Proof of Theorem 1.3.

Let $C:=C_G(M)$ and $\overline {G}=G/C$ . Then, $\overline {G}$ has a unique minimal normal subgroup $\overline {M}$ and $M\cong \overline {M}=MC/C$ . It follows that there exists a nonabelian simple group $\overline {S}$ such that $\overline {M}=\overline {S_1}\times \cdots \times \overline {S_n}$ and $\overline {S_i}\cong \overline {S}$ for every i.

(i) Let $p \in \pi (M)$ . Suppose first that $\overline {S}$ is a group of Lie type or $p\geqslant 5$ . From [Reference Dolfi, Pacifici, Sanus and Spiga6, Proposition 2.1], there exists an irreducible character $\theta $ of $\overline {S}$ of p-defect zero. Then, the irreducible character $\psi =\theta \times \cdots \times \theta $ of $\overline {M}$ is of p-defect zero. Let $x \in M$ be an element of order p. It follows from the proof of [Reference Erkoç, Bozkurt Güngör and Akar7, Lemma 2.4] that $\overline {G}$ has a strongly monolithic character $\chi $ such that $Z(\chi )=\ker \chi $ and $\chi (\overline {x})=0$ . Let $\tilde {\chi }\in \mathrm {Irr}(G)$ be the corresponding character of $\chi $ . Then, $\tilde {\chi } \in \mathrm {Irr}_{\mathrm {sm}^*}(G)$ and $\tilde {\chi }(x)=0$ ; that is, $x \in M$ is an $\mathrm {SM}^*$ -vanishing p-element of G.

Suppose now that $p\in \{2,3\}$ and that $\overline {S}$ is a sporadic simple group or $\overline {S}\cong A_n$ for some $n\geqslant 7$ . From [Reference Dolfi, Pacifici, Sanus and Spiga6, Lemma 2.3 and Proposition 2.4], there exists an irreducible character $\theta $ of $\overline {S}$ that extends to $\mathrm {Aut}(\overline {S})$ and $\theta (\overline {x})=0$ for some p-element $\overline {x}\in \overline {S}$ . Also, there exists a p-element g of M such that $\overline {g}=\overline {x}\times \cdots \times \overline {x} \in \overline {S_1}\times \cdots \times \overline {S_n} = \overline {M}$ . Thus, $\overline {g}$ is a p-element of $\overline {M}$ . However, it follows from [Reference Bianchi, Chillag, Lewis and Pacifici2, Lemma 5] that $\theta \times \cdots \times \theta \in \mathrm {Irr}(\overline {M})$ extends to $\overline {G}$ . Now, let $\chi \in \mathrm {Irr}(\overline {G})$ be such that $\chi _{\overline {M}}=\theta \times \cdots \times \theta $ . Then, $\chi (\overline {g})=(\theta (\overline {x}))^n=0$ . Since $\chi $ is a faithful irreducible character of $\overline {G}$ and $\mathrm {Z}(\overline {G})=1$ , it follows that $\chi $ is a strongly monolithic character of $\overline {G}$ with the property that $Z(\chi )=\ker \chi $ . Similarly, let $\tilde {\chi }\in \mathrm {Irr}(G)$ be the corresponding character of $\chi $ . Then, $\tilde {\chi } \in \mathrm {Irr}_{\mathrm {sm}^*}(G)$ and $\tilde {\chi }(g)=0$ . This completes the proof of statement (i).

(ii) From [Reference Dolfi, Malle and Navarro4, Theorem D], $\overline {M}$ has a real element $\overline {m}$ of order $5$ . Since $\overline {S}$ has an irreducible character of $5$ -defect zero, $\overline {M}$ has an irreducible character of $5$ -defect zero. Thus, from [Reference Erkoç, Bozkurt Güngör and Akar7, Lemma 2.4], $\overline {m} \in \overline {M}$ is a real $\mathrm {SM}^*$ -vanishing element of order $5$ of $\overline {G}$ . Also, by [Reference Navarro, Sanus and Tiep13, Lemma 3.2], there is a real element $n \in G$ such that $\overline {m}=mC=nC$ . Since $n \in MC$ , we can choose a real element $n_0\in M$ such that $\overline {m}=n_0C$ . Hence, $n_0\in M$ is a real $\mathrm {SM}^*$ -vanishing $5$ -element of G.

