Proof. Clearly, the ‘if’ parts of both cases are true. To prove the ‘only if’ parts, assume that the assertions are false and let G be a minimal counterexample for both cases. First, we will show that G has a unique minimal normal subgroup. To see why this is true, assume that G has two distinct minimal normal subgroups $E_1$ and $E_2$ . Then, for all strongly monolithic characters $\chi $ of $G/E_i$ ,

$$ \begin{align*} (NE_i/E_i)(\ker\chi /E_i)/(\ker\chi /E_i)\cong N\ker\chi/\!\ker\chi. \end{align*} $$

By the minimality of G, we obtain ${{\mathrm h}}(NE_i/E_i) \leq m\ (i=1,2)$ . Note that the 1–1 homomorphism $\theta : G \rightarrow G/E_1 \times G/E_1$ given by $g \mapsto (gE_1, gE_2)$ allows us to see N as a subgroup of $NE_1/E_1 \times NE_2/E_2$ , which yields

$$ \begin{align*} {\mathrm h}(N) \leq {\mathrm h}(NE_1/E_1 \times NE_2/E_2) \leq \max\{{\mathrm h}(NE_1/E_1), {\mathrm h}(NE_2/E_2)\} \leq m, \end{align*} $$

which is a contradiction. Thus, G has a unique minimal normal subgroup, say M, and so there exists a faithful irreducible character of G.

Now assume that $1< {\mathrm Z}(G)$ . Then, ${\mathrm h}(N)= {{\mathrm h}}(NZ(G)/Z(G)) \leq m$ , which contradicts the choice of G. Thus, G is centreless and so all faithful irreducible characters of G are strongly monolithic by [Reference Isaacs3, Lemma 2.27]. By the hypothesis of the theorem, we obtain the contradiction ${\mathrm h}(N) \leq m$ and this completes the proof of item (a).

Now let us assume that G is solvable and prove the ‘only if’ part of item (b). Now, G has a monomial strongly monolithic irreducible character. If $1<\Phi (G)$ , then ${\mathrm h}(N)= {\mathrm h}(N \Phi (G)/\Phi (G)) \leq m$ , which contradicts the choice of G. Thus, $\Phi (G)=1$ and so $ \mathrm F(G)=M $ has a complement H in G. Let $ 1\neq \lambda $ be an irreducible character of M. Note that M is abelian and so $\lambda (1)=1 $ . Then, $K:=I_{G}(\lambda )= MI_{H}(\lambda )$ and $M\cap I_H(\lambda )=1$ since M is complemented by H in G. By [Reference Isaacs3, Problem 6.18], there exists an irreducible character $\alpha $ of K such that $\alpha _M=\lambda $ . Note that $\alpha (1)=\lambda (1)=1$ . By [Reference Isaacs3, Theorem 6.11], $\alpha ^G $ is irreducible and faithful since $\lambda \neq 1$ is an irreducible constituent of $(\alpha ^G)_M$ . Thus, $\chi := \alpha ^G$ is faithful and monomial. We have already seen that all faithful irreducible characters of G are strongly monolithic. Thus, $\chi $ is a monomial, strongly monolithic character of G and so by hypothesis, ${\mathrm h}(N)= {\mathrm h}(N\ker \chi /\!\ker \chi ) \leq m$ , which is a contradiction. This contradiction completes the proof.

Proof. Let G be a minimal counterexample. Assume that G has no faithful, monomial, strongly monolithic character. By the minimality of G,

$$ \begin{align*} \mathrm{h}(G/\!\ker\chi) \leq |\mathrm{cod}_{\mathrm{msm}}(G/\!\ker\chi)|+1\leq |\mathrm{cod}_{\mathrm{msm}}(G)|+1 \end{align*} $$

for all $\chi \in \mathrm {Irr}_{\mathrm {msm}}(G)$ . However now, by using Lemma 2.1, $\mathrm {h}(G) \leq |\mathrm {cod}_{\mathrm {msm}}(G)|+1$ , which is a contradiction. Thus, G has at least one faithful, monomial, strongly monolithic character and so G has a unique minimal normal subgroup, say M.

Now assume that $N\lhd G$ and ${\mathrm h}(G/N)={\mathrm h}(G)$ . We argue that $N=1$ . Otherwise, by the minimality of G,

$$ \begin{align*} {\mathrm h}(G)={\mathrm h}(G/N)\leq |\mathrm{cod}_{\mathrm{msm}}(G/N)|+1\leq |\mathrm{cod}_{\mathrm{msm}}(G)|+1, \end{align*} $$

which is a contradiction. Thus, $N=1$ . This implies that $\Phi (G)=1={\mathrm Z}(G)$ and so $ \mathrm F(G)=M $ has a complement H in G.

