Proof of Theorem 1.1.

We will prove the theorem by way of contradiction. To do this, we assume that the assertions of (i) and (ii) given in the theorem are not true and let G be a minimal counterexample to all these statements. Suppose that M and N are distinct minimal normal subgroups of G. Since $\textrm {Irr}_{mm}(G/M)$ and $\textrm {Irr}_{mm}(G/N)$ are subsets of $\textrm {Irr}_{mm}(G)$ , both factor groups $G/M$ and $G/N$ have a normal p-complement by induction. Since G is isomorphic to a subgroup of $G/M\times G/N$ , it follows that G has a normal p-complement, which is a contradiction. Obviously, if both $G/M$ and $G/N$ have a normal Sylow p-subgroup, then we again reach a contradiction.

From now on, we assume that G has a unique minimal normal subgroup M for the assertions (i) and (ii). Since G is solvable, M is an elementary abelian q-group for some prime q. First of all let us suppose that p divides all monomial monolithic character degrees of G. Since $G/M$ has a normal p-complement by induction, we see that $p=q$ . It follows that $A:=MH$ is a normal subgroup of G, where H is a Hall $p'$ -subgroup of G. By using Theorem 1 of [Reference Revin and Vdovin11], we see that $G=A\textrm {N}_G(H)=MH\textrm {N}_G(H)=M\textrm {N}_G(H)$ . Since M is abelian, it is well known that $M\cap \textrm {N}_G(H)\unlhd G$ , which is possible only when $M\leq \textrm {N}_G(H)$ or $M\cap \textrm {N}_G(H)=1$ . If $M\leq \textrm {N}_G(H)$ , then $H\unlhd G$ , which is a contradiction. Therefore, we can assume that $G=M\rtimes \textrm {N}_G(H)$ , that is, G splits over the unique minimal normal subgroup M. On the other hand, there exists a Sylow p-subgroup P of G satisfying $M\leq P$ . Since $M\unlhd P$ , we observe that P acts by conjugation on M and so P acts on $\textrm {Irr}(M)$ . Also, we know that $\textrm {C}_P(x)=P$ for $ x\in M\cap \textrm {Z}(P)>1$ . This implies that there exists a nontrivial linear character $\lambda $ of M with inertia subgroup $\textrm {I}_P(\lambda )=P$ and hence we conclude that $P\leq \textrm {I}_G(\lambda )$ for some nontrivial linear character $\lambda $ of M. Since G splits over M and $M\leq \textrm {I}_G(\lambda )$ , the group $\textrm {I}_G(\lambda )$ splits over M and also the character $\lambda $ is invariant in $\textrm {I}_G(\lambda )$ . It follows from Lemma 2.3 that $\varphi _M=\lambda $ for a linear character $\varphi $ of $\textrm {I}_G(\lambda )$ . By the Clifford correspondence, $\varphi ^G=\chi \in \textrm {Irr}(G)$ . Notice that $\chi $ is a monomial character. If $\textrm {ker}\chi>1$ , then $M\leq \textrm {ker}\chi $ . Therefore, $M\leq \textrm {ker}\varphi ^G\leq \textrm {ker}\varphi $ and hence $M\leq \textrm {ker}\varphi _M=\textrm {ker}\lambda $ . This leads to a contradiction as $\lambda $ is a nontrivial character of M. Thus, $\textrm {ker}\chi =1$ , that is, $\chi $ is a monomial monolithic character of G. By the hypothesis of (i), $p \mid \chi (1)$ . By considering $P\leq \textrm {I}_G(\lambda )$ , we reach a contradiction that $p\nmid |G:\textrm {I}_G(\lambda )|=\varphi ^G(1)=\chi (1)$ . This completes the proof of (i).