Proof. Assume that G is a nonsolvable group. Let M be the maximal solvable normal subgroup of G and $N/M$ be a minimal normal subgroup of $G/M$ . Clearly, $N/M$ is nonabelian. Let $p,q \in \pi (N/M)$ such that $p\neq q$ . From Theorem 1.3(i), $G/M$ has strongly monolithic characters $\chi _1$ and $\chi _2$ such that $Z(\chi _i)=\ker \chi _i$ for $i=1,2$ and these characters vanish on a p-element $x_1M$ and a q-element $x_2M$ of $N/M$ , respectively. Moreover, $x_1$ and $x_2$ can be chosen as a p-element and a q-element of N, respectively. Let $\tilde {\chi _i}\in \mathrm {Irr}(G)$ be the corresponding characters of $\chi _i$ for $i=1,2$ . Then, $\tilde {\chi _i}$ is a strongly monolithic character of G with the property that $Z(\tilde {\chi _i})=\ker \tilde {\chi _i}$ and $\tilde {\chi _i}(x_i)=0$ for each i. Since $|x_1|$ and $|x_2|$ both lie in $\mathrm {SMVo}(G)$ and gcd $(\vert x_1\vert ,\,\vert x_2\vert )=1$ , it follows that $\Gamma _{\mathrm {sm}}(G)$ is not a complete graph.

Proof. This is an easy consequence of Corollary 2.1.

Proof. Suppose that G is not a solvable group. Then, $\mathrm {SMVo}_{p'}(G)\neq \emptyset $ ; otherwise, by Corollary 2.2, G would be solvable since every element of $\mathrm {SMVo}(G)$ is divisible by p. Let M be the maximal solvable normal subgroup of G and $N/M$ be a minimal normal subgroup of $G/M$ . Clearly, $N/M$ is nonabelian. Let $q,r \in \pi (N/M)-\{p\}$ . It follows from Theorem 1.3(i) that $G/M$ has strongly monolithic characters $\chi _1$ and $\chi _2$ such that $Z(\chi _i)=\ker \chi _i$ for $i=1,2$ and they vanish on a q-element $x_1M$ and an r-element $x_2M$ of $N/M$ , respectively. Hence, $|\mathrm {SMVo}_{p'}(G)|\geq 2$ , which is a contradiction. This contradiction completes the proof.

Proof. See the proof of [Reference Erkoç, Bozkurt Güngör and Akar7, Lemma 2.6].

Proof of Theorem 1.4.

Let G be a minimal counterexample. Assume that P is a Sylow p-subgroup of G. Thus, $P\neq 1$ . Moreover, G has at least one strongly monolithic character $\chi $ with the property that $Z(\chi )=\ker \chi $ ; otherwise, G would be nilpotent by [Reference Özkan and Erkoç14, Theorem B(i)], which is a contradiction. Now, let M be a minimal normal subgroup of G. Since G is a minimal counterexample and there is no p-element $gM$ of $G/M$ such that $\chi (g)=0$ , where $\chi \in \mathrm {Irr}_{\mathrm {sm}^*}(G)$ and $M\leq \ker \chi $ , we know that $G/M$ has a normal Sylow p-subgroup. Suppose that p divides $|M|$ . Then, M must be nonsolvable. However, this gives a contradiction with Theorem 1.3(i). Therefore, p does not divide $|M|$ .

Let N be a minimal normal subgroup of G such that $N\neq M$ . Similarly, $G/N$ has a normal Sylow p-subgroup by the minimality of G. Since G embeds into $(G/M)\times (G/N)$ , it follows that G has a normal Sylow p-subgroup, which is a contradiction. Therefore, G has a unique minimal normal subgroup M. Moreover, $PM/M$ is a normal Sylow p-subgroup of $G/M$ by the minimality of G. Let $N/M=\mathrm {Z}(PM/M)$ . Then, we have $N\triangleleft G$ and $M<N$ . Since $O_p (N)\leq O_p (G)=1$ , from the Hall–Higman theorem (see [Reference Huppert10, Lemma 1.2.3]), $C_N (M)\leq M$ . It follows from Lemma 2.4 that G has a faithful strongly monolithic character $\chi $ with the property that $Z(\chi )=\mathrm {ker}\chi $ and $\chi (x)=0$ for every $x\in N-M$ . Hence, G has an $\mathrm {SM}^*$ -vanishing p-element. This contradiction completes the proof.

Proof. Since $p \nmid \chi (1)$ for every $\chi \in \mathrm {Irr}_{\mathrm {sm}^*}(G)$ , from Remark 2.5, G has no $\mathrm {SM}^*$ -vanishing p-element. Thus, G has a normal Sylow p-subgroup by Theorem 1.4.