Let $\psi $ be an irreducible character of $G/M$ with codegree as large as possible. Since $\mathrm {F}(G)=M \leq \ker \psi $ , G has an irreducible character $\theta $ with $a(\psi )< a(\theta )$ by [Reference Lu and Meng4, Lemma 2.11]. Note that $\theta $ is faithful since it does not lie in $\mathrm {Irr}(G/M)$ . Now let $\chi $ be a faithful irreducible character of G with the largest possible codegree among the faithful irreducible characters of G. We claim that $\chi $ is monomial. To see why this is true, let $\lambda $ be an irreducible constituent of $\chi _M$ . Clearly, $\lambda \neq 1$ . Now $K:=I_{G}(\lambda )= MI_{H}(\lambda )$ and $M\cap I_H(\lambda )=1$ since M is complemented by H in G. Thus, there exists an irreducible character $\alpha $ of K such that $\alpha _M=\lambda $ . Note that $\alpha (1)=\lambda (1)=1$ . From [Reference Isaacs3, Theorem 6.11], $\alpha ^G $ is irreducible and faithful since $\lambda \neq 1$ is an irreducible constituent of $(\alpha ^G)_M$ . Moreover, $\chi =\beta ^G$ for some $\beta \in \mathrm {Irr}(K)$ . By the choice of $\chi $ ,

$$ \begin{align*} |G|/|G:K|=a(\alpha^G)\leq a(\chi)=|G|/\chi(1)=|G|/\beta^G(1)=|G|/\beta(1)|G:K|, \end{align*} $$

which forces $\beta (1)=1$ . This means $\chi $ is monomial as desired. Therefore, $\chi $ is a monomial strongly monolithic character of G and so $\mathrm {a}(\chi )\in \mathrm {cod}_{\mathrm {msm}}(G)$ . However, ${a(\chi )\notin \mathrm {cod}_{\mathrm {msm}}(G/M)}$ since $a(\psi )<a(\theta ) \leq a(\chi )$ which means $|\mathrm {cod}_{\mathrm {msm}}(G/M)| \leq |\mathrm {cod}_{\mathrm {msm}}(G)|-1$ . Thus,

$$ \begin{align*} {\mathrm h}(G)={\mathrm h}(G/F(G))+1={\mathrm h}(G/M)+1 \leq |\mathrm{cod}_{\mathrm{msm}}(G/M)|+2 \leq |\mathrm{cod}_{\mathrm{msm}}(G)|+1. \end{align*} $$

This final contradiction completes the proof.

Proof. Let G be a minimal counterexample and note that the hypothesis is inherited by factor groups. Assume that G has no faithful strongly monolithic character. By the minimality of G,

$$ \begin{align*} \mathrm{h}(G/\!\ker\chi) \leq |\mathrm{cod}_{\mathrm{sm}}(G/\!\ker\chi)|+1\leq |\mathrm{cod}_{\mathrm{sm}}(G)|+1 \end{align*} $$

for every $\chi \in \mathrm {Irr}_{\mathrm {sm}}(G)$ . However now, by using Lemma 2.1, we obtain $\mathrm {h}(G) \leq |\mathrm {cod}_{\mathrm {sm}}(G)|+1$ , which is a contradiction. Thus, G has at least one faithful strongly monolithic character and this implies that all faithful irreducible characters of G are strongly monolithic. We also deduce that G has a unique minimal normal subgroup, say M.

Let $\chi $ be an irreducible character of G which does not contain M in its kernel. Then, $\chi $ is strongly monolithic since it is faithful and so p does not divide $\mathrm {a} (\chi )$ by hypothesis. Therefore, p does not divide the order of M and the action of P on M is Frobenius by [Reference Qian, Wang and Wei6, Theorem A], where P is a Sylow p-subgroup of G. Hence, G is solvable since M is nilpotent and $G/M$ is solvable by the minimality of G.

Now we argue that $\Phi (G)=1$ . Otherwise, we would have

$$ \begin{align*} {\mathrm h}(G)={\mathrm h}(G/\Phi(G))\leq |\mathrm{cod}_{\mathrm{sm}}(G/\Phi(G))|+1\leq |\mathrm{cod}_{\mathrm{sm}}(G)|+1, \end{align*} $$

which is a contradiction. Thus, $\Phi (G)=1$ which yields $\mathrm F(G)=M$ . By using [Reference Lu and Meng4, Lemma 2.11], we deduce that $|\mathrm {cod}_{\mathrm {sm}}(G/M)| \leq |\mathrm {cod}_{\mathrm {sm}}(G)|-1 $ and so

$$ \begin{align*} {\mathrm h}(G)={\mathrm h}(G/F(G))+1={\mathrm h}(G/M)+1 \leq |\mathrm{cod}_{\mathrm{sm}}(G/M)|+2 \leq |\mathrm{cod}_{\mathrm{sm}}(G)|+1, \end{align*} $$

which is the final contradiction completing the proof.