Now, we assume that p does not divide all monomial monolithic character degrees of G. Since $G/M$ has a normal Sylow p-subgroup by induction, we see that $p\neq q$ . Thus, $K:=MP$ becomes a normal subgroup of G for $P\in \textrm {Syl}_p(G)$ . It follows by a Frattini argument that $G=K\textrm {N}_G(P)=MP\textrm {N}_G(P)=M\textrm {N}_G(P)$ . Also, we deduce that $M\cap \textrm {N}_G(P)=1$ , that is, G splits over the minimal normal subgroup M. Let $1_M\neq \lambda \in \textrm {Irr}(M)$ and $T:=\textrm {I}_G(\lambda )$ . It can be easily seen that T also splits over M and $\lambda $ is invariant in T. By Lemma 2.3, we conclude that there exists a linear character $\theta $ of T such that $\theta _M=\lambda $ . Since T is the inertia group of $\lambda $ , by the Clifford correspondence we deduce that $\theta ^G=\chi $ for an irreducible character $\chi $ of G. If $\textrm {ker}\chi>1$ , then $M\leq \textrm {ker}\theta _M=\textrm {ker}\lambda $ , which is a contradiction. It follows that $\chi $ is a monomial and monolithic character of G and so $p\nmid \chi (1)$ from the hypothesis of (ii). On the other hand, since M is an abelian and normal subgroup of K, we know by Ito’s theorem that $\xi (1)$ divides $|K:M|=|P|$ for every irreducible character $\xi $ of K. Since $p\nmid \chi (1)$ , we conclude from Clifford’s theorem that every irreducible constituent of $\chi _K$ must be linear and so $K=M\times P$ is an abelian group. This yields a contradiction $P\unlhd G$ and completes the proof of the theorem.

Proof. Let $N\unlhd G$ be the maximal counterexample among normal subgroups of G. First suppose that $G/N$ has two distinct minimal normal subgroups $M_1/N$ and $M_2/N$ . Since $M_1/N\cap M_2/N=N$ and both the normal subgroups $M_1$ and $M_2$ are the intersections of the kernels of some monolithic characters of G, we arrive at a contradiction that N is the intersection of the kernels of some monolithic characters of G. Thus, we can assume that $G/N$ has a unique minimal normal subgroup. This implies that $G/N$ has a monolithic character $\chi $ such that $N=\textrm {ker}\chi $ , which contradicts our assumption. The conclusion follows.

Proof of Theorem 1.2.

Let $\chi $ be the unique imprimitive monolithic character of G. If G has a unique nonlinear monolithic character, then G is one of the Berkovich groups in Theorem 2.2. By the hypothesis of the theorem, we may assume that G has at least one nonlinear primitive monolithic character $\theta $ . Let $\textrm {ker}\theta $ be maximal among the kernels of nonlinear primitive monolithic characters of G. By [Reference Erkoç and Çınarcı5, Lemma 2.2], $\textrm {ker}\theta <\textrm {ker}\chi $ . It can be seen immediately that $G/\textrm {ker}\chi $ is either an extraspecial $2$ -group or a Frobenius group as in Theorem 2.1. First suppose that $G/\textrm {ker}\chi $ is an extraspecial $2$ -group and set $\textrm {Z}(G/\textrm {ker}\theta )=:\textrm {Z}(\overline {G})$ . By [Reference Isaacs8, Corollary 6.13], $\textrm {Z}(\overline {G})>1$ . Since $\overline {G}/\textrm {Z}(\overline {G})$ has exactly one nonlinear monolithic character $\chi $ , we see that $\overline {G}/\textrm {Z}(\overline {G})$ is the nilpotent group in Theorem 2.2(i). This leads to a contradiction that $G/\textrm {ker}\theta $ is a nilpotent group.

