Non-solvable groups whose non-linear character degrees have the same number of different prime divisors

Abstract.

By a result of Noritzsch, a finite solvable group whose non-linear character degrees have the same set of prime divisors is meta-abelian. In this note we investigate finite non-solvable groups whose non-linear character degrees have the same number of different prime divisors, and show that up to an abelian direct factor, such groups are exactly $L_{2}(4),L_{2}(8),A_{7},S_{7}$, the central product of a cyclic $3$-group with $3.A_{7}$, or the semi-direct product of $A_{7}$ by a cyclic $2$-group $\langle a\rangle$ such that $a$ non-trivially acts on $A_{7}$ by conjugation. As consequence, we show that only the primes $2,3,5,7$ may occur as prime divisors of their irreducible character degrees, and that Huppert’s $\rho$-$\sigma$ conjecture holds for them.

1. Introduction

The study of character degrees is an interesting topic and has a long history in the representation theory of finite groups. It was Isaacs and Passman who first started the project on the structure of finite groups all of whose non-linear complex irreducible characters have prime degrees, see [6, 5]. Along the line of research, O. Manz dealt with finite groups all of whose non-linear complex irreducible characters have prime power degrees and classified such non-solvable groups, see [9, 8].

As a further step, we investigate finite groups all of whose non-linear complex irreducible characters have the same number of different prime divisors. Relevantly, Noritzsch studied finite solvable groups all of whose non-linear complex irreducible characters have the same set of prime divisors and showed that such groups are meta-abelian, see [3, Proposition 27.9] or [11]. In this note, we focus on finite non-solvable groups.

Let $G$ be a finite group, ${\rm Irr}(G)$ the set of complex irreducible characters of $G$, and ${\rm cd}(G)$ the degree set of ${\rm Irr}(G)$. For a positive integer $n$, we denote by $\pi(n)$ the set of prime divisors of $n$ and by $|\pi(n)|$ the number of prime divisors of $n$.

Definition 1.1.

A finite group $G$ is called a SNPD-group if for all $1\neq\chi(1)\in{\rm cd}(G)$, $\chi(1)$ has the same number of different prime divisors.

We first classify almost simple SNPD-groups.

Theorem 1.2.

Let $G$ be an almost simple group. Then $G$ is a SNPD-group if and only if $G$ is isomorphic to $L_{2}(4)$, $L_{2}(8)$, $A_{7}$ or $S_{7}$.

In general, we have

Theorem 1.3.

Let $G$ be a finite non-solvable group. Then $G$ is a SNPD-group if and only if $G\cong A\times B$, where $A$ is an abelian group and $B$ satisfies one of the following:

1. $B=L_{2}(4),L_{2}(8),A_{7}$ or $S_{7}$.

2. $B$ is the central product of a cyclic $3$-group with $3.A_{7}$.

3. $B$ is the semi-direct product of $A_{7}$ by a cyclic $2$-group $\langle a\rangle$ such that $B/\langle a^{2}\rangle\cong S_{7}$.

Let

$$\sigma(G)=\max\{|\pi(\chi(1))|:\chi\in\operatorname{Irr}(G)\}$$

and

$$\rho(G)=\{p\text{ prime}:p\mid\chi(1)\text{ for some }\chi\in\operatorname{Irr}(G)\}.$$

Huppert conjectured that $|\rho(G)|$ can be bounded in terms of $\sigma(G)$ and, if $G$ is solvable, then $|\rho(G)|\leq 2\sigma(G)$. Nowadays it is often called Huppert’s $\rho$-$\sigma$ conjecture, and is a problem of central importance in character theory. According to Theorem 1.3 and its proof, we have

Corollary 1.4.

Let $G$ be a finite non-solvable SNPD-group. Then $\rho(G)\subseteq\{2,3,5,7\}$, and $\sigma(G)=1$ or 2.

In particular, Huppert’s $\rho$-$\sigma$ conjecture holds for non-solvable SNPD-groups.

2. Proof of Theorems

We first prove Theorem 1.2.

Proof of Theorem 1.2.