Proof. Since $m\in M$ is an $\mathrm {SM}^*$ -vanishing element of G, there exists a character $\chi \in \mathrm {Irr}_{\mathrm {sm}^*}(G)$ such that $\chi (m)=0$ . Let $\psi (\neq 1_M)$ be an irreducible constituent of $\chi _M$ . Then,

$$ \begin{align*} 0=\chi(m)=\chi_M(m) =e\sum\limits_{i=1}^{n}\psi^{g_i}(m), \end{align*} $$

where $e=[\chi _M, \psi ]\neq 0$ and $\{g_1, \ldots , g_n\}$ is a right transversal for the inertia subgroup $I_G(\psi )$ of $\psi $ in G. Moreover, $\psi \times 1_C \in \mathrm {Irr}(M\times C)$ and $I_G(\psi \times 1_C)=I_G(\psi )$ . Also, there exists an irreducible character $\varphi $ of $ I_G(\psi )$ such that $\xi := \varphi ^G \in \mathrm {Irr(G)}$ . Thus, $mc$ is a vanishing element of G for every $c\in C$ because

$$ \begin{align*} \xi(mc)=\xi_{M\times C}(mc)=f\sum\limits_{i=1}^{n}(\psi\times 1_C)^{g_i}(mc)=f\sum\limits_{i=1}^{n}\psi^{g_i}(m)=0 , \end{align*} $$

where $f=[\xi _{MC}, \psi \times 1_C]$ . Note that $\ker \xi =C$ and $MC/C (\cong M)$ is the unique minimal (nonabelian) normal subgroup of $G/C$ . Thus, $\xi \in \mathrm {Irr}_{\mathrm {sm}^*}(G)$ . Therefore, $mc$ is an $\mathrm {SM}^*$ -vanishing element of G for every $c\in C$ .

Proof of Theorem 1.5.

Let G be a minimal counterexample and let M be a nonabelian minimal normal subgroup of G. Assume that G does not have any $\mathrm {SM}^*$ -vanishing element whose order is divisible by $p\in \pi (G)$ . Write $\overline {G}=G/C$ , where $C=C_G(M)$ . Thus, $\overline {G}$ has a unique minimal normal subgroup $\overline {M}$ . Suppose that $C\neq 1$ . Then, we have $p\notin \pi (\overline {G})$ ; otherwise, G would have an $\mathrm {SM}^*$ -vanishing element whose order is divisible by p, since $\overline {G}$ has an $\mathrm {SM}^*$ -vanishing element $\overline {x}$ whose order is divisible by p by the minimality of G. Therefore, $p\in \pi (C)$ and so there exists an element $c\in C$ such that $p\mid |c|$ . However, by Theorem 1.3(i), ${G}$ has an $\mathrm {SM}^*$ -vanishing q-element y in M, where $q\in \pi (M)$ . It follows from Lemma 2.7 that $cy$ is an $\mathrm {SM}^*$ -vanishing element of G. However, this is a contradiction since $p\mid |cy|$ . Then, C must be trivial and so M is the unique minimal normal subgroup of G. Since G is a minimal counterexample, G has no $\mathrm {SM}^*$ -vanishing p-element. Thus, from Theorem 1.4, G has a normal Sylow p-subgroup, which is a contradiction. This contradiction completes the proof.

Proof of Theorem 1.6.

Let G be a minimal counterexample. Then, $\mathcal {K}_p(G)\neq \emptyset $ . Otherwise, G would have a normal p-complement by Theorem 1.4, and so the theorem would hold. Also, $|\mathcal {K}_p(G/N)|\leq |\mathcal {K}_p(G)|$ for every normal subgroup N of G. Hence, $O_{p'}(G)$ must be trivial. Otherwise, since G is a minimal counterexample, $\ell _p(G)=\ell _p(G/O_{p'}(G))\leq |\mathcal {K}_p(G/O_{p'}(G))|+1 \leq |\mathcal {K}_p(G)|+1$ , which is a contradiction. Furthermore, $\mathrm {F}(G)=O_p (G)$ is the unique minimal normal subgroup of G. Since G is solvable, $\mathrm {Z}(G)<\mathrm {F}(G)$ and so $\mathrm {Z}(G)$ must be trivial. Therefore, every faithful irreducible character $\chi $ of G is a strongly monolithic character of G with the property that $\mathrm {Z}(\chi )=\ker \chi $ . Now, let $N/O_p(G)$ be a chief factor of G. Then, $N/O_p(G)$ is an elementary abelian q-group and $q\neq p$ . From Lemma 2.4, there exists a faithful $\chi \in \mathrm {Irr}_{\mathrm {sm}^*}(G)$ such that $\chi (x)=0$ for every $x\in N-O_p(G)$ . Thus, $|\mathcal {K}_p(G/O_{p}(G))|<|\mathcal {K}_p(G)|$ . Finally, $\ell _p(G)=\ell _p(G/O_{p}(G))+1\leq |\mathcal {K}_p(G/O_{p}(G))|+2 \leq |\mathcal {K}_p(G)|+1$ , which is a contradiction. This contradiction completes the proof.

