Proof. (i) and (iii) follow from [Reference Kondratev and Mazurov15, lemma 1] and [Reference Vasiliev and Vdovin17], respectively and (ii) is straightforward. Working towards a contradiction, let $H$ be a nilpotent $2$-complement of $S$. Let $t \in \pi (S)-\{2\}$, $T \in {\rm Syl}_t(H)$ and $1 \neq x \in Z(T)$. Then, $|cl_S(x)|$ is a power of $2$, contradicting Burnside's theorem [Reference Huppert11, 15.2]. Obviously, ${\rm Alt}_l$ contains no nilpotent Hall $(\pi (S)-\{2,\,3\})$-subgroup and also, by [Reference Conway, Curtis, Norton, Parker and Wilson4], ${\rm Suz}$ and ${\rm Co}_3$ contain no nilpotent Hall $(\pi (S)-\{2,\,3\})$-subgroup. So, (iv) follows.

Proof. The proof of (i) is straightforward. For proving (ii), let $T_1 \in {\rm Syl}_t(C_G(x))$. Then, $T_1N/N\leq C_G(x)N/N \leq C_{G/N}(xN)$. Thus, $|T_1/(T_1\cap N)| \mid |C_{G/N}(xN)|$. So, $|C_G(x)|_t=|T_1|$ divides $|N|_t|C_{G/N}(xN)|_t$. The remaining claim of (ii) is straightforward. Also, (iii) and (iv) are taken from [Reference Ahanjideh1, lemma 2.5(iv) and theorem 1.1]. Finally, (v) and (vi) follow from [Reference Casolo, Dolfi and Jabara3, lemma 2.7] and [Reference Gorenstein9, lemma 5.2.3].

Proof. Let $x \in G - {\rm Fit}(G)$ and $P$ be a $p$-subgroup of $G$ such that $PN/N \in {\rm Syl}_p(C_{G/N}(xN))$. Then, $xN \in C_{G/N}(PN/N)$. By lemma 2.3(v), $C_{G/N}(PN/N) =C_G(P)N/N.$ So, $x=yn$, for some $y \in C_G(P)$ and $n \in N$. As, $N \leq {\rm Fit}(G)$ and $x\not \in {\rm Fit}(G)$, $y \not \in {\rm Fit}(G)$. Hence, $|cl_G(y)|=m$. Since $P \leq C_G(y)$, $|m|_p=|cl_G(y)|_p \leq |G|_p/|P|.$ So, $|cl_{G/N}(xN)|_p=|G/N|_p/|C_{G/N}(xN)|_p$ $=|G/N|_p/|PN/N|=|G|_p/|P| \geq |m|_p$. By lemma 2.3(ii), $|cl_{G/N}(xN)|_p \leq | m|_p$. Hence, $|cl_{G/N}(xN)|_p = | m|_p$.

Proof. Let $i$ be the smallest number such that $0 \leq i \leq t$ and $G/M_i = N/M_i \times M/M_i$, for some subgroup $M_i \leq M \unlhd G$. Then, $M/M_i \cong G/N$. Working towards a contradiction, let $i>0$. Then, $(M/M_{i-1})' \not \leq M_i/M_{i-1}$, because $M/M_i$ is non-abelian. So, $\{M_i/M_{i-1}\} \neq \frac {(M/M_{i-1})' M_i/M_{i-1}}{M_i/M_{i-1}} \unlhd \frac {M/M_{i-1}}{M_i/M_{i-1}} \cong M/M_{i} \cong G/N .$ Thus, $\frac {(M/M_{i-1})' M_i/M_{i-1}}{M_i/M_{i-1}}= \frac {M/M_{i-1}}{M_i/M_{i-1}}$. Since $M_{i}/M_{i-1}$ is a minimal normal subgroup of $G/M_{i-1}$, $(M/M_{i-1})'\cap M_i/M_{i-1}=M_i/M_{i-1}$ or $\{M_{i-1}\}$. Therefore, either $(M/M_{i-1})'=M/M_{i-1}$ or $M /M_{i-1}=(M/M_{i-1})' \times M_i/M_{i-1}$. In the former case, since $M_i/M_{i-1} \leq Z(M/M_{i-1})$ and $M/M_i \cong G/N$, we get that $|M_i/M_{i-1}|$ divides the order of the Schur multiplier of $G/N$, a contradiction, because $M_i/M_{i-1}$ is a $p$-group. In the latter case, regarding $M/M_{i-1} \cap N/M_{i-1}=M_i/M_{i-1}$, we have $(M/M_{i-1})' \cap N/M_{i-1}=(M/M_{i-1})' \cap M_{i}/M_{i-1}=\{M_{i-1}\}$. Also, $(M/M_{i-1})' \cong G/N$, hence $|G/M_{i-1}|=|G/N||N/M_{i-1}|=|(M/M_{i-1})'||N/M_{i-1}|$. Consequently, $G/M_{i-1}=(M/M_{i-1})' \times N/M_{i-1},$ a contradiction with minimality of $i$. Therefore, $i=0$. Now, the lemma follows.

Proof. (i) follows immediately from the facts that $P=P/(P\cap N) \cong PN/N \in {\rm Syl}_2(G/N)$ and $G/N \cong {\rm Alt}_5$. Since $N$ is a $3$-group and $M \unlhd N$, $Z(N) \cap M \neq \{1\}$. Hence, we get from minimality of $M$ that $M \cap Z(N) =M$. Consequently, $M \leq Z(N)$. Now, let $M \not \leq Z(G)$. Since $M \unlhd G$, $P$ acts on $M$. By lemma 2.3(vi), $M=C_M(P) \times T$, where $T=[M,\,P]$. If $T=\{1\}$, then $M=C_M(P)$. So $N,\,P \leq C_G(M)$. Therefore, $\{N\} \neq PN/N \leq C_G(M)/N \trianglelefteq G/N \cong {\rm Alt}_5$. By simplicity of $G/N$, $C_G(M)=G$, a contradiction with $M \not \leq Z(G)$. This guarantees that $T \neq \{1\}$. We observe that $P$ acts on $T$ by conjugation and $C_T(x) \times C_T(y) \times C_T(xy) \leq T$. Taking into account the fact that ${\rm gcd}(|P|,\,|T|)=1$, Maschke's theorem yields the existence of a $P$-invariant subgroup $T_1$ of $T$ such that $T=C_T(x) \times C_T(y) \times C_T(xy) \times T_1$. If $T_1 \neq \{1\}$, then since $(C_T(x) \times C_T(y) \times C_T(xy)) \cap T_1 =\{1\}$, we get that $P$ acts fixed point freely on $T_1$. Hence, $P$ is cyclic, a contradiction. This shows that $T_1=\{1\}$ and $M=C_M(P) \times C_T(x) \times C_T(y) \times C_T(xy)$, as needed in (ii).

Since $G/N \cong {\rm Alt}_5$ and $PN/N \in {\rm Syl}_2(G/N)$, we get that $N_G(P)N/N =N_{G/N}(PN/N)$ is a non-abelian group of order $12$. Thus, $N_G(P)$ contains a $3$-element $\sigma$ such that $\sigma \not \in N \cup C_G(P)$. Hence, $\sigma$ permutes the elements of $P-\{1\}$. Without loss of generality, we can assume that $x^{\sigma }=y$, $y^{\sigma }=xy$ and $(xy)^{\sigma }=x$. As $\sigma \in N_G(P)$, we can see $T^{\sigma }=T$. Hence, (iii) follows.