Notice that $A_{5}\cong L_{2}(4)\cong L_{2}(5)$. The “if” part is clear by Table 1, where character degrees of $L_{2}(4)$, $L_{2}(8)$, $A_{7}$ and $S_{7}$ are listed.

Table 1. Degrees of $L_{2}(4)$, $L_{2}(8)$, $A_{7}$ and $S_{7}$

$S$	${\rm cd}(S)$
$L_{2}(4)$	$\{1,3,4,5\}$
$L_{2}(8)$	{1,7,8,9}
$A_{7}$	$\{1,6,10,14,15,21,35\}$
$S_{7}$	$\{1,6,20,14,15,21,35\}$

For the “only if” part, we let $S$ be the socle of $G$ and note that, by the Classification Theorem of Finite Simple Groups, $S$ is an alternating group, a simple group of Lie type, a sporadic simple group or the Tits simple group. It is easy to check by GAP [14] that sporadic simple groups and the Tits simple group are not a SNPD-group (see also Table 4).

Let $S$ be a simple group of Lie type. Then $S$ has the Steinberg character, whose degree is a prime power and which extends to $G$ (see [12, 13]). So, if $G$ is a SNPD-group then all complex irreducible characters of $G$ have prime power degrees. By [15, Proposition B], $G$ is isomorphic to $L_{2}(4)$ or $L_{2}(8)$.

Finally, let $S$ be the alternating group $A_{n}$ with $n\geq 5$. Then ${\rm Aut}(A_{n})=S_{n}$ if $n\neq 6$. The conclusion is true for $n=5,6$ or $7$ by the GAP libray [14]. So we may assume that $n>7$ in the following. The complex irreducible characters of the symmetric group $S_{n}$ are naturally labeled by the partitions of $n$. Let Irr$(S_{n})=\{[\lambda]\mid\lambda\vdash n\}$, where $\lambda\vdash n$ denotes a partition of $n$. Notice that the degree of $[\lambda]$ is $\frac{n!}{\prod_{i,j}h_{i,j}}$, where $h_{i,j}$ is the $(i,j)$-hook number, and that the restriction $[\lambda]_{A_{n}}$ of $[\lambda]\in$ Irr$(S_{n})$ to $A_{n}$ is irreducible if and only if $\lambda$ is not self-conjugate. See [7].

Let $\lambda_{1}=(n-1,1),\lambda_{2}=(n-2,1^{2})$ and $\lambda_{3}=(n-3,1^{3})$ so that the degrees $d_{1},d_{2}$ and $d_{3}$ of $[\lambda_{1}],[\lambda_{2}]$ and $[\lambda_{3}]$ are $n-1$, $\frac{(n-1)(n-2)}{2}$ and $\frac{(n-1)(n-2)(n-3)}{6}$, respectively. Clearly, $\lambda_{i}$ is not self-conjugate for all $1\leq i\leq 3$ and $n>7$, and so $\{d_{1},d_{2},d_{3}\}\subset{\rm cd}(S)$.

If $n$ is even, then $\frac{n-2}{2}$ is an integer. Clearly, $n-1$ and $\frac{n-2}{2}$ are coprime to each other. Hence

$$|\pi(d_{1})|=|\pi(n-1)|<|\pi(\frac{(n-1)(n-2)}{2})|=|\pi(d_{2})|.$$

Therefore, if $G$ is a SNPD-group then $n$ is odd. Furthermore, since for $4\mid(n-1)$,

$$|\pi(d_{1})|=|\pi(n-1)|=|\pi(\frac{n-1}{2})|<|\pi(\frac{n-1}{2}\cdot(n-2))|=|\pi(d_{2})|,$$

it follows that $\frac{n-1}{2}$ is odd and that $n-2$ is a prime power.

