of generality, we may assume that

where S is a nonabelian simple group with $p||S|$.

By Lemma 2.2, there exists $\mu \in \mathrm{\text{Irr}}(N)$ such that $p|\mu (1)$ and $\mu$ is extendible to $I_G(\mu )$. It follows from Lemma 2.1 that $n_1(G)\le n_d(G)|G:I_G(\mu )|$, where $d=\mu (1)|G:I_G(\mu )|$. Since $d=\mu (1)|G:I_G(\mu )|\ge p|G:I_G(\mu )|$,

It follows that

Let ${\mathrm{\text{Irr}}}_p^1(G)=\{\chi \in \mathrm{\text{Irr}}(G):\chi (1)=1\mathrm{\text{or}}d\}$. Since ${\mathrm{\text{R}}}_p(G)\le p$, by Lemma 2.3

This contradicts our assumption if $p\ge 5$. So it remains to consider the case $p=3$. From the classification of finite irreducible subgroups of ${\mathrm{\text{GL}}}_3(ℂ)$ in [1], ${\mathrm{\text{A}}}_5$ and ${\mathrm{\text{PSL}}}_2(7)$ are the only nonabelian simple groups possessing an irreducible character of degree 3. We know that ${\mathrm{\text{PSL}}}_2(7)$ has an irreducible character of degree 6 which is extendible to $\mathrm{\text{Aut}}({\mathrm{\text{PSL}}}_2(7))$. Hence if $N\ncong {\mathrm{\text{A}}}_5$, then by Lemma 2.2 there exists $\mu \in \mathrm{\text{Irr}}(N)$ such that $3|\mu (1)$ with $\mu (1)\ge 6$, and $\mu$ is extendible to $I_G(\mu )$. By Lemma 2.1, $n_1(G)\le n_d(G)|G:I_G(\mu )|$, where $d=\mu (1)|G:I_G(\mu )|$. Since $d=\mu (1)|G:I_G(\mu )|\ge 6|G:I_G(\mu )|$, or equivalently, Now by Lemma 2.3, a contradiction.

If $N\cong {\mathrm{\text{A}}}_5$, then by [2], we deduce that N has two distinct irreducible characters of degree 3, say ${𝜃}_1$ and ${𝜃}_2$, and these two characters fuse to give a single character of ${\mathrm{\text{S}}}_5$. Since $|\mathrm{\text{Aut}}(N):N|=2$ and $N{C}_G(N)\le I_G({𝜃}_1)$, $|G:I_G({𝜃}_1)|\le 2$. If $|G:I_G({𝜃}_1)|=2$, then $I_G({𝜃}_1)=N\times C_G(N)$ and so ${𝜃}_1$ is extendible to $I_G({𝜃}_1)$. By Lemma 2.1, $n_1(G)\le 2n_6(G)$. This implies that

It follows from Lemma 2.3 that ${\mathrm{\text{R}}}_3(G)\ge \frac{19}{4}$, a contradiction. If $|G:I_G({𝜃}_1)|=1$, then G does not induce an outer automorphism of N. Hence $G\cong N\times C_G(N)$ and so ${𝜃}_1$ and ${𝜃}_2$ are extendible to G. Since $N\le G^{\prime }$, by [6, Corollary 6.17] $2n_1(G)\le n_3(G)$. This implies that It follows from Lemma 2.3 that ${\mathrm{\text{R}}}_3(G)\ge \frac{19}{7}$, a contradiction.

The proof is now complete. □

Remark 3.1. Let N be a normal subgroup of a group G and p a prime number. Assume that $\{{𝜗}_1,\dots ,{𝜗}_t\}$ is a complete set of representatives of the action of G on $\mathrm{\text{Irr}}(N)$; now $\mathrm{\text{Irr}}(G)={\bigcup }_{i=1}^t\mathrm{\text{Irr}}(G|{𝜗}_i)$ is a disjoint union. Hence there exists some $i\in \{1,\dots ,t\}$ such that ${\mathrm{\text{Irr}}}_p(G|{𝜗}_i)\ne \varnothing$.