Finally, suppose that $t \in P- \{1\}$ and $u \in P- \{1,\,t\}$. Let $1\neq n \in C_T(t)$. Note that $n^{u} \in C_{T^{u}}(t^{u})=C_T(t)$ and regarding $o(u)=2$, $(n^{u}n)^{u}=n^{u}n$. Thus, $n^{u}n \in C_T(t) \cap C_T(u)=C_T(P)=C_M(P) \cap T=\{1\}$. This gives $n^{u}n=1$. Since $T \leq M$, $o(n)=3$. It follows that $n^{u}=n^{2}$, as desired in (iv).

Proof. Set $\mathfrak {A}=\{1\leq i \leq t: |C_{M_i/M_{i-1}}(x_5M_{i-1})| \geq |C_{M_i/M_{i-1}}(x_2M_{i-1})|\}$, for some $1\neq x_2 \in P$. Since $G/N \cong {\rm Alt}_5$, $|C_{G/N}(x_5N)|_3=|C_{G/N}(x_2N)|_3=1$. So, corollary 2.4 yields that $\mathfrak {A} \neq \emptyset$. Working towards a contradiction, let for every $i \in \mathfrak {A}$, $M_i /M_{i-1} \leq Z(G/M_{i-1})$, which gives that $C_{M_i/M_{i-1}}(x_5M_{i-1})=M_i/M_{i-1}=C_{M_i/M_{i-1}}(x_2M_{i-1})$. If there exists an integer $i \in \{1,\,\ldots,\,t\}\!-\!\mathfrak {A}$, then $|C_{M_i/M_{i-1}} (x_5M_{i-1})|<|C_{M_i/M_{i-1}}(x_2M_{i-1})|$. Hence, corollary 2.4 forces $|C_G(x_5)|_3< |C_G(x_2)|_3$, a contradiction. Therefore, $\mathfrak {A} =\{1,\,\ldots,\,t\}$. So, for every $i \in \{1,\,\ldots,\,t\}$, $M_i /M_{i-1} \leq Z(G/M_{i-1})$. By lemma 2.6, $G = M \times N$, where $M \cong {\rm Alt}_5$. Let $x_3 \in M$ be of order $3$. Then, $x_3 \in G - N$ and $3^{e}=|cl_G(x_3)|_3=1$. Hence, $|cl_G(x_5)|_3 =1$. By lemma 2.3(v), $3 \mid |C_{G/N}(x_5N)|$, a contradiction, because $G/N\cong {\rm Alt}_5$. Thus, there is an $i \in \mathfrak {A}$ such that $M_i/M_{i-1} \not \leq Z(G/M_{i-1})$, as wanted.

Proof. Let $p$ be a prime divisor of $o(\bar {x})/r$ and $P \in {\rm Syl}_p(N)$. By the Frattini argument, $G=NN_G(P)$. Thus, $x=nx'$, for some $n \in N$ and $x' \in N_G(P)$. First suppose that $n=1$. Then, $x \in N_G(P)$. Set $T=\langle P,\,x\rangle$. For every $y \in T-\langle T \cap N,\, x^{o(\bar {x})/r}\rangle$, let $C_y =C_T(C_G(y))$. If $z \in T-\langle T \cap N,\, x^{o(\bar {x})/r}\rangle$, then there exist an element $n \in T \cap N$ and an integer $\alpha$ such that $1 \leq \alpha < o(\bar {x})$ and $z=n x^{\alpha }$. Taking into account the facts that every element of $\langle \bar {x} \rangle$ whose order divides $r$ lies in $\langle \bar {x}^{o(\bar {x})/r} \rangle$ and $\bar {x}^{\alpha }=\bar {z}\not \in \langle \bar {x}^{o(\bar {x})/r} \rangle$, we get that $o(\bar {x}^{\alpha }) \nmid r$ and the assumption yields that $z \in G_m$. Hence, $T-\langle T \cap N,\, x^{o(\bar {x})/r}\rangle \subseteq G_m$. Thus, for every $u,\,v \in T-\langle T \cap N,\, x^{o(\bar {x})/r}\rangle$, $|C_G(u)|=|C_G(v)|$. If there exists an element $w \in (C_u \cap C_v)-(\langle T \cap N,\, x^{o(\bar {x})/r}\rangle )$, then $C_G(u),\,C_G(v) \leq C_G(w)$. On the other hand, $w \in T-\langle T \cap N,\, x^{o(\bar {x})/r} \rangle$. Hence, $w \in G_m$. So, $|C_G(w)|=|C_G(u)|=|C_G(v)|$. Therefore, $C_G(u)=C_G(w)=C_G(v)$. Consequently, $C_u=C_v$. Now, we claim that $T$ is abelian. If not, then for every $y \in T-\langle T \cap N,\, x^{o(\bar {x})/r}\rangle$, $C_y \neq T$, because $C_y \leq Z(C_G(y))$ is abelian. This yields that $T$ is partitioned relative to $\langle T \cap N,\, x^{o(\bar {x})/r}\rangle$. Since $P \leq T \cap N$, $T/\langle T \cap N,\, x^{o(\bar {x})/r}\rangle \leq \langle x\langle T \cap N,\, x^{o(\bar {x})/r}\rangle \rangle$, which is abelian. Consequently, $T/\langle T \cap N,\, x^{o(\bar {x})/r}\rangle$ is abelian. Also, $x \in N_G(P)$ and hence, $T \leq N_G(P)$. Thus, a Sylow $p$-subgroup of $T$ is normal in $T$. By lemma 2.9, $T/\langle T \cap N,\, x^{o(\bar {x})/r}\rangle$ is an elementary abelian $p$-group. This forces $o(\bar {x})/r$ to be prime, a contradiction. Therefore, $T$ is abelian. Thus, $P$ is abelian and $x \in C_G(P)$, as desired in (i). Now let $n \neq 1$. Set $T_1=\langle P,\,x'\rangle$ and $H=\langle x',\, N \rangle$. Since $\bar {x}=\bar {x}'$, substituting $x$ with $x'$ in the above argument shows that $T_1$ is abelian. Also, $T_1$ contains a Sylow $p$-subgroup of $H$. Hence, the Sylow $p$-subgroups of $H$ are abelian. On the other hand, $x=x'n$ for some $n\in N$. Thus, $H=\langle x,\, N \rangle$. Note that $p \mid o(\bar {x})$. Let $x_p$ be the $p$-part of $x$. Then, $C_G(x_p)$ contains a Sylow $p$-subgroup of $N$. By our assumption, $x,\,x_p \in G_m$ and since, $C_G(x) \leq C_G(x_p)$, we have $C_G(x) =C_G(x_p)$. Therefore, $C_G(x)$ contains a Sylow $p$-subgroup of $N$. So, (i) follows. By (i), there exists a $P \in {\rm Syl}_p(N)$ such that $T=\langle P,\,x\rangle$ is abelian. Hence, $T \leq C_G(x)$. Therefore, $|T|_p=|N|_p|o(\bar {x})|_p \mid |C_G(x)|$, as desired in (ii).

Proof. By our assumption, every $S_i$ has an irreducible character $\theta _i$ of $p$-defect zero, because $S_i \cong S$. Thus, $\theta _1 \times \cdots \times \theta _l \in {\rm Irr}(N)$ is of $p$-defect zero. So, (a) follows from [Reference Dolfi, Pacifici, Sanus and Spiga7, lemma 2.7]. Also, (b) can be concluded from lemma 2.11 and (a).