If $3\mid(n-1)$ then we let $\lambda_{4}=(n-3,3)$ so that $d_{4}:=[\lambda_{4}](1)=\frac{n(n-1)(n-5)}{6}$. It is clear that $n$ and $\frac{n-1}{3}$ are coprime. Since $\frac{n-1}{2}$ and $\frac{n-5}{2}$ are adjacent odd numbers, it follows that $2\cdot\frac{n-1}{2}$ and $\frac{n-5}{2}$ are coprime. Hence $\frac{n-1}{3}$ and $\frac{n-5}{2}$ are coprime. In addition, since $n-(2\cdot\frac{n-5}{2})=5$, we have that $n\cdot\frac{n-5}{2}$ can not be a prime power (since otherwise both $n$ and $\frac{n-5}{2}$ are powers of 5, which is not possible). Therefore,

$$|\pi(n-1)|<|\pi(\frac{n-1}{3}\cdot n\cdot\frac{n-5}{2})|=|\pi(d_{4})|.$$

If $3\mid(n-2)$, then $n-2$ is a power of 3, and so

$$|\pi(n-1)|<|\pi((n-1)\cdot\frac{n-2}{3}\cdot\frac{n-3}{2})|=|\pi(d_{3})|.$$

Finally, if $3\mid(n-3)$, then

$$|\pi(n-1)|<|\pi((n-1)\cdot(n-2)\cdot\frac{n-3}{6})|=|\pi(d_{3})|.$$

This finishes the proof. ∎

In the following we prove Theorem 1.3, starting with a result of Bianchi, Chillag, Lewis and Pacifici.

Lemma 2.1.

Let $N$ be a minimal normal subgroup of $G$ such that $N=S_{1}\times\cdots\times S_{t}$, where $S_{i}\cong S$, a nonabelian simple group. If $\sigma\in\operatorname{Irr}(S)$ extends to ${\rm Aut}(S)$, then $\sigma\times\cdots\times\sigma\in\operatorname{Irr}(N)$ extends to $G$.

Proof.

This is [1, Lemma 5]. ∎

The following results about the decomposition of abelian $p$-groups might be well known. However, we add their proofs for the purpose of completeness.

Lemma 2.2.

Let $P$ be an abelian $p$-group and $x\in P$ of order $p$, and $y\in P$ an element of maximal order such that $\langle y\rangle$ contains $x$. Then $\langle y\rangle$ is a direct factor of $P$.

Proof.

If $P$ has only one subgroup of order $p$, then $P$ is cyclic, so the result trivially holds. So we may assume that $P$ has at least two subgroups of order $p$. Let $Q$ be one of these subgroups different from $\langle x\rangle$. Write $\overline{P}=P/Q$. We have $o(\overline{x})=p$, $o(\overline{y})=o(y)$, and $\langle\overline{y}\rangle=\overline{\langle y\rangle}$. By the inductive hypothesis, $\langle\overline{y}\rangle$ is a direct factor of $\overline{P}$, i.e., $\overline{P}=\langle\overline{y}\rangle\times\overline{H}$ for some $\overline{H}\leq\overline{P}$. Let $H$ be the pre-image of $\overline{H}$ under the natural epimorphism from $P$ to $\overline{P}$, so that $(\langle y\rangle Q)\cap H=Q$. Now $P=\langle y\rangle QH=\langle y\rangle H=\langle y\rangle\times H$, finishing the proof. ∎

Lemma 2.3.

Let $P$ be an abelian $p$-group, $Q$ be a maximal subgroup of $P$, and $a\in P\backslash Q$ be of maximal order among $P\backslash Q$. Then $\langle a\rangle$ is a direct factor of $P$.

Proof.

We first claim that $a$ is of maximal order in the whole group $P$. Assume that $b$ has order greater than $o(a)$. Then we have $b\in Q$. However, it follows that $ab\in P\backslash Q$ with $o(ab)>o(a)$. This contradicts the maximality of the order of $a$ among $P\backslash Q$, proving the claim. Now the lemma follows by the structural theorem of abelian $p$-groups. ∎

Lemma 2.4.

Let $N$ be a normal solvable subgroup of $G$ such that $G/N\cong A_{7}$ or $S_{7}$. If $G$ is a SNPD-group, then $G\cong A\times B$, where $A$ is an abelian group and $B$ satisfies one of the following:

1. $B=A_{7}$ or $S_{7}$.