Corollary 3.2. Let N be a normal subgroup of a group G and p an odd prime. Assume that $𝜗\in \mathrm{\text{Irr}}(N)$such that ${\mathrm{\text{Irr}}}_p(G|𝜗)\ne \varnothing$. If $G/N$is non-p-solvable, then ${\mathrm{\text{R}}}_p(G|𝜗)\ge f(p)$.

Proof. We prove the corollary by induction on $|G/N|$. Let $M⊴G$ with $N\le M$ be maximal such that $G/M$ is non-p-solvable. We claim that $G/M$ has a unique minimal normal subgroup $L/M$ and $L/M$ is non-p-solvable. Suppose by contradiction that $L_1/M$ and $L_2/M$ are distinct minimal normal subgroups of $G/M$. Then $G/L_1$ and $G/L_2$ are p-solvable by the choice of M. Hence $G/M=G/(L_1\cap L_2)\lesssim G/L_1\times G/L_2$ is p-solvable. This contradiction proves the uniqueness. Observe that $(G/M)/(L/M)\cong G/L$ is p-solvable. Thus, if $L/M$ is p-solvable, then $G/M$ is p-solvable; this contradiction proves the claim.

If $N\lneqq M$, then by induction for every $\lambda \in \mathrm{\text{Irr}}(M|𝜗)$, either ${\mathrm{\text{Irr}}}_p(G|\lambda )=\varnothing$ or ${\mathrm{\text{R}}}_p(G|\lambda )\ge f(p)$. Note that

where $e_{\lambda }=[{𝜗}^M,\lambda ]$ is a positive integer. Since ${\mathrm{\text{Irr}}}_p(G|𝜗)\ne \varnothing$, there must exist some $𝜃\in \mathrm{\text{Irr}}(M|𝜗)$ such that ${\mathrm{\text{Irr}}}_p(G|𝜃)\ne \varnothing$, and therefore ${\mathrm{\text{R}}}_p(G|𝜃)\ge f(p)$. It follows that ${\mathrm{\text{R}}}_p(G|𝜗)\ge f(p)$, as desired.

Now we may assume that $N=M$. If $𝜗$ does not extend to G, then ${\mathrm{\text{Irr}}}_p(G|𝜗)$ contains no linear character, so ${\mathrm{\text{R}}}_p(G|𝜗)\ge p>f(p)$, whence the result. So we may assume that $𝜗$ is extendible to $\chi \in \mathrm{\text{Irr}}(G)$, then by [6, Corollary 6.17], the $\beta \chi$ for $\beta \in \mathrm{\text{Irr}}(G/N)$ are irreducible, distinct for distinct $\beta$ and are all of the irreducible constituents of $\mathrm{\text{Irr}}(G|𝜗)$. If $p|\chi (1)$, then ${\mathrm{\text{Irr}}}_p(G|𝜗)=\mathrm{\text{Irr}}(G|𝜗)$. Thus

as required. If $p\nmid \chi (1)$, then ${\mathrm{\text{Irr}}}_p(G|𝜗)=\{\chi \beta :\beta \in {\mathrm{\text{Irr}}}_p(G/N)\}$. Hence Since $G/N$ is non-p-solvable, ${\mathrm{\text{R}}}_p(G/N)\ge f(p)$ by Theorem 1.1. Therefore ${\mathrm{\text{R}}}_p(G|𝜗)=\chi (1){\mathrm{\text{R}}}_p(G/N)\ge f(p)$, as claimed. The proof is complete. □

Proof of Theorem 1.2. Suppose that the theorem is false and let G be a counterexample of minimal order. In particular, $O^{p^{\prime }}(G)$ is not solvable and ${\mathrm{\text{R}}}_p(G)<f(p)$. By Theorem 1.1, G is p-solvable. Let $N/L$ be a nonabelian chief factor of G below $O^{p^{\prime }}(G)$ such that $|N|$ is smallest possible. Hence L is solvable and $N=N^{\prime }$, thereby $L\le G^{\prime }$. By Lemma 2.3, ${\mathrm{\text{R}}}_p(G/L)\le {\mathrm{\text{R}}}_p(G)<f(p)$, and so $L=1$ by the choice of G. Thus N is a nonabelian minimal normal subgroup of G. Without loss of generality, we may assume that

where all of the $N_i$’s are isomorphic to a nonabelian simple group S. Since G is p-solvable, N is a $p^{\prime }$-group and so is S. Note that $O^{p^{\prime }}(G)C_G(N)$ since $NC_G(N)$. Hence $p||G/C_G(N)|$. Let K be the kernel of the action of G on $\{N_1,\dots ,N_n\}$. Then $C_G(N)\le K$ and so either $p||G/K|$ or $p||K/C_G(N)|$.