Proof of Theorem 1.7.

Assume that G is not a solvable group. Let M be a maximal solvable normal subgroup of G and $N/M$ be a minimal normal subgroup of $G/M$ . Clearly, $N/M$ is nonabelian. Now, let $R/M$ be the largest normal subgroup of $G/M$ such that $(N/M) \cap (R/M)=1$ . Thus, $NR/R$ is the unique (nonabelian) minimal normal subgroup of $G/R$ . Therefore, there exists a nonabelian simple group S such that $NR/R=S_1/R\times \cdots \times S_n/R$ and $S_i/R\cong S$ for every i. From [Reference Dolfi, Navarro and Tiep5, Proposition 6.4], $S_1/R$ has a real element $xR$ of odd order. Assume that the order of $xR$ is divisible by a prime $p\geq 5$ . From [Reference Granville and Ono9, Corollary 2], $S_1/R$ has an irreducible character $\theta $ of p-defect zero. By [Reference Isaacs11, Theorem 8.17], $\theta (xR)=0$ . Thus, $xR$ is a real $\mathrm {SM}^*$ -vanishing element of $S_1/R$ . Now, assume that $xR$ is a $3$ -element. The simple groups that do not have an irreducible character of $3$ -defect zero are given in [Reference Granville and Ono9, Corollary 2]. These groups are isomorphic to Suz, $Co_3$ or some alternating groups $A_n$ with $n\geq 7$ . From [Reference Conway, Curtis, Norton, Parker and Wilson3], each of the groups Suz and $Co_3$ has a real $\mathrm {SM}^*$ -vanishing $5$ -element. Moreover, since $A_5$ has a real element of order $5$ , it follows that any alternating group with $n\geq 5$ has a real $\mathrm {SM}^*$ -vanishing $5$ -element. Therefore, $S_1/R$ has a real element $xR$ whose order is odd and divisible by a prime $p\geq 3$ and $S_1/R$ has an irreducible character $\theta $ of p-defect zero, so that $\theta (xR)=0$ . Let $\psi =\theta \times \cdots \times \theta \in \mathrm {Irr}(NR/R)$ . Then, $\psi $ is a character of p-defect zero. Take $gR=xR\times R \times \cdots \times R \in NR/R$ . Of course, $gR$ is a real element of $G/R$ whose order is odd and divisible by p. By [Reference Navarro, Sanus and Tiep13, Lemma 3.2], there is a real element h of odd order in G such that $gR=hR$ . Now, let $\chi $ be an irreducible character of $G/R$ such that $[\psi ^{G/R} ,\chi ]\neq 0 $ . By Frobenius reciprocity, $[\psi ^{G/R} ,\chi ]=[\psi , \chi _{NR/R}]\neq 0 $ . Observe that $\chi $ is a faithful irreducible character of $G/R$ . Also, $Z(G/R)=1$ and so $\chi \in \mathrm {Irr}_{\mathrm {sm}^*}(G/R)$ . By Clifford’s theorem, $\chi _{NR/R}=e \sum _{i} \psi _i $ , where $\psi _i$ is a $G/R$ -conjugate of $\psi $ for every i and $e=[\psi , \chi _{NR/R}]$ . As $\psi _i (1)=\psi (1)$ , every $\psi _i$ is an irreducible character of p-defect zero of $NR/R$ . It follows from [Reference Isaacs11, Theorem 8.17] that $\psi _i(gR)=0$ for every i. Let $\tilde {\chi }\in \mathrm {Irr}(G)$ be the corresponding character of $\chi $ . Then, $\tilde {\chi } \in \mathrm {Irr}_{\mathrm {sm}^*}(G)$ . Thus,

$$ \begin{align*} \tilde{\chi}(h)=\chi(hR)=\chi(gR)=\chi_{NR/R}(gR)=e\sum\limits_{i}\psi_i (gR)=0. \end{align*} $$

Hence, we have a real $\mathrm {SM}^*$ -vanishing element h of odd order in G. Thus, every nonsolvable group has a real $\mathrm {SM}^*$ -vanishing element of odd order. This completes the proof.