2. $B$ is the central product of a cyclic $3$-group with $3.A_{7}$.

3. $B$ is the semi-direct product of $A_{7}$ by a cyclic $2$-group $\langle a\rangle$ such that $B/\langle a^{2}\rangle\cong S_{7}$.

Proof.

Let $\theta\in{\rm Irr}(N)$ and $T$ the stabilizer of $\theta$ in $G$.

We first suppose that $G/N\cong A_{7}$. According to the Atlas [2], the maximal subgroups of $A_{7}$ up to isomorphism are $L_{2}(7),S_{5},A_{6}$ or $\left(A_{4}\times C_{3}\right):C_{2}$ of index $3\cdot 5,3\cdot 7,7$ or $5\cdot 7$, respectively (see also Table 2).

Table 2. Maximal subgroups of $A_{7}$ and their indices

Maximal subgroup	$L_{2}(7)$	$S_{5}$	$A_{6}$	$\left(A_{4}\times C_{3}\right):C_{2}$
Index	$3\cdot 5$	$3\cdot 7$	$7$	$5\cdot 7$

(i) Suppose that $T<G$ and that $T$ is not a maximal subgroup of $G$. Let $M$ be a maximal subgroup of $G$ containing $T$. If $M/N\cong L_{2}(7)$ then since $L_{2}(7)$ only has maximal subgroups of index $7$ or $8$, it follows that $|G:T|$ is divisible by at least 3 different prime divisors. The conclusion is also true for $M/N\cong S_{5}$ or $A_{6}$ where $S_{5}$ (resp. $A_{6}$) only has maximal subgroups of index $2,5,2\cdot 3$ or $2\cdot 5$ (resp. $2\cdot 3,2\cdot 5$ or $3\cdot 5$). Now let $M/N\cong\left(A_{4}\times C_{3}\right):C_{2}$. If $2\mid|M:T|$ then $|G:T|$ is divisible by at least 3 different prime divisors. For the case where $2\nmid|M:T|$, we have that $T/N$ is a Sylow 2-subgroup of $M/N$ and is isomorphic to $D_{8}$. So there is $\widetilde{\theta}\in{\rm Irr(T\mid\theta)}$ with $2\theta(1)\mid\widetilde{\theta}(1)$, and thus there is a $\chi\in{\rm Irr(G\mid\theta)}$ such that $\chi(1)$ is divisible by at least 3 different prime divisors. Therefore, in any case $G$ has an irreducible character of degree with at least 3 different prime divisors, contradicting $2\cdot 3\in{\rm cd}(G)$ and that $G$ is a SNPD-group.

(ii) We now suppose that $T$ is a maximal subgroup of $G$. If $T/N\cong L_{2}(7)$, then using the Atlas [2], either $\operatorname{cd}(T\mid\theta)=\{1,3,2\cdot\left.3,7,2^{3}\right\}\cdot\theta(1)$ or $\left\{2^{2},2\cdot 3,2^{3}\right\}\cdot\theta(1)$ (where $\left\{2^{2},2\cdot 3,2^{3}\right\}\cdot\theta(1):=\left\{2^{2}\theta(1),2\cdot 3\theta(1),2^{3}\theta(1)\right\}$). Both cases implies that $G$ is not a SNPD-group, a contradiction.

If $T/N\cong S_{5}$, we have either $\operatorname{cd}(T\mid\theta)=\left\{1,2^{2},5,2\cdot 3\right\}\cdot\theta(1)$ or $\left\{2^{2},2\cdot 3\right\}\cdot\theta(1)$ by the Atlas [2]. Also, both cases imply that $G$ is not a SNPD-group, a contradiction.