If $p||G/K|$, then $n\ge 2$. By Lemma 2.4, there exists $\mu \in \mathrm{\text{Irr}}(N)$ such that $\mu (1)\ge 12$, $\mu$ is extendible to a character of $I_G(\mu )$, and $p||G:I_G(\mu )|$. Lemma 2.1 implies that $n_1(G)\le n_d(G)|G:I_G(\mu )|$, where $d=\mu (1)|G:I_G(\mu )|\ge 12|G:I_G(\mu )|\ge 12p$ is divisible by p. Hence

and It follows that and By Lemma 2.3, we get a contradiction.

So we may assume that $p||K/C_G(N)|$. Arguing as in the proof of [4, Theorem 3.5], we find $1_N\ne \mu \in \mathrm{\text{Irr}}(N)$ such that $\mu$ is extendible to $I_G(\mu )$ and $p||G:I_G(\mu )|$. By Lemma 2.1, $n_1(G)\le n_d(G)|G:I_G(\mu )|$, where $d=\mu (1)|G:I_G(\mu )|\ge p\mu (1)$ is divisible by p. Hence

and This implies that and By Lemma 2.3, we get the final contradiction. □

Corollary 3.3. Let p be an odd prime and $1\ne N⊴G$with ${\mathrm{\text{Irr}}}_p(G|N)\ne \varnothing$. If ${\mathrm{\text{R}}}_p(G|N)<f(p)$, then G is p-solvable and $O^{p^{\prime }}(G)$is solvable.

Proof. Since ${\mathrm{\text{R}}}_p(G|N)<f(p)<p$, ${\mathrm{\text{Irr}}}_p(G|N)$ contains some linear character $\lambda$. Thus ${\lambda }_N\ne 1_N$ and ${\lambda }_N\in \mathrm{\text{Irr}}(N)$. Let $\{{𝜗}_1,\dots ,{𝜗}_t\}$ be the set of all linear characters of N such that every ${𝜗}_i\ne 1_N$ is extendible to ${\chi }_i\in \mathrm{\text{Irr}}(G)$. By [6, Corollary 6.17], the $\beta {\chi }_i$ for $\beta \in \mathrm{\text{Irr}}(G/N)$ are irreducible, distinct for distinct $\beta$ and are all of the irreducible constituents of $\mathrm{\text{Irr}}(G|{𝜗}_i)$. Hence ${\mathrm{\text{Irr}}}_p(G|{𝜗}_i)=\{\beta {\chi }_i:\beta \in {\mathrm{\text{Irr}}}_p(G/N)\}$. Since ${\chi }_i\in \mathrm{\text{Irr}}(G|N)$, ${\mathrm{\text{Irr}}}_p(G|{𝜗}_i)\subseteq {\mathrm{\text{Irr}}}_p(G|N)$. Note that ${𝜗}_i$ and ${𝜗}_j$ are not G-conjugate for $1\le i\ne j\le t$. If $1\le i\ne j\le t$ and there exists $\chi \in \mathrm{\text{Irr}}(G|{𝜗}_i)\cap \mathrm{\text{Irr}}(G|{𝜗}_j)$, then ${𝜗}_i$ and ${𝜗}_j$ are irreducible constituents of ${\chi }_N$. By [6, Theorem 6.2], ${𝜗}_i$ and ${𝜗}_j$ are G-conjugate, a contradiction. This means that ${\bigcup }_{i=1}^t{\mathrm{\text{Irr}}}_p(G|{𝜗}_i)$ is a disjoint union. Therefore