Proof. The proof is by induction on $|G|$. For every non-trivial normal subgroup $M$ of $G$, $NM/M$ is normal in $G/M$ and since $NM/M \cong N/(N\cap M)$, $NM/M$ is solvable. By induction, for every $\{1\} \neq M \unlhd G$, we get that $xM \in {\rm Fit}(G/M)$. Now as mentioned in the proof of [Reference Dolfi, Navarro, Pacifici, Sanus and Tiep5, theorem A], one of the following cases occurs:

Case 1. Let $M_1 \neq M_2$ be minimal normal subgroups of $G$. Then, the function $\phi : G \rightarrow \hat {G} = G/M_1 \times G/M_2$, defined by $\phi (g) = (gM_1,\, gM_2)$ for $g \in G$, is an injective homomorphism. By induction, $\phi (x) \in {\rm Fit}(G/M_1)\times {\rm Fit}(G/M_2) = {\rm Fit}(\hat {G} )$. So, $\phi (x) \in \phi (G) \cap {\rm Fit}(\hat {G} ) \leq {\rm Fit}(\phi (G))$. Since $\phi$ induces an isomorphism between $G$ and $\phi (G)$, we get that $x \in {\rm Fit}(G)$, as wanted.

Case 2. Assume that $G$ has the unique minimal normal subgroup $M$. By our assumption on $N$, $M\leq N$. Hence, $M \leq {\rm Fit}(G)$. Let $\Phi (G)$ denote the Frattini subgroup of $G$. If $\Phi (G) \neq \{1\}$, then $x \Phi (G) \in {\rm Fit}(G/\Phi (G))$, by induction. However, ${\rm Fit}(G/\Phi (G))={\rm Fit}(G)/\Phi (G)$. So, $x \in {\rm Fit}(G)$, as wanted. Now let $\Phi (G)=\{1\}$ and $x\not \in M$. By [Reference Huppert12, III, lemma 4.4], $M$ has a complement as $H$ in $G$, because $M \unlhd G$ is abelian. Since $C_H(M) \unlhd G$, the uniqueness of $M$ forces $C_H(M) =\{1\}$. So, $C_G(M)=M$. Let $V$ be the group of the irreducible characters of $M$. We can check that $V$ is a faithful and irreducible $G/M$-module. On the other hand, by lemma 2.13(vi), $xM$ fixes some element of each orbit of $G/M$ on $V$ and by induction, $xM \in {\rm Fit}(G/M)$. So, [Reference Isaacs, Navarro and Wolf14, theorem 4.2] forces $x^{2} \in M$. Since $M \leq {\rm Fit}(G)$, $x^{2} \in {\rm Fit}(G)$. However, $o(x{\rm Fit}(G))$ is odd. Hence, $x \in {\rm Fit(G)}$, as desired.

Proof. Since $M \unlhd N$ and $N$ is a $p$-group, $\{1\} \neq M \cap Z(N) \unlhd G$. As $M$ is a minimal normal subgroup of $G$, $M \cap Z(N) =M$. So, (i) follows. If $I_G(\chi )=G$, then it is easy to see that $M$ is a cyclic group of order $p$. By (i), $M \leq Z(N)$. Therefore, $\frac {G/N}{C_G(M)/N}\cong G/C_G(M)=N_G(M)/C_G(M)\lesssim {\rm Aut}(M)$ is cyclic. Hence, $G/N =C_G(M)/N$. Consequently, $C_G(M)=G$, so $M \leq Z(G)$, as wanted in (ii).

Proof. Let $P \in {\rm Syl}_7(G)$ and $1_M=\lambda _1,\,\ldots,\,\lambda _t$ be the representatives of the action of $P$ on ${\rm Irr}(M)$. If $\mathcal {O}_i$ is the $P$-orbit of $\lambda _i$, then $1+\Sigma _{i=2}^{t}|\mathcal {O}_i|\lambda _i(1)^{2}=\Sigma _{\lambda \in {\rm Irr}(M)}\lambda (1)^{2}=|M| \equiv _7 0$. Thus, there exists an $i >1$ such that $7 \nmid |\mathcal {O}_i|=[P:I_P(\lambda _i)]$. Therefore, $|\mathcal {O}_i|=1$ and hence $P \leq I_G(\lambda _i)$. On the other hand, $M \leq Z(N)$, by lemma 2.15(i). So, $N \leq I_G(\lambda _i)$. This yields that $\{N\} \neq PN/N \leq I_G(\lambda _i)/N \leq G/N \cong {\rm Alt}_7$. Lemma 2.15(ii) shows that $I_G(\lambda _i)/N < G/N \cong {\rm Alt}_7$. Note that the only maximal subgroup of ${\rm Alt}_7$ whose order is divisible by $7$ is isomorphic to $PSL_2(7)$. This signifies that

(2.1)\begin{equation} I_{G}(\lambda_i)/N \text{ is isomorphic to a subgroup of }PSL_2(7). \end{equation}

So, $I_{G}(\lambda _i)/N$ does not contain any element of order $6$ and neither does $I_{G}(\lambda _i)$. It follows from lemma 2.13(vi) that every element of $G$ of order $6$ is vanishing in $G$.

Now, let $\phi$ be a group isomorphism from $G/N$ to ${\rm Alt}_7$. It is known that ${\rm Alt}_7$ contains exactly two conjugacy classes containing $3$-elements. Let $\phi (x_3N)$ and $\phi (y_3N)$ be the representatives of these classes, for some $x_3N,\,y_3 N\in G/N$ of orders $3$. Since $N$ is a $7$-group, we can assume that $o(x_3)=o(y_3)=3$. By [Reference Conway, Curtis, Norton, Parker and Wilson4], we can assume that $\phi (x_3N) \in {\rm Van}({\rm Alt}_7)$, $\phi (x_3N)$ normalizes some Sylow $7$-subgroup of ${\rm Alt}_7$ and $\phi (y_3N)$ does not normalize any Sylow $7$-subgroup of ${\rm Alt}_7$. So,

(2.2)\begin{equation} x_3N \in {\rm Van}(G/N), \end{equation}

$x_3N$ normalizes some Sylow $7$-subgroup of $G/N$ and $y_3 N$ does not normalize any Sylow $7$-subgroup of $G/N$. By (2.1), $I_G(\lambda _i)/N$ is isomorphic to a subgroup of $PSL_2(7)$. However, $PSL_2(7)$ has only one conjugacy class containing $3$-elements and every element of this class normalizes some Sylow $7$-subgroup of $PSL_2(7)$. Note that a Sylow $7$-subgroup of $I_G(\lambda _i)/N$ is a Sylow $7$-subgroup of $G/N$. This shows that no conjugate of $y_3N$ lies in $I_G(\lambda _i)/N$ and if $I_G(\lambda _i)/N$ contains a $3$-element $uN$, then $uN \in cl_{G/N}(x_3N)$. This yields that no conjugate of $y_3$ lies in $I_G(\lambda _i)$. Thus, for every $3$-element $w \in G$, either $wN \in cl_{G/N}(x_3N)$ or no conjugate of $w$ lies in $I_G(\lambda _i)$. In the former case, (2.2) and lemma 2.13(ii) show that $w \in {\rm Van}(G)$. In the latter case, $w \in {\rm Van}(G)$, by lemma 2.13(vi). Now, the proposition follows.