If $T/N\cong A_{6}$, then $\operatorname{cd}(T\mid\theta)$ is one of the four sets $\left\{1,5,2^{3},3^{2},2\cdot 5\right\}\cdot\theta(1),\left\{2^{2},2^{3},2\cdot 5\right\}\cdot\theta(1),\left\{3,2\cdot 3,3^{2},3\cdot 5\right\}\cdot\theta(1)$ or $\left\{2\cdot 3,2^{2}\cdot 3\right\}\cdot\theta(1)$ by the Atlas [3]. The former three cases obviously lead to a contradiction. For the last case, since $|G:T|=7$ it follows that $G$ has an irreducible character of degree with at least 3 different prime divisors, contradicting $2\cdot 3\in{\rm cd}(G)$ and that $G$ is a SNPD-group.

Finally, if $T/N\cong\left(A_{4}\times C_{3}\right):C_{2}$, then $|G:T|=5\cdot 7$ and $T$ contains a normal subgroup $N_{1}$ of $G$ such that $T/N_{1}\cong S_{4}$. Consider a character $\widetilde{\theta}\in\operatorname{Irr}\left(N_{1}\mid\theta\right)$. Then $\operatorname{cd}(T\mid\widetilde{\theta})=\{1,2,3\}\cdot\widetilde{\theta}(1)$ or $\left\{2,2^{2}\right\}\cdot\widetilde{\theta}(1)$. In any case, $G$ has an irreducible character of degree with at least 3 different prime divisors, contradicting $2\cdot 3\in{\rm cd}(G)$ and that $G$ is a SNPD-group.

Now we have proved that all irreducible characters of $N$ are $G$-invariant. According to the Atlas [2], the Schur multiplier of $A_{7}$ is $6$ and ${\rm cd}(G)$ contains the subset

$$\{1,2\cdot 3,2\cdot 5,2\cdot 7,3\cdot 5,3\cdot 7,5\cdot 7\}\cdot\theta(1),\{2^{2},2\cdot 7,2^{2}\cdot 5,2^{2}\cdot 3^{2}\}\cdot\theta(1),$$

$$\{2\cdot 3,3\cdot 5,3\cdot 7,2^{3}\cdot 3\}\cdot\theta(1),{\rm or}\ \{2\cdot 3,2^{3}\cdot 3,2^{2}\cdot 3^{2}\}\cdot\theta(1).$$

Since $G$ is a SNPD-group, we deduce that the second possibility above can not occur, and so $2.A_{7}$ and $6.A_{7}$ is not a quotient group of $G$. It is clear that all irreducible characters of $N$ have degree $1$, i.e., $N$ is abelian. In addition, by a theorem of Brauer [4, Theorem 6.32], we have $N\leq Z(G)$. Now, let $E:=[G,G]$. Then $E\cong A_{7}$ or $3.A_{7}$. If the former case occurs, then $G\cong A_{7}\times N$. For the latter case, we have $G\cong EP\times H$, where $H$ is a 3-complement of $N$, $P$ is a Sylow 3-subgroup of $N$, and $EP$ is the central product of $E$ and $P$ with $|E\cap P|=3$. Let $y\in P$ be an element of maximal order such that $\langle y\rangle$ contains $E\cap P$. By Lemma 2.2, $P=\langle y\rangle\times R$ for some subgroup $R$ of $P$. Therefore,

$$G=EP\times H=E(\langle y\rangle\times R)\times H=(E\langle y\rangle)\times(R\times H),$$

and thus (2) holds (with $A=R\times H$ and $B=E\langle y\rangle$).

Finally, we suppose that $G/N\cong S_{7}$. The character degrees of $S_{7},2.S_{7},3.S_{7},6.S_{7}$ can be found in the Atlas [2], see also Table 3.