We may now assume that $G/\textrm {ker}\chi $ is a Frobenius group of order $p^a(p^a-1)$ for some prime p and integer a. Suppose that every nonlinear primitive monolithic character of G is faithful. We will show that $\textrm {ker}\chi =\textrm {Z}(G)$ . Since we know that $\textrm {Z}(G)$ is the unique maximal abelian normal subgroup of G from [Reference Isaacs8, Corollary 6.13], it is obvious that $\textrm {Z}(G)=\textrm {Z}(\textrm {ker}\chi )$ . Since $G/\textrm {Z}(G)$ is the Berkovich group in Theorem 2.2(ii), we see that $\textrm {ker}\chi /\textrm {Z}(G)$ is abelian and so $\textrm {ker}\chi $ is nilpotent. Since $p\nmid \chi (1)=p^a-1$ , we deduce from Theorem 1.1 that the Sylow p-subgroup of G is normal. Since G has a unique minimal normal subgroup M, $\textrm {F}(G)$ is a p-group and hence $\textrm {ker}\chi <\textrm {F}(G)=G'\textrm {ker}\chi $ . Since $M<G'\leq \textrm {F}(G)$ and $\textrm {F}(G)/M$ is abelian, we obtain $\textrm {F}'(G)=G''=M$ . Notice that $|M|=p$ because $\textrm {Z}(G)$ is cyclic. Observing that $1\neq \textrm {Z}(G')\leq \textrm {ker}\chi \cap G'$ and $M\leq \textrm {ker}\chi \cap G'$ , we obtain from Lemma 2.4 that $\textrm {Z}(G')=M$ . This shows that $G'$ is an extraspecial p-group. Since $\theta _{G'}\in \textrm {Irr}(G')$ , we have $\theta (1)^2=|G':M|$ . Since $G/\textrm {F}(G)$ is abelian, we also get $\theta _{\textrm {F}(G)}\in \textrm {Irr}(\textrm {F}(G))$ . From [Reference Isaacs8, Theorem 2.31], $\theta (1)^2=|\textrm {F}(G):\textrm {Z}(G)|$ and hence $|\textrm {F}(G)/\textrm {Z}(G)|=|G'/M|$ . Since $\textrm {F}(G)/\textrm {Z}(G)=\textrm {ker}\chi /\textrm {Z}(G)\times G'\textrm {Z}(G)/\textrm {Z}(G)$ and $G'\textrm {Z}(G)/\textrm {Z}(G)\cong G'/M$ , from Lemma 2.4, we obtain $\textrm {ker}\chi =\textrm {Z}(G)$ . On the other hand, from [Reference Manz and Wolf9, Corollary 16.3], the cyclic group $G/\textrm {F}(G)$ acts on $\textrm {F}(G)/\textrm {Z}(G)$ preserving the symplectic form and $\textrm {F}(G)/\textrm {Z}(G)$ is an irreducible and faithful $G/\textrm {F}(G)$ -module. From Lemma 2.6, $p^a-1=|G:\textrm {F}(G)|\leq \sqrt {p^a}+1$ . It follows that $p^a=3$ or $4$ . Since $\theta (1)^2=|\textrm {F}(G):\textrm {Z}(G)|=p^a$ , we conclude $p^a=4$ and hence $G/\textrm {Z}(G)\cong \textrm {A}_4$ , as desired.

Now, we may assume that G has at least one nonfaithful monolithic character $\theta $ and that $\textrm {ker}\theta $ is maximal among the kernels of nonlinear primitive monolithic characters of G. Since all nonlinear primitive monolithic characters of $G/\textrm {ker}\theta $ are faithful and $1<\textrm {ker}\theta <\textrm {ker}\chi $ by the first paragraph of the proof, $G/\textrm {ker}\chi \cong \textrm {A}_4$ and $\theta (1)=2$ . Thus, the Sylow $2$ -subgroup P of G is normal and nonabelian by Theorem 1.1. Since $1\neq P'\leq \Phi (P)$ and $P'\unlhd G$ , we have $P'\leq \Phi (G)$ . Notice that $\theta _P\in \textrm {Irr}(P)$ for every nonlinear primitive character $\theta $ of G because $G/P$ is abelian. Thus, $P'\leq \Phi (G)\nleq \textrm {ker}\theta $ for every nonlinear primitive monolithic character $\theta $ of G. Therefore, $G/\Phi (G)$ has exactly one nonlinear monolithic character $\chi $ . It follows from Theorem 2.2(ii) that $\textrm {F}(G/\Phi (G))=\textrm {F}(G)/\Phi (G)=G'\textrm {ker}\chi /\Phi (G)$ and this yields $G'\textrm {ker}\chi =\textrm {F}(G)$ .

Suppose that there exists a primitive monolithic character $\varphi $ of G with $\textrm {ker}\varphi <\textrm {ker}\theta $ . Let $\widetilde {M}:=M/\textrm {ker}\varphi $ be the unique minimal normal subgroup of $\widetilde {G}:=G/\textrm {ker}\varphi $ . Since $\textrm {Z}(\widetilde {G})$ is cyclic and the unique maximal abelian subgroup among normal subgroups of $\widetilde {G}$ , there exists $\widetilde {E}\unlhd \widetilde {G}$ satisfying $\textrm {F}(\widetilde {G})=\widetilde {E} \: \textrm {Z}(\widetilde {G})$ and $\widetilde {E}\cap \textrm {Z}(\widetilde {G})=\widetilde {M}$ from [Reference Manz and Wolf9, Theorems 1.4(vii) and 1.9]. Therefore, $\textrm {F}(\widetilde {G})/\widetilde {M}=\widetilde {E}/\widetilde {M}\times \textrm {Z}(\widetilde {G})/\widetilde {M}$ becomes an abelian group. This shows that $\widetilde {M}=\textrm {F}'(\widetilde {G})\leq \widetilde {\textrm {ker}}\theta \leq \textrm {F}(\widetilde {G})$ , but this is a contradiction because $\theta _{\textrm {F}(\widetilde {G})}\in \textrm {Irr}(\textrm {F}(\widetilde {G}))$ . Thus, we conclude that every nonlinear primitive monolithic character of G has the maximal kernel among the kernels of nonlinear primitive monolithic characters. Hence, these primitive monolithic characters cannot be faithful.