Proof. Since $|Q|=5$, there is an element $x_5 \in G - N$ such that $o(x_5)=5$ and $Q=\langle x_5 \rangle$. Also, regarding $G/N \cong {\rm Alt}_5$, we get that $N_{G/N}(QN/N) =N_G(Q)N/N$ is a dihedral group of order $10$. Thus, $N_G(Q)$ contains an element $x$ such that $o(x)=2$ and $x \not \in N\cup C_G(Q)$. Let $P \in {\rm Syl}_2(G)$ such that $x \in P$. By lemma 2.7 (i,iii), $P = \{1,\,x,\,y,\,xy\}$ such that $o(y)=o(xy)=2$ and there is a $3$-element $\sigma \in N_G(P) - N$ such that $x^{\sigma }=y,\,~ y^{\sigma }=xy,\,~(xy)^{\sigma }=x.$ Put $\bar {G}=G/N$ and for every $H \leq G$ and $g \in G$, let $\bar {H}=HN/N$ and $\bar {g}$ denote the image of $g$ in $\bar {G}$. As $\bar {G} \cong {\rm Alt}_5$, we observe that

(2.3)\begin{equation} \bar{G}=\langle \bar{x}_5\rangle N_{\bar{G}}(\bar{P})=\langle \bar{x}_5\rangle \bar{P}\langle \bar{\sigma} \rangle. \end{equation}

Since $\bar {G} \cong {\rm Alt}_5$, $(2~5)(3~4) \in N_{{\rm Alt}_5}(\langle (1~2~3~4~5)\rangle )$, $U=\langle (2~5)(3~4),\, (2~3)(4~5) \rangle \in {\rm Syl}_2({\rm Alt}_5)$ and $(2~3~4) \in N_{{\rm Alt}_5}(U)$, there exists a group isomorphism $\phi$ from $\bar {G}$ to ${\rm Alt}_5$ which sends $\bar {x}_5$ to $(1~2~3~4~5)$, $\bar {x}$ to $(2~5)(3~4)$, $\bar {y}$ to $(2~3)(4~5)$ and $\bar {\sigma }$ to $(2~3~4)$. Let $\bar {u} \in \langle \bar {x}_5\rangle \bar {P}$. If $o(\bar {u})=t \in \{2,\,3,\,5\}$, then $o(\phi (\bar {u}))=t$. So, considering the $t$-elements of ${\rm Alt}_5$ lying in $\phi (\langle \bar {x}_5\rangle \bar {P})$ shows that if $t=2$, then $\phi (\bar {u}) \in N_{{\rm Alt}_5}(\langle (1~2~3~4~5)\rangle ) - \langle (1~2~3~4~5)\rangle =N_{\phi (\bar {G})}(\langle \phi (\bar {x}_5)\rangle ) - \langle \phi (\bar {x}_5)\rangle$ or $\phi (\bar {u}) \in \{(2~3)(4~5),\,(2~4)(3~5)\}=\{\phi (\bar {y}),\,\phi (\bar {x}\bar {y})\}$, if $t=3$, then $\phi (\bar {u}) \in \{(1~2~4),\,(1~5~3),\,(1~3~2),\,(1~4~5)\}$ and if $t=5$, then $\phi (\bar {u}) \in \langle (1~2~3~4~5)\rangle =\langle \phi (\bar {x}_5)\rangle$ or $\phi (\bar {u}) \in \{(1~3~4~2~5),\,(1~4~3~5~2),\,(1~2~5~4~3),\,(1~5~2~3~4)\}$. Thus, if $t=3$, then $\phi (\bar {u}^{-1}) \in$ $\{ (1\; 4\; 2) = \phi (\bar{x}_5^3 \bar{x}{\bar{\sigma }}^2),{\mkern 1mu} (1\; 3\; 5) = \phi (\bar{x}_5^2 \bar{y}{\bar{\sigma }}^2),{\mkern 1mu} (1\; 2\; 3)$ $_5^4 \bar{\sigma })\}$. Hence, $\bar {u}^{-1} \in \{\bar {x}_5^{3}\bar {x}\bar { \sigma }^{2},\,\bar {x}_5^{2}\bar {y}\bar { \sigma }^{2},\,\bar {x}_5\bar {y}\bar { \sigma },\,\bar {x}_5^{4}\bar { \sigma }\}$. Consequently, $\bar {u}^{-1} \not \in \langle \bar {x}_5\rangle \bar {P}$. Similarly, if $t=5$, then either $\bar {u}^{-1} \not \in \langle \bar {x}_5\rangle \bar {P}$ or $\bar {u} \in \langle \bar {x}_5\rangle$. In addition, we get that

(2.4)\begin{align} & \text{if }~\bar{u},\bar{u}^{{-}1} \in \langle \bar{x}_5\rangle\bar{P}, \text{ then either }o(\bar{u})=5 \text{ and }\bar{u} \in \langle \bar{x}_5\rangle\\ & \text{or }o(\bar{u})=2~{\rm and}~\bar{u} \in \{\bar{x}_5^{i}\bar{x},\bar{y},\bar{x}\bar{y}:1\leq i \leq 5\}. \nonumber \end{align}

On the other hand, by lemma 2.7(ii,iii),

(2.5)\begin{align} & M=C_M(P) \times C_T(x) \times C_T(y) \times C_T(xy), \end{align}

(2.6)\begin{align} & C_T(x)^{\sigma}=C_T(y),\quad C_T(y)^{\sigma}=C_T(xy),\ C_T(xy)^{\sigma}=C_T(x), \end{align}

where $T=[M,\,P]$. If $C_T(x) =\{1\}$, then (2.6) shows that $C_T(y)=C_T(xy)=\{1\}$. Thus, $M=C_M(P)$. So, $P \leq C_G(M)$. However, $N \leq C_G(M)$, by lemma 2.15(i). Hence, $\{\bar {1}\} \neq \bar {P} \leq C_G(M)/N \unlhd G/N=\phi ^{-1}({\rm Alt}_5)$. Therefore, $C_G(M)=G$, because $G/N$ is simple. Consequently, $M \leq Z(G)$, a contradiction. Thus, $C_T(x) \neq \{1\}$. By our assumption, $M$ is an elementary abelian $3$-group, so is $C_T(x)$. Hence, there exist subgroups $A_1,\, \ldots,\, A_t$ of $C_T(x)$ such that $|A_1|=\cdots =|A_t|=3$ and

(2.7)\begin{equation} C_T(x) = A_1 \times \cdots\times A_t,\quad C_T(y) = B_1 \times \cdots \times B_t,~ C_T(xy)= C_1 \times \cdots \times C_t, \end{equation}

where $B_i=A_i^{\sigma }$ and $C_i=A_i^{\sigma ^{2}}$, for every $1\leq i \leq t$, by (2.6). Let $n\in M$. By (2.5),

(2.8)\begin{equation} n=n_1n_2n_3n_4, \end{equation}

where $n_1 \in C_M(P)$, $n_2 \in C_T(x)$, $n_3 \in C_T(y)$ and $n_4 \in C_T(xy)$. Also, by (2.7),

(2.9)\begin{equation} \text{for every } j \in \{2,3,4\},\quad n_j=n_{j1} \ldots n_{jt}, \end{equation}

where for every $1\leq i \leq t$, $n_{2i} \in A_i$, $n_{3i} \in B_i$ and $n_{4i} \in C_i$.