Proof. Assume that the theorem is false and let G be a minimal counterexample. Let $1< N \lhd G$ . Then, since $G/N$ is solvable by the minimality of G, we conclude that N is not solvable. In particular, N cannot be abelian. Thus, ${\mathrm Z}(G)=1$ .

It is not difficult to see that G has a unique minimal normal subgroup, say M. Note that M is not abelian. Let $1\neq \lambda $ be an irreducible character of M and choose an irreducible character $\chi $ of G with $[\chi _M, \lambda ]\neq 0$ . Note that $\chi $ is faithful and so strongly monolithic. Since $4\nmid \mathrm {a}(\chi )$ , we see that $4\nmid \mathrm a(\lambda )$ by [Reference Qian, Wang and Wei6, Lemma 2.1(2)], which means M also satisfies the hypothesis of the theorem. It turns out that $M=G$ is a simple group. From the equality

$$ \begin{align*} |G|=\sum_{\chi \in \mathrm{Irr}(G)}\chi(1)^2=1+\sum_{1 \neq \chi \in \mathrm{Irr}(G)}\chi(1)^2, \end{align*} $$

we deduce that G has a nonprincipal irreducible character, say $\chi $ , with odd degree. Since G is a simple group, we see that $\chi $ is a strongly monolithic character and so $4$ does not divide $\mathrm a(\chi )= |G|/\chi (1)$ by hypothesis. This forces the order of the Sylow 2-subgroup of G to be 2 since $\chi (1)$ is odd. This implies that G has a normal 2-complement. However, this contradicts the simplicity of G.

Proof. Note that all vertices of the graph $\Gamma (G)$ in [Reference Qian, Wang and Wei6] are isolated and so G has at most two prime divisors by [Reference Qian, Wang and Wei6, Theorem E(2)]. Hence, G is solvable.

Proof. Assume that the theorem is false and let G be a minimal counterexample. It is not difficult to see that G has a unique minimal normal subgroup, say M.

Let $1< N \lhd G$ . Then, since $G/N$ is solvable by the minimality of G, we conclude that N is not solvable. Thus, ${\mathrm Z}(G)=1$ and M is nonsolvable. It follows that G has a faithful irreducible character and all such characters are strongly monolithic.

Let $1\neq \lambda $ be an irreducible character of M and choose an irreducible character $\chi $ of G with $[\chi _M, \lambda ]\neq 0$ . Note that $\chi $ is faithful and so strongly monolithic since M is the unique minimal normal subgroup of G. Thus, $\mathrm a(\chi )$ is a prime power by hypothesis. Now, $\mathrm a(\lambda )$ is a prime power too, since $\mathrm a(\lambda ) \mid \mathrm a(\chi )$ by [Reference Qian, Wang and Wei6, Lemma 2.1(2)]. Thus, M also satisfies the hypothesis of the theorem which means $G=M$ is a simple group. However, this contradicts [Reference Qian, Wang and Wei6, Lemma 2.3].

Proof. Assume that the theorem is false and let G be a minimal counterexample. First, we argue that G has a unique minimal normal subgroup. To see why this is true, let M and N be two different minimal normal subgroups of G. By the minimality of G, the factor groups $G/M$ , $G/N$ and so $G/M \times G/N$ have normal Sylow p-subgroups. Thus, G, which is isomorphic to a subgroup of $G/M \times G/N$ , also has a normal Sylow p-subgroup, which is a contradiction with the choice of G. Thus, G has a unique minimal normal subgroup, say M, and so has a faithful irreducible character.

Now we claim that ${\mathrm Z}(G)=1$ . Otherwise, M is contained in ${\mathrm Z}(G)$ and so normalises P, where P is a Sylow p-subgroup of G. Since G is a minimal counterexample, we obtain $PM \lhd G$ and so, by using a Frattini argument, we see that $G=N_G(P)M$ , which means P is normal in G which is not the case. Thus, ${\mathrm Z}(G)=1$ as desired. This means all faithful irreducible characters of G are strongly monolithic. It turns out that G has a faithful strongly monolithic character, say $\chi $ . Then, $|G|/\chi (1)=\mathrm a(\chi )$ is a power of p by hypothesis. Thus, $\mathrm O_p(G) \neq 1$ by [Reference Chillag, Mann and Manz1, Theorem 4] and it follows that $M\leq \mathrm O_p(G) \leq P $ , which means $P/M$ is a Sylow p-subgroup of $G/M$ . By the minimality of G, we see that $P/M \lhd G/M$ , which is equivalent to $P \lhd G$ . However, this is a contradiction.