Let $\mathrm{\Omega }={\mathrm{\text{Irr}}}_p(G|N)\setminus {\bigcup }_{i=1}^t{\mathrm{\text{Irr}}}_p(G|{𝜗}_i)$. If $\mu \in \mathrm{\Omega }$, then $p|\mu (1)$. Otherwise, there exists some $1\le k\le t$ such that $\mu \in {\mathrm{\text{Irr}}}_p(G|{𝜗}_k)$, a contradiction. Thus ${\mathrm{\text{R}}}_p(\mathrm{\Omega })\ge p>f(p)$. If ${\mathrm{\text{R}}}_p({\bigcup }_{i=1}^t{\mathrm{\text{Irr}}}_p(G|{𝜗}_i))\ge f(p)$, then ${\mathrm{\text{R}}}_p(G|N)\ge f(p)$, a contradiction. So we may assume that ${\mathrm{\text{R}}}_p({\bigcup }_{i=1}^t{\mathrm{\text{Irr}}}_p(G|{𝜗}_i))<f(p)$, that is, ${\mathrm{\text{R}}}_p(G/N)<f(p)$. Therefore ${\mathrm{\text{R}}}_p(G)<f(p)$, so G is p-solvable and $O^{p^{\prime }}(G)$ is solvable by Theorems 1.1 and 1.2, respectively. The proof is now complete. □

4. Proof of Theorem 1.4

Theorem 4.1. Let p be a prime and $G=N⋊H$, where N is abelian. Assume that, in the action of H on $\mathrm{\text{Irr}}(N)\setminus \{1_N\}$, there exists an orbit of size 1 or divisible by p. If ${\mathrm{\text{R}}}_p(G)\le p$, then there exists an orbit ${𝒪}$, in the action of H on $\mathrm{\text{Irr}}(N)\setminus \{1_N\}$, such that $|{𝒪}|=1$or $p||{𝒪}|$, and $\frac{|{𝒪}|r+1}{r+1}\le {\mathrm{\text{R}}}_p(G)$, where r is the number of H-orbits on $\mathrm{\text{Irr}}(N)\setminus \{1_N\}$whose sizes are 1 or divisible by p.

Proof. Let $\{{\alpha }_0=1_N,{\alpha }_1,\dots ,{\alpha }_r,{\alpha }_{r+1},\dots ,{\alpha }_t\}$ be a set of representatives of the action of H on Irr$(N)$, where $\{{\alpha }_0,{\alpha }_1,\dots ,{\alpha }_r\}$ are representatives of those orbits whose sizes are 1 or divisible by p. By assumption, $r\ge 1$. For $0\le i\le t$, let $I_i=I_G({\alpha }_i)$ and we write

Since ${\alpha }_i$ is linear and $I_i$ splits over N, ${\alpha }_i$ extends to a linear character ${\beta }_i$ of $I_i$. By [6, Corollary 6.17 and Theorem 6.11], there is a bijection $\lambda \mapsto {(\lambda {\beta }_i)}^G$ from $\mathrm{\text{Irr}}(I_i/N)$ onto $\mathrm{\text{Irr}}(G|{\alpha }_i)$. Note that ${(\lambda {\beta }_i)}^G(1)=|G:I_i|\lambda (1)$, so $p|{(\lambda {\beta }_i)}^G(1)$ if and only if exactly one of the following holds: (i) $p|\lambda (1)$ or (ii) $p||G:I_i|$ and $p\nmid \lambda (1)$. Since $\mathrm{\text{Irr}}(G)$ is the disjoint union of the sets $\mathrm{\text{Irr}}(G|{\alpha }_i)$ for $0\le i\le t$, and Since ${\sum }_{\chi \in {\mathrm{\text{Irr}}}_p(G)}\chi {(1)}^2={\mathrm{\text{R}}}_p(G){\sum }_{\chi \in {\mathrm{\text{Irr}}}_p(G)}\chi (1)$ and ${\mathrm{\text{R}}}_p(G)\le p$, Notice that $S_{i,p}\ge p{T}_{i,p}$, so Since $p{\sum }_{i=0}^t|G:I_i{|}^2{T}_{i,p}+{\sum }_{p||G:I_i|}|G:I_i{|}^2{S}_{i,p^{\prime }}\ge p{\sum }_{i=0}^t|G:I_i|T_{i,p}+p{\sum }_{p||G:I_i|}|G:I_i|T_{i,p^{\prime }}$, Since $I_0=G$, Since $n_1(G/N)=|G/N:{(G/N)}^{\prime }|$ and $n_1(I_i/N)=|I_i/N:{(I_i/N)}^{\prime }|$, for every $1\le i\le r$. Therefore Combining inequalities (4.1) and (4.2), Therefore, there exists some $1\le j\le r$ such that that is, where ${𝒪}$ is the orbit that contains ${\alpha }_j$. The proof is now complete. □