If $C_M(x_5)=\{1\}$, then $|C_M(x)| \geq |C_T(x)|>1=|C_M(x_5)|$. So, (ii) follows. Next, let $1 \neq n \in C_M(x_5)$. For every $1 \leq i \leq 5$, (2.8) and lemma 2.7(iii,iv) yield that

(2.10)\begin{align} n^{x_5^{i}}& =n;~n^{x_5^{i}x}=n_1n_2n_3^{2}n_4^{2};~n^{x_5^{i}y}=n_1n_2^{2}n_3n_4^{2};~n^{x_5^{i}xy}=n_1n_2^{2}n_3^{2}n_4; \\ n^{x_5^{i}\sigma}& =n_1^{\sigma} n_4^{\sigma}n_2^{\sigma}n_3^{\sigma};~ n^{x_5^{i}x\sigma}= n_1^{\sigma} (n_4^{2})^{\sigma}n_2^{\sigma}(n_3^{2})^{\sigma};~ n^{x_5^{i}y\sigma}= n_1^{\sigma} (n_4^{2})^{\sigma}(n_2^{2})^{\sigma}(n_3)^{\sigma};\nonumber\\ n^{x_5^{i}xy\sigma}& =n_1^{\sigma} n_4^{\sigma}(n_2^{2})^{\sigma}(n_3^{2})^{\sigma};~n^{x_5^{i}\sigma^{2}}= n_1^{\sigma^{2}}n_3^{\sigma^{2}}n_4^{\sigma^{2}}n_2^{\sigma^{2}};~ n^{x_5^{i}x\sigma^{2}}= n_1 (n_3^{2})^{\sigma^{2}}(n_4^{2})^{\sigma^{2}}(n_2)^{\sigma^{2}};\nonumber\\ n^{x_5^{i}y\sigma^{2}}& = n_1 (n_3)^{\sigma^{2}}(n_4^{2})^{\sigma^{2}}(n_2^{2})^{\sigma^{2}};~n^{x_5^{i}xy\sigma^{2}}= n_1 (n_3^{2})^{\sigma^{2}}(n_4)^{\sigma^{2}}(n_2^{2})^{\sigma^{2}}. \nonumber \end{align}

Let check one of the above equalities in details. For instance, $n^{x_5^{i}xy\sigma }= (x_5^{i}xy\sigma )^{-1}n(x_5^{i}xy\sigma )=((xy)^{-1}((x_5^{i})^{-1}nx_5^{i})xy)^{\sigma }= ((xy)^{-1}nxy)^{\sigma }$ $=(n_1n_2^{2}n_3^{2}n_4)^{\sigma } = n_1^{\sigma } n_4^{\sigma }(n_2^{2})^{\sigma }(n_3^{2})^{\sigma }$. Similarly, we can check the other ones. Note that

(2.11)\begin{equation} \text{for every}\ i \in \{2,3,4\},\quad n_i^{\sigma}=n_{i1}^{\sigma}n_{i2}^{\sigma}\ldots n_{it}^{\sigma}, \end{equation}

by (2.7). We continue the proof in the following cases:

Case 1. Assume that $n_2=n_3=n_4=1$. Then, $n=n_1 \in C_M(P)$. Regarding the facts that $n \in C_M( x_5)$ and $M \leq Z(N)$, we have $P,\, \langle x_5\rangle,\,N \leq G_G(n)$. Therefore, $\bar {P},\,\langle \bar {x}_5\rangle \leq C_G(n)/N \leq G/N=\phi ^{-1}({\rm Alt}_5)$. Since the only subgroup of ${\rm Alt}_5$ whose order is divisible by $20$ is ${\rm Alt}_5$, we get that $C_G(n)/N=G/N$. Thus, $C_G(n)=G$. Consequently, $n \in Z(G)$. Therefore, $M=\langle n\rangle \leq Z(G)$, a contradiction.

Case 2. Assume that $n_{2i} \neq 1$, for some $1 \leq i \leq t$. Without loss of generality, let $i=1$. Since $x \in N_G(\langle x_5\rangle )$ and $n \in C_M(x_5)$, we have $n^{x} \in C_M(\langle x_5\rangle )^{x}=C_M(\langle x_5\rangle )=C_M(x_5)$. By lemma 2.7(iv), $n_3^{x}=n_3^{2}$ and $n_4^{x}=n_4^{2}$. Thus, $(n^{x})_{31}=n_{31}^{2}$ and $(n^{x})_{41}=n_{41}^{2}$. Also, $n_2^{x}=n_2$, because $n_2 \in C_T(x)$. Hence, $(nn^{x})_{21}=(n_{21})^{2} \neq 1$, $(nn^{x})_{31}=n_{31}(n_{31})^{2} = 1$ and $(nn^{x})_{41}=n_{41}(n_{41})^{2} = 1$, because $M$ is an elementary abelian $3$-group and $n_{21},\,n_{31},\,n_{41} \in M$. As $1 \neq nn^{x} \in C_M(x_5)$, by substituting $n$ with $nn^{x}$, we can assume that $n_{31}=n_{41}=1$. Set $\chi = 1_{C_{M}(P)} \times 1_{C_T(x)} \times (\theta _3 \times 1_{B_2} \times \cdots \times 1_{B_t})\times (\theta _4 \times 1_{C_2}\times \cdots \times 1_{C_t}),\,$ where $\theta _3 \in {\rm Irr}(B_1)-\{1_{B_1}\}$, $\theta _4 \in {\rm Irr}(C_1)-\{1_{C_1}\}$ and $\theta _4(m^{\sigma })=\theta _3(m^{2})$, for every $m \in B_1$. Then, $1_M \neq \chi \in {\rm Irr}(M)$. As $M \leq Z(N)$, $N \leq I_G(\chi )$. Let $u \in I_{G}(\chi )- N$. Then $\bar {u} \in \bar {G}$. By (2.3), $\bar {u}=\bar {x}_5^{i}\bar {x}^{j}\bar {y}^{k}\bar {\sigma }^{l}$, for some non-negative integers $i,\,j,\,k$ and $l$. Working towards a contradiction, let $l \neq 0$. Since $o(\phi (\bar {\sigma }))=3$, $o(\bar {\sigma })=3$ and hence, we can assume that $l \in \{1,\,2\}$. Also, $j,\,k \in \{0,\,1\}$. Regarding $u \in I_{G}(\chi )$, $u^{-1} \in I_G(\chi )$. Therefore, $\chi ^{u^{-1}}(n)=\chi (n)$. Also, $n \in M \leq Z(N)$. By (2.10), $\chi ^{u^{-1}}(n) = \chi (u^{-1}nu)= \chi ({\sigma ^{-l}(n_1n_2^{2^{k}}n_3^{2^{j}}n_4^{2^{|j-k|}})\sigma ^{l}})$, so

(2.12)\begin{equation} \chi^{u^{{-}1}}(n)= \left\{\begin{array}{@{}ll}\chi(n_1^{\sigma}(n_4^{2^{|j-k|}})^{\sigma}(n_2^{2^{k}})^{\sigma}(n_3^{2^{j}})^{\sigma}), & \text{ if }l=1 \\ \chi(n_1^{\sigma^{2}}(n_3^{2^{j}})^{\sigma^{2}}(n_4^{2^{|j-k|}})^{\sigma^{2}}(n_2^{2^{k}})^{\sigma^{2}}), & \text{ if }l=2 \end{array}\right.. \end{equation}

It follows that either $l=1$ and $\chi ^{u^{-1}}(n)=\theta _3((n_{21}^{2^{k}})^{\sigma })\theta _4((n_{31}^{2^{j}})^{\sigma })$ or $l=2$ and $\chi ^{u^{-1}}(n)=\theta _3((n_{41}^{2^{|j-k|}})^{\sigma ^{2}})\theta _4((n_{21}^{2^{k}})^{\sigma ^{2}})$. Since $n_{31}=n_{41}=1$, $n_{21} \neq 1$ and $\theta _4((n_{21}^{2^{k}})^{\sigma ^{2}})=\theta _3(((n_{21}^{2^{k}})^{\sigma })^{2})$, we have $\chi ^{u^{-1}}(n)=\theta _3((n_{21}^{2^{k}})^{\sigma })^{l} \neq 1$. However, $\chi (n)=\theta _3(n_{31})\theta _4(n_{41})=1$, a contradiction. This forces $\bar {u} \in \langle \bar {x}_5\rangle \bar {P}$. Consequently,