Table 3. Character degrees of $S_{7}$ and its central extensions

Group	Character dergees
$S_{7}$	$1,2\cdot 3,2\cdot 7,3\cdot 5,2^{2}\cdot 5,3\cdot 7,5\cdot 7$
$2.S_{7}$	$1,2\cdot 3,2^{3},2\cdot 7,3\cdot 5,2^{2}\cdot 5,3\cdot 7,2^{2}\cdot 7,5\cdot 7,2^{2}\cdot 3^{2}$
$3.S_{7}$	$1,2\cdot 3,2^{2}\cdot 3,2\cdot 7,3\cdot 5,2^{2}\cdot 5,3\cdot 7,2\cdot 3\cdot 5,5\cdot 7,2\cdot 3\cdot 7,2^{4}\cdot 3$
$6.S_{7}$	$1,2\cdot 3,2^{3},2^{2}\cdot 3,2\cdot 7,3\cdot 5,2^{2}\cdot 5,3\cdot 7,2^{2}\cdot 7,2\cdot 3\cdot 5,5\cdot 7,2^{2}\cdot 3^{2},2\cdot 3\cdot 7,2^{4}\cdot 3,2^{3}\cdot 3^{2}$

Let $M/N$ be the socle of $G/N$ so that $M/N\cong A_{7}$ and $|G:M|=2$. With the previous conclusion on $\operatorname{cd}(M\mid\theta)$ and Clifford’s theory, it is easy to see that $\theta$ is $M$-invariant, and so $\theta(1)=1$ and all irreducible characters of $N$ are $M$-invariant. Hence $N\leq Z(M)$. Since $\{2\cdot 3,2^{3}\}\subset{\rm cd}(2.S_{7})\cap{\rm cd}(6.S_{7})$ and $\{2\cdot 3,2\cdot 3\cdot 5\}\subset{\rm cd}(3.S_{7})$, we see that all non-trivial central extensions of $S_{7}$ are not quotient groups of $G$. This implies that $M\cong A_{7}\times N$.

Let $L\lhd G$ be such that $L\cong A_{7}$ and $L<L_{1}\lhd G$ such that $L_{1}\cong S_{7}$. If $G/L$ is not abelian, then we have $2\in{\rm cd}(G)$. But it then follows that $G$ is not a SNPD-group, a contradiction. Hence $G/L$ is abelian, and so $N\leq Z(G)$. Let $N_{0}$ be the unique Sylow $2$-subgroup of $N$, $N_{2^{\prime}}$ the unique 2-complement of $N$, and $\widehat{N_{0}}$ be the subgroup of $G$ such that $G=L_{1}N=L\widehat{N_{0}}\times N_{2^{\prime}}$ and $|\widehat{N_{0}}:N_{0}|=2$. In particular, $N_{0}$ is a maximal subgroup of $\widehat{N_{0}}$. Let $a\in\widehat{N_{0}}\backslash N_{0}$ be of maximal order among $\widehat{N_{0}}\backslash N_{0}$. By Lemma 2.3, $\langle a\rangle$ is a direct factor of $\widehat{N_{0}}$, say $\widehat{N_{0}}=\langle a\rangle\times N_{a}$ with $N_{a}\leq\widehat{N_{0}}$. Now we have

$$G=L\widehat{N_{0}}\times N_{2^{\prime}}=L(\langle a\rangle\times N_{a})\times N_{2^{\prime}}=L\langle a\rangle\times(N_{a}\times N_{2^{\prime}}),$$

and thus (3) holds (with $A=N_{a}\times N_{2^{\prime}}$ and $B=L\langle a\rangle$). This finishes the proof. ∎

Finally, we prove Theorem 1.3.

Proof of Theorem 1.3.

The “if” part follows from Table 1 and the facts that ${\rm cd}(B)={\rm cd}(S_{7})$ (resp. ${\rm cd}(3.A_{7})=\{1,6,10,14,15,21,24,35\}$) if Theorem 1.3 (3) (resp. (2)) occurs. For the “only if” part, we let $N$ be the largest normal solvable subgroup of $G$ and $M/N$ a nonabelian chief factor of $G$. Then $M/N\cong S_{1}\times\cdots\times S_{t}$, where $S_{i}\cong S$, a nonabelian simple group. Notice that ${\rm cd}(G/N)\subseteq{\rm cd}(G)$.

We first suppose that $S$ is a sporadic simple group or the Tits simple group. By the Atlas [2] or the GAP library [14], $S$ has two ${\rm Aut}(S)$-extendible irreducible characters of degrees as listed in Table 4. By