Proof of Theorem 1.4. The case $p=2$ was proved in [15, Theorem A], therefore we assume that p is odd. Suppose that the theorem is false and let G be a counterexample of minimal order. In particular, ${\mathrm{\text{R}}}_p(G)<\frac{p+1}{2}<f(p)$. Hence $O^{p^{\prime }}(G)$ is solvable by Theorem 1.2. If $G^{\prime }\cap O^{p^{\prime }}(G)=1$, then $O^{p^{\prime }}(G)$ is abelian. Thus the Sylow p-subgroup of $O^{p^{\prime }}(G)$ is normal in G, a contradiction.

Therefore, we may assume that $G^{\prime }\cap O^{p^{\prime }}(G)\ne 1$. Let N be a minimal normal subgroup of G such that $N\le G^{\prime }\cap O^{p^{\prime }}(G)$. Since $O^{p^{\prime }}(G)$ is solvable, N is an elementary abelian q-group for some prime q. It follows from Lemma 2.3 and the choice of G that $G/N$ has a normal Sylow p-subgroup, say $P_1/N$. If $q=p$, then $P_1$ is a normal Sylow p-subgroup of G, a contradiction. So it remains to consider the case $q\ne p$. By the Schur–Zassenhaus Theorem [5, Kapitel I, Hauptsatz 18.1], $P_1=P⋉N$, where P is a Sylow p-subgroup of $P_1$ and also of G. By Frattini’s argument [5, Kapitel I, Satz 7.8], $G=P_1{N}_G(P)=N{N}_G(P)$ and so $NN_G(P)$. By the minimality of N, $N\cap N_G(P)=1$, that is, $G=N⋊N_G(P)$. Assume that there exists $1_N\ne \lambda \in \mathrm{\text{Irr}}(N)$ such that $\lambda$ is $N_G(P)$-invariant. Then for arbitrary $g\in N_G(P)$ and $n\in N$, $\lambda (n)={\lambda }^g(n)$. Hence $[g,n]\in \mathrm{\text{Ker}}(\lambda )$, that is, $[N_G(P),N]\le \mathrm{\text{Ker}}(\lambda )<N$ since $\lambda \ne 1_N$. By [5, Kapitel III, Hilfssatz 1.10], we deduce that

The minimality of N implies that $[N_G(P),N]=1$, and thereby $P⊴G$, a contradiction. So we may assume that no nonprincipal irreducible character of N is $N_G(P)$-invariant. By Theorem 4.1, all orbits in the action of $N_G(P)$ on $\mathrm{\text{Irr}}(N)\setminus \{1_N\}$ have sizes coprime to p. Otherwise, there exists an orbit ${𝒪}$ such that $p||{𝒪}|$ and where $r\ge 1$ is the number of $N_G(P)$-orbits on $\mathrm{\text{Irr}}(N)\setminus \{1_N\}$ whose sizes are divisible by p. This contradicts our assumption. Hence all orbits in the action of P on $\mathrm{\text{Irr}}(N)$ have sizes coprime to p. Therefore, P acts trivially on $\mathrm{\text{Irr}}(N)\cong N$, and so $P_1=P\times N$. The final contradiction completes the proof. □

Acknowledgments

We thank the referee for careful reading of this paper and for valuable comments which helped improve our paper. This research was supported by NSF of China (Nos. 12071484 and 12271527).