(2.13)\begin{equation} I_G(\chi)/N \subseteq \langle \bar{x}_5 \rangle \bar{P}. \end{equation}

Also, for $1 \neq \gamma \in B_1$ and $1 \neq \beta \in C_1$, $\chi ^{xy}(\gamma )=\chi ^{x}(\gamma )=\chi (\gamma ^{2})=\theta _3(\gamma ^{2}) \neq \theta _3(\gamma )=\chi (\gamma )$ and $\chi ^{y}(\beta )=\chi (\beta ^{2})=\theta _4(\beta ^{2}) \neq \theta _4(\beta )=\chi (\beta )$. Therefore,

(2.14)\begin{equation} x,y,xy \not \in I_G(\chi). \end{equation}

Now, assume that $u \in I_G(\chi )$. Then, $\bar {u},\,\bar {u}^{-1} \in I_G(\chi )/N$. So, in view of (2.4), (2.13) and (2.14), one of the following sub-cases holds:

Sub-case a. Assume that $o(\bar {u})=2$. If $4$ or $5 \mid |I_G(\chi )/N|$, then taking the elements mentioned in (2.4) into account, we conclude that $I_G(\chi )/N=\bar {P}$ or $I_G(\chi )/N=N_{\bar {G}}(\langle \bar {x}_5\rangle )=\langle \bar {x}_5\rangle \langle \bar {x}\rangle$, contradicting (2.14). Thus, $|I_G(\chi )/N|=o(\bar {u})=2$ and $\bar {u} \in \{\bar {x}_5^{i}\bar {x} :1\leq i \leq 5\}$. So, $\mathcal {B}=\{(x_5^{j}y^{l}\sigma ^{k})^{-1}: 1\leq j\leq 5,\, 0 \leq l \leq 1 ~{\rm and}~ 0\leq k \leq 2\}$ is a transversal set of $I_G(\chi )$ in $G$. Hence, for every $\psi \in {\rm Irr}(G|\chi )$, $\psi (n)= e\Sigma _{g \in \mathcal {B}} \chi (n^{g^{-1}})=e\Sigma _{i=1}^{5}[\chi (n^{(x_5^{i})})+\chi (n^{(x_5^{i}\sigma )})+\chi (n^{(x_5^{i}\sigma ^{2})})]+e\Sigma _{i=1}^{5}[\chi (n^{(x_5^{i}y)})]$ $+\chi (n^{(x_5^{i}y\sigma )})+\chi (n^{(x_5^{i}y\sigma ^{2})})]$, for some positive integer $e$. By (2.10), $\psi (n)= e\Sigma _{i=1}^{5}[ \chi (n)+\chi (n_1^{\sigma }n_4^{\sigma }n_2^{\sigma }n_3^{\sigma })+ \chi (n_1^{\sigma ^{2}}n_3^{\sigma ^{2}}n_4^{\sigma ^{2}}n_2^{\sigma ^{2}})]$ $~+~e\Sigma _{i=1}^{5}[ \chi (n_1n_2^{2}n_3n_4^{2})+\chi (n_1(n_4^{2})^{\sigma } (n_2^{2})^{\sigma }n_3^{\sigma })+ \chi (n_1n_3^{\sigma ^{2}} (n_4^{2})^{\sigma ^{2}} (n_2^{2})^{\sigma ^{2}})]$. Thus, $\psi (n) ~=~e\Sigma _{i=1}^{5}[ 1+ \theta _3((n_{21})^{\sigma }) \theta _4 ((n_{31})^{\sigma }) +\theta _3((n_{41})^{\sigma ^{2}}) \theta _4((n_{21})^{\sigma ^{2}})] +e\Sigma _{i=1}^{5}[ \theta _3(n_{31}) ]$ $\theta _4(n_{41}^{2})+\theta _3((n_{21}^{2})^{\sigma })\theta _4((n_{31})^{\sigma }) +\theta _3((n_{41}^{2})^{\sigma ^{2}})]$ $\theta _4((n_{21}^{2})^{\sigma ^{2}})]=5e[2( 1+\theta _3((n_{21})^{\sigma }) +\theta _3(((n_{21})^{\sigma })^{2}))]$. Note that $n_{21} \neq 1$. Therefore, $\theta _3((n_{21})^{\sigma })\neq 1$ is a primitive $3$rd root of unitary. Hence,

(2.15)\begin{equation} \theta_3((n_{21})^{\sigma})^{2}+\theta_3((n_{21})^{\sigma})+1=0 \end{equation}

It follows that $\psi (n)=0$, as wanted in (i).

Sub-case b. Assume that $I_G(\chi ) / N \leq \langle \bar {x}_5\rangle$. So, either $I_G(\chi ) / N = \langle \bar {x}_5\rangle$ or $I_G(\chi )=N$. If $I_G(\chi )/N=\langle \bar {x}_5\rangle$, let $b=1$ and if $I_G(\chi )=N$, let $b=5$. So, $\mathcal {B}=\{(x_5^{i}x^{j}y^{k}\sigma ^{l})^{-1} : 0 \leq i \leq b-1,\,j,\,k \in \{0,\,1\},\,l\in \{0,\,1,\,2\}\}$ is a transversal set of $I_G(\chi )$ in $G$. Hence, for every $\psi \in {\rm Irr}(G|\chi )$, $\psi (n)=e\Sigma _{g \in \mathcal {B}} \chi ^{g}(n)= e\Sigma _{i=0}^{b-1}[\chi (n^{x_5^{i}})+\chi (n^{x_5^{i}\sigma })+\chi (n^{x_5^{i}\sigma ^{2}})] $ $ +~ e\Sigma _{i=0}^{b-1}[\chi (n^{x_5^{i}x})+\chi (n^{x_5^{i}x\sigma })+\chi (n^{x_5^{i}x\sigma ^{2}})]$ $+~ e\Sigma _{i=0}^{b-1}[\chi (n^{x_5^{i}y})+\chi (n^{x_5^{i}y\sigma })+\chi (n^{x_5^{i}y\sigma ^{2}})] + e\Sigma _{i=0}^{b-1}[\chi (n^{x_5^{i}xy})]$ $+\chi (n^{x_5^{i}xy\sigma })+\chi (n^{x_5^{i}xy\sigma ^{2}})]$, for some positive integer $e$. By (2.10) and (2.15), we can check at once that $\psi (n)= 4be[1+\theta _3(n_{21}^{\sigma })+\theta _3(n_{21}^{\sigma })^{2} ] =0$, as wanted in (i).

Case 3. Assume that $n_2=1$ and there exists an $1 \leq i \leq t$ such that $n_{3i} \neq 1$ and $n_{4i}=1$. Without loss of generality, let $i=1$ and set $\chi = 1_{C_{M}(P)} \times (\theta _2 \times 1_{A_2} \times \cdots \times 1_{A_t}) \times 1_{C_T(y)}\times (\theta _4 \times 1_{C_2}\times \cdots \times 1_{C_t}),\,$ where $\theta _2 \in {\rm Irr}(A_1)-\{1_{A_1}\}$, $\theta _4 \in {\rm Irr}(C_1)-\{1_{C_1}\}$ and $\theta _4(m^{\sigma ^{2}})=\theta _2(m^{2})$, for every $m \in A_1$. Also, if $n_2=1$ and there exists an $1 \leq i \leq t$ such that $n_{4i} \neq 1$ and $n_{3i}=1$, then without loss of generality, let $i=1$ and set $\chi = 1_{C_{M}(P)} \times (\theta _2 \times 1_{A_2} \times \cdots \times 1_{A_t}) \times (\theta _3 \times 1_{B_2}\times \cdots \times 1_{B_t}) \times 1_{C_T(xy)} ,\,$ where $\theta _2 \in {\rm Irr}(A_1)-\{1_{A_1}\}$, $\theta _3 \in {\rm Irr}(B_1)-\{1_{B_1}\}$ and $\theta _3(m^{\sigma })=\theta _2(m^{2})$, for every $m \in A_1$. Then, $1_M \neq \chi \in {\rm Irr}(M)$ and arguing by analogy as Case 2 shows that for every $\psi \in {\rm Irr}(G|\chi )$, $\psi (n)=0$, as wanted in (i).

Case 4. Assume that $C_M(x_5)$ does not contain any element satisfying cases 1–3. Let $\alpha,\, \beta \in C_M(x_5)$. By (2.8), $\alpha =\alpha _1\alpha _2\alpha _3\alpha _4$ and $\beta =\beta _1\beta _2\beta _3\beta _4$, where $\alpha _1,\,\beta _1 \in C_M(P)$, $\alpha _2,\,\beta _2 \in C_T(x)$, $\alpha _3,\,\beta _3 \in C_T(y)$ and $\alpha _4,\,\beta _4 \in C_T(xy)$ are uniquely determined. (2.9) shows that for every $j \in \{2,\,3,\,4\}$, $\alpha _j=\alpha _{j1} \ldots \alpha _{jt}$ and $\beta _j=\beta _{j1} \ldots \beta _{jt}$, where for every $1\leq i \leq t$, $\alpha _{2i},\,\beta _{2i} \in A_i$, $\alpha _{3i},\,\beta _{3i} \in B_i$ and $\alpha _{4i},\,\beta _{4i} \in C_i$. By our assumption, $\alpha _2=\beta _2=1$, $\alpha _3,\,\beta _3,\,\alpha _4,\,\beta _4 \neq 1$ and for every $1 \leq i \leq t$, $\alpha _{3i} \neq 1$ if and only if $\alpha _{4i} \neq 1$. Also, $\beta _{3i} \neq 1$ if and only if $\beta _{4i} \neq 1$. If $\alpha _{3i}=\beta _{3i}\neq 1$ and $\alpha _{4i}=\beta _{4i}^{2}\neq 1$, for some $1\leq i \leq t$, then $(\alpha \beta )_{3i}=\alpha _{3i} \beta _{3i}=\alpha _{3i}^{2} \neq 1$ and $(\alpha \beta )_{4i}=\alpha _{4i} \beta _{4i}=\alpha _{4i}^{3}= 1$. However, $\alpha \beta \in C_M(x_5)$. So, $\alpha \beta$ satisfies the assumption of case 3, a contradiction. This shows that for an element $\alpha \in C_M(x_5)$ and an integer $1 \leq i \leq t$, if $\alpha _{3i}\neq 1$, then $\alpha _{4i} \neq 1$ and

(2.16)\begin{equation} \text{for every }\beta \in C_M(x_5),\quad (\beta_{3i},\beta_{4i}) \in \{(1,1), (\alpha_{3i}, \alpha_{4i}),(\alpha_{3i}^{2}, \alpha_{4i}^{2}) \}. \end{equation}

Now, working towards a contradiction, let $\alpha _1\neq 1$. Then, since $x \in N_{G}(\langle x_5\rangle )$, $\alpha ^{x} \in C_M({x}_5)^{x}=C_M(x_5)$. By lemma 2.7(iv), $\alpha ^{x}=\alpha _1\alpha _2\alpha _3^{2}\alpha _4^{2}$. Thus, $\alpha \alpha ^{x}=\alpha _1^{2}\alpha _2^{2}\alpha _3^{3}\alpha _4^{3}$. Note that $\alpha _2=1$ and $o(\alpha _3)=o(\alpha _4)=3$. Therefore, $1 \neq \alpha \alpha ^{x}=\alpha _1^{2} \in C_M(P)$. On the other hand, $\alpha,\, \alpha ^{x}\in C_M(x_5)$. So, $1 \neq \alpha ^{x}\alpha =\alpha _1^{2} \in C_M(x_5) \cap C_M(P)$, which is a contradiction with case 1. This shows that for every $\alpha \in C_M(x_5)$, $\alpha _1=1$. It follows from (2.16) that $|C_M(x_5)|\leq |C_T(y)|$. If $C_M(P) \neq \{1\}$, then we get that $|C_M(x_5)| < |C_M(P)||C_T(y)|=|C_M(P)||C_T(x)|=|C_M(x)|$, so (ii) follows. Next, let $C_M(P)=\{1\}$. Then, $T=M=C_M(x) \times C_M(y) \times C_M(xy)$ and $|C_M(x_5)| \leq |C_T(y)|=|C_M(y)|=|C_M(x)|$, by (2.16). If $|C_M(x_5)| <|C_M(x)|$, then (ii) follows. Otherwise, $|C_M(x_5)|=|C_M(x)|$. If $|C_M(x)|=3$, then $|M|=27$ and $|[M,\,\langle x_5\rangle ]|=9$. However, $\langle x_5\rangle$ acts fixed point freely on $[M,\,\langle x_5\rangle ]$. So, $5 \mid |[M,\,\langle x_5\rangle ]|-1=8$, which is impossible. This forces $|C_M(x)| \geq 9$. Consequently, $t \geq 2$ ($t$ was fixed in (2.7)). Since $|C_M(x_5)|=|C_M(x)|=|C_M(y)|$, we get from (2.16) that for every $1 \leq i \leq t$ and $m \in B_i$, there exists an element $\alpha \in C_M(x_5)$ such that $\alpha _{3i}=m$. So, for $1 \neq n \in C_M(x_5)$, we can assume that $n_{31},\,n_{32} \neq 1$. Consequently, $n_{41},\,n_{42} \neq 1$. As was mentioned above, $n_2=1$. Set $\chi = 1_{C_{M}(x)} \times (\theta _3 \times \theta _3'\times 1_{B_3} \times \cdots \times 1_{B_t}) \times (\theta _4 \times \theta _4' \times 1_{C_3}\times \cdots \times 1_{C_t}),\,$ where $\theta _3 \in {\rm Irr}(B_1)-\{1_{B_1}\}$, $\theta _3' \in {\rm Irr}(B_2)-\{1_{B_2}\}$, $\theta _4\in {\rm Irr}(C_1)-\{1_{C_1}\}$ and $\theta _4' \in {\rm Irr}(C_2)-\{1_{C_2}\}$ such that $\theta _3'(n_{32})=\theta _3(n_{31})^{2}$ and $\theta _4'(n_{42})=\theta _4(n_{41})^{2}$. Moreover, suppose that $\theta _4(n_{41})=\theta _3(n_{31})$. Then, $1_{M}\neq \chi \in {\rm Irr}(M)$. Note that $n_{31}$ and $n_{32}$ are generators of $B_1$ and $B_2$, respectively. Also, $n_{41}$ and $n_{42}$ are generators of $C_1$ and $C_2$, respectively.