Next, we shall show by induction on $|G|$ that $\textrm {ker}\chi =\textrm {Z}(G)$ . If $\textrm {ker}\lambda>1$ for every nonlinear $\lambda \in \textrm {Irr}(G)$ , then $\textrm {Z}(G/\textrm {ker}\lambda )=\textrm {ker}\chi /\textrm {ker}\lambda =\textrm {Z}(\lambda )/\textrm {ker}\lambda $ by induction because $\textrm {ker}\lambda <\textrm {ker}\chi $ . It follows from [Reference Isaacs8, Corollary 2.28] that $\textrm {Z}(G)=\textrm {ker}\chi $ , as desired. Now, we suppose that G has a faithful irreducible character $\mu $ and then we will determine ${\mu (1)=2}$ . First let us consider the case where $\textrm {F}(G)\neq P$ for the normal and nonabelian Sylow $2$ -subgroup P of G. Recall that $G/P$ is abelian, which implies that $F:=\textrm {F}(G)=P\times A$ , where A is an abelian group of odd order. Since $F/A$ is nonabelian and $A\unlhd G$ , we have $A\leq \textrm {ker}\chi $ . By induction, $\textrm {Z}(G/A)=\textrm {ker}\chi /A$ . Let $K/A$ be maximal among abelian normal subgroups of $F/A$ . It is well known that $K/A$ is self centralising, that is, $\textrm {Z}(F/A)=\textrm {Z}(G/A)< K/A$ , and hence $|F:K|=2$ . Thus, applying [Reference Isaacs8, Lemma 12.12] for the group $F/A$ , we find $2=|(F/A)'|$ . Since $F'\leq P$ , we see that $(F/A)'\cong F'$ of order two. Moreover, $G/\textrm {ker}\theta $ has a unique minimal normal subgroup. Therefore, $F/\textrm {ker}\theta $ is a $2$ -group and so $A\leq \textrm {ker}\theta $ for every nonlinear primitive monolithic character $\theta $ of G. Also, by Lemma 2.5, $A\leq \textrm {Z}(G)$ and hence $F'A\leq \textrm {Z}(G)$ . On the other hand, since $F'\leq \textrm {Z}(F)$ and $\theta _{F}\in \textrm {Irr}(F)$ , we have $4=\theta (1)^2=|F:\textrm {Z}(F)|$ from [Reference Isaacs8, Theorem 2.31]. Thus, every nonlinear irreducible character degree of F is equal to 2. Since $|G:F|=3$ , we see by [Reference Isaacs8, Corollary 11.29] that every nonlinear irreducible character degree of G may be 2, 3 or 6. Since P is nonabelian, by Clifford’s theorem and Frobenius reciprocity, $2$ divides the faithful character degrees of G. This shows that the faithful character degree $\mu (1)=2$ or $6$ . Since G has an irreducible character $\chi $ of degree 3 and $\mu $ is faithful, $\mu $ must be primitive when $\mu (1)=2$ and, by [Reference Isaacs8, Corollary 6.13], $\textrm {Z}(F)\leq \textrm {Z}(G)\leq \textrm {ker}\chi <F$ . Thus, $\textrm {Z}(G)=\textrm {ker}\chi $ , as desired, because $4=|F:\textrm {Z}(F)|$ . Suppose instead that $\mu (1)=6$ . Since $F'\leq G'\cap \textrm {ker}\chi $ and $G''\leq G'\cap \textrm {ker}\chi $ , we have $F'=G''=G'\cap \textrm {ker}\chi $ by Lemma 2.4. Also, since $G'/G''\cong G'\textrm {ker}\chi /\textrm {ker}\chi =F/\textrm {ker}\chi $ is of order four, $G'\cong \textrm {Q}_8$ . Since $|G:F|=3$ , we can pick an element x of G such that $|x|=3$ and $x\notin A$ . Then we have a normal subgroup $L:=A\times (G'\rtimes \langle x\rangle )$ of G because $A\leq \textrm {Z}(G)$ . Notice that $L<G$ since G has an irreducible character $\mu $ of degree 6 and that $G'\rtimes \langle x\rangle \cong \textrm {SL}(2, 3)$ . If $\delta $ is an irreducible constituent of $\mu _L$ , then, by [Reference Isaacs8, Corollary 11.29], $\mu (1)/\delta (1)$ divides $|G:L|$ , which is a power of two. Therefore, considering $\textrm {cd}(L)=\{1, 2, 3\}$ , we get $\delta (1)=3$ . Since the kernels of odd degree irreducible characters of L contain the group $L''$ of order two, we have a contradiction $L''\leq \textrm {ker}\mu _L=1$ . This contradiction shows that G cannot have a faithful irreducible character of degree 6.