In the following, we first assume that $(n_{31}^{\sigma },\, n_{32}^{\sigma }) =(n_{41},\,n_{42})$ or $(n_{31}^{\sigma },\, n_{32}^{\sigma }) =(n_{41}^{2},\,n_{42}^{2})$. If $(n_{31}^{\sigma },\, n_{32}^{\sigma }) =(n_{41},\,n_{42})$, let $u=1$ and otherwise let $u=2$. We note that $N \leq I_G(\chi )$. By (2.10), we can check that for every $g\in G$, $\chi (n^{g})=1$, for instance, $\chi (n^{x_5^{i}x\sigma ^{2}})= \chi ((n_3^{2})^{\sigma ^{2}}(n_4^{2})^{\sigma ^{2}}n_2^{\sigma ^{2}})= [\theta _3((n_{41}^{2})^{\sigma ^{2}})\theta _3'((n_{42}^{2})^{\sigma ^{2}})] [\theta _4(1)\theta _4'(1)]=\theta _3(n_{31})^{6u}=1$. So, for every $\psi \in {\rm Irr}(G|\chi )$, $\psi (n) =\psi (1)$. Therefore, $1 \neq n \in {\rm ker}\psi \cap M$. Thus, $\{1\} \neq M \cap {\rm ker}\psi \unlhd G$. Since $M$ is a minimal normal subgroup of $G$, $M \cap {\rm ker}\psi =M$. Therefore, $\psi _M =\psi (1)1_M$, a contradiction.

Next, suppose that $(n_{31}^{\sigma },\,n_{32}^{\sigma }) \in \{(n_{41},\,n_{42}^{2}),\,(n_{41}^{2},\,n_{42})\}$. If $u=x_5^{i}x^{j}y^{k}\sigma ^{l} h \in I_G(\chi )$, where $h \in N$, $i \in \{1,\,\ldots,\,5\}$, $j,\,k \in \{0,\,1\}$ and $l \in \{1,\,2\}$, then $u^{-1} \in I_G(\chi )$ and we get from (2.12) that if $l=1$, then $\chi ^{u^{-1}}(n)\in \{(\theta _4(n_{41})^{2})^{2^{j}},\,(\theta _4(n_{41}))^{2^{j}}\}$ and if $l=2$, then $\chi ^{u^{-1}}(n)\in \{(\theta _3(n_{31})^{2})^{2^{|j-k|}},\,(\theta _3(n_{31}))^{2^{|j-k|}}\}$. Hence, $\chi ^{u^{-1}}(n) \neq 1$. However, $\chi (n)=1$, a contradiction. Consequently, $\bar {u} \in \langle \bar {x}_5\rangle \bar {P}$. Therefore, $I_G(\chi )/N \subseteq \langle \bar {x}_5\rangle \bar {P}.$ Let $\beta =n_{41} \in C_1-\{1\}$. By lemma 2.7(iv), $\chi (\beta )=\theta _4(n_{41})$, $\chi ^{x^{-1}}(\beta )=\chi (\beta ^{x})=\chi (\beta ^{2})=\theta _4(n_{41})^{2}$ and $\chi ^{y^{-1}}(\beta )=\chi (\beta ^{y})=\chi (\beta ^{2})=\theta _4(n_{41})^{2}$. Since $n_{41} \neq 1$, $\theta _4(n_{41})^{2} \neq \theta _4(n_{41})$. Consequently, $\chi ^{x^{-1}},\,\chi ^{y^{-1}} \neq \chi$. Hence, $x,\,y \not \in I_G(\chi ).$ By (2.16) and since $|C_M(y)|=|C_M(x_5)|$, we can assume that there exists an element $\alpha \in C_M(x_5)$ such that $\alpha _2=1$, $\alpha _{31}=n_{31}\neq 1$ and for every $j \in \{2,\,\ldots,\,t\}$, $\alpha _{3j}=1$. Then, (2.16) guarantees that $\alpha _{41}=n_{41} \neq 1$ and for every $j\in \{2,\,\ldots,\,t\}$, $\alpha _{4j}=1$. However, $\chi (\alpha )=\theta _3(\alpha _{31})\theta _4(\alpha _{41})=\theta _3(n_{31})\theta _4(n_{41})= \theta _3(n_{31})^{2}$ and $\chi ^{(x_5^{i}x)^{-1}}(\alpha )=\chi (\alpha ^{x_5^{i}x})=\chi (\alpha _2\alpha _3^{2}\alpha _4^{2})= \theta _3(\alpha _{31}^{2})\theta _4(\alpha _{41}^{2})=\theta _3(n_{31})^{4}=\theta _3(n_{31})$. Since $\theta _3(n_{31}) \neq 1$, $\chi ^{(x_5^{i}x)^{-1}}(\alpha ) \neq \chi (\alpha )$. This shows that $x_5^{i}x \not \in I_G(\chi )$. Taking the elements mentioned in (2.4) into account, we conclude that $I_G(\chi ) / N \leq \langle \bar {x}_5\rangle$. So, either $I_G(\chi ) / N = \langle \bar {x}_5\rangle$ or $I_G(\chi )=N$. If $I_G(\chi )/N=\langle \bar {x}_5\rangle$, let $b=1$ and if $I_G(\chi )=N$, let $b=5$. So, $\mathcal {B}=\{(x_5^{i}x^{j}y^{k}\sigma ^{l})^{-1} : 0 \leq i \leq b-1,\,j,\,k \in \{0,\,1\},\,l\in \{0,\,1,\,2\}\}$ is a transversal set of $I_G(\chi )$ in $G$. Hence, for every $\psi \in {\rm Irr}(G|\chi )$, $\psi (n)=e\Sigma _{g \in \mathcal {B}} \chi ^{g}(n)= e\Sigma _{i=0}^{b-1}[\chi (n^{x_5^{i}})+\chi (n^{x_5^{i}\sigma })+\chi (n^{x_5^{i}\sigma ^{2}})]$ $ + e\Sigma _{i=0}^{b-1}[\chi (n^{x_5^{i}x})+\chi (n^{x_5^{i}x\sigma })+\chi (n^{x_5^{i}x\sigma ^{2}})] + e\Sigma _{i=0}^{b-1}[\chi (n^{x_5^{i}y})+\chi (n^{x_5^{i}y\sigma })+\chi (n^{x_5^{i}y\sigma ^{2}})]$ $+ e\Sigma _{i=0}^{b-1}[\chi (n^{x_5^{i}xy})+\chi (n^{x_5^{i}xy\sigma })+\chi (n^{x_5^{i}xy\sigma ^{2}})]$, for some positive integer $e$. We note that $\chi (n_2n_3n_4)=\chi (n_2n_3^{2}n_4^{2})=\chi (n_2^{2}n_3n_4^{2})=\chi (n_2^{2}n_3^{2}n_4)=1$. Thus, if $(n_{31}^{\sigma },\,n_{32}^{\sigma })=(n_{41},\,n_{42}^{2})$, then by (2.10), $\psi (n) =e\Sigma _{i=0}^{b-1}[1+ \theta _3(n_{31})^{2}+\theta _3(n_{31})^{2}] + e\Sigma _{i=0}^{b-1}[1+ \theta _3(n_{31})$ $\theta _3(n_{31})] + e\Sigma _{i=0}^{b-1}[1+\theta _3(n_{31})+ \theta _3(n_{31})^{2} ] +e\Sigma _{i=0}^{b-1} [1+ \theta _3(n_{31})^{2}$ $+ \theta _3(n_{31})] = 4be(1+\theta _3(n_{31})+\theta _3(n_{31})^{2}).$ Also, if $(n_{31}^{\sigma },\,n_{32}^{\sigma })=(n_{41}^{2},\,n_{42})$, then similarly $\psi (n)= 4be(1+\theta _3(n_{31})+\theta _3(n_{31})^{2}).$ However, $n_{31} \neq 1$. So, $o(\theta _3(n_{31}))=3$. Therefore, $1 \neq \theta _3(n_{31})$ is a primitive third root of unitary. It follows that $(\theta _3(n_{31}))^{2}+\theta _3(n_{31})+1=0$. Thus, we get that $\psi (n)=4be(1+\theta _3(n_{31})+\theta _3(n_{31})^{2})=0,\,$ as desired in (i). Now, the proof is complete.