Now, we restrict our attention to the case where $\textrm {F}(G)= P$ for the normal nonabelian Sylow $2$ -subgroup P of G. Assume that there exists a minimal normal subgroup M of G such that $M\leq \textrm {ker}\theta $ but $M\nleq F'$ . By induction, we have $\textrm {Z}(G/M)=\textrm {ker}\chi /M=\textrm {Z}(F/M)$ . Applying [Reference Isaacs8, Theorem 12.12] for the group $F/M$ , we get $|(F/M)'|=2$ since $|F/M:K/M|=2$ for the self-centralising normal subgroup $K/M$ of $F/M$ . Hence, $|F'|=2$ because $F'\cap M=1$ . From [Reference Isaacs8, Theorem 2.31], $\theta (1)^2=4=|F:\textrm {Z}(F)|$ for every nonlinear primitive monolithic character $\theta $ of G. By a similar argument to that used in the case $\textrm {F}(G)\neq P$ , we can show that if the faithful character degree $\mu (1)$ is 2, then $\textrm {ker}\chi =\textrm {Z}(G)$ . Using the same argument again, we get a normal subgroup $R:=G'\rtimes \langle x\rangle $ , where $G'\cong \textrm {Q}_8$ and $x\in G$ of order three. If $\mu (1)=6$ , then by [Reference Isaacs8, Corollary 11.29] we reach a contradiction that $\textrm {Z}(R)=1$ as $\mu _R=2\delta $ for $\delta \in \textrm {Irr}(R)$ of degree 3.

From now on suppose, by way of contradiction, that if $M\leq \textrm {ker}\theta $ for any nonlinear primitive monolithic character $\theta $ and minimal normal subgroup M of G, then $M\leq F'$ . Recall that G has no faithful primitive monolithic character as we have seen before. Let N be the intersection of the kernels of nonlinear monolithic characters of G. If $N>1$ , then $N\cap F'\leq N\cap G'=1$ from Lemma 2.5, which contradicts our assumption. Hence, N must be trivial. It follows that G has a minimal normal subgroup S and a nonlinear primitive monolithic character $\tilde {\theta }$ such that $S\leq \textrm {ker}\theta $ but $S\nleq \textrm {ker}\tilde {\theta }$ . Since primitive characters are homogeneous and $S\cap \textrm {ker}\tilde {\theta }=1$ , we see that $\tilde {\theta }_{S}=2\alpha $ for $1_S\neq \alpha \in \textrm {Irr}(S)$ . Since the elementary abelian group S has a faithful irreducible character, we get $|S|=2$ . Again by induction and [Reference Isaacs8, Theorem 12.12], we obtain $|(F/S)'|=2$ . From our assumption $S\leq F'$ , we already know that $|F'|=4$ . Since $G'$ must be nonabelian, we see that $G''=F'<G'$ from Lemma 2.4. Since $G'/G''\cong G'\textrm {ker}\chi /\textrm {ker}\chi $ has order four, we obtain $|G'|=16$ and so $|\textrm {Z}(G')|=2$ . Therefore, $G'\cong \textrm {D}_8, \textrm {Q}_{16}$ or $\textrm {SD}_{16}$ whose automorphism groups are $2$ -groups. Consider $G'\rtimes \textrm {C}_3=:V\unlhd G$ . Then $\textrm {N}_V(G')/\textrm {C}_V(G')=V/\textrm {C}_V(G')$ is a subgroup of the $2$ -group $\textrm {Aut}(G')$ and so $\textrm {C}_3\leq \textrm {C}_V(G')$ . This shows that $G'\times \textrm {C}_3=V\unlhd G$ , giving $\textrm {C}_3\unlhd G$ , which is a contradiction since the Fitting subgroup F is a $2$ -group. From this contradiction, it follows that $G/\textrm {Z}(G)\cong \textrm {A}_4$ .

Observe that since $G'$ must be nonabelian, $1<G''=G'\cap \textrm {ker}\chi $ from Lemma 2.4. On the other hand, since G has a unique monomial monolithic character $\chi $ of degree 3, we conclude by Theorem 1.1 that $G\cong N\rtimes Q$ , where $Q\in \textrm {Syl}_3(G)$ and N is the nilpotent Hall $3'$ -subgroup of G. By [Reference Berkovich and Zhmud2, Proposition 30.18(c)], the Sylow p-subgroups of N are abelian for every prime p different from two. We observe that $N\nleq \textrm {ker}\chi $ , which implies that $G/N\cong Q$ is an abelian group, as desired in (iii) of the theorem.

For the converse of the theorem, it is easy to check that G has a unique imprimitive monolithic character if (i) or (ii) of the theorem holds. Suppose that $G/\textrm {Z}(G)\cong \textrm {A}_4$ . By [Reference Isaacs8, Corollaries 2.28 and 2.30], it is immediate that $\textrm {Z}(G)=\textrm {Z}(\varphi )$ for every nonlinear irreducible character $\varphi $ of G and $\textrm {cd}(G)=\{1, 2, 3\}$ . Let $\textrm {ker}\chi =\textrm {Z}(G)$ for a nonlinear irreducible character $\chi $ of G. Thus, $\chi $ is imprimitive and monolithic of degree 3. If $\chi $ is the unique nonlinear monolithic character of G, then G is a group as in (i) and (ii) of the theorem. Therefore, assume that G has another nonlinear monolithic character $\varphi $ and that $\varphi $ is imprimitive too. We know that $\textrm {ker}\varphi <\textrm {Z}(G)=\textrm {ker}\chi $ . If $\varphi (1)=2$ , then there exist a proper subgroup $H/\textrm {ker}\varphi $ of $G/\textrm {ker}\varphi $ and $\lambda \in \textrm {Lin}(H/\textrm {ker}\varphi )$ such that $\lambda ^G=\varphi $ . Since $\varphi $ is a faithful character of $G/\textrm {ker}\varphi $ and $|G:H|=2$ , the factor group $H/\textrm {ker}\varphi $ must be an abelian normal subgroup of $G/\textrm {ker}\varphi $ . It follows from Ito’s theorem that every nonlinear character degree of $G/\textrm {ker}\varphi $ is equal to 2, which contradicts $\chi (1)=3 $ . Thus, we may assume that $\varphi (1)=3$ . Since $G/\textrm {ker}\varphi $ has a unique minimal normal subgroup and $\varphi (1)=3$ , by Clifford’s theorem the Fitting subgroup $\textrm {F}(G/\textrm {ker}\varphi )=\textrm {F}(G)/\textrm {ker}\varphi $ becomes an abelian Sylow $2$ -subgroup of $G/\textrm {ker}\varphi $ . Since $G/\textrm {ker}\varphi $ has the unique minimal normal subgroup of order two and $\textrm {ker}\varphi <\textrm {Z}(G)$ , we get a contradiction that $2\nmid |(G/\textrm {ker}\varphi )'\cap \textrm {Z}(G)/\textrm {ker}\varphi |$ by applying [Reference Isaacs8, Theorem 5.6]. Thus, if $G/\textrm {Z}(G)\cong \textrm {A}_4$ , then G has exactly one imprimitive monolithic character. This completes the proof.

Proof. By the last paragraph of the proof of Theorem 1.2, it is easy to see that $\textrm {cd}(G)=\textrm {cd}_m(G)=\{1, 2, 3\}$ since $G/\textrm {Z}(G)\cong \textrm {A}_4$ . In fact, G has a unique monolithic character of degree 3 and the kernel of this character is equal to $\textrm {Z}(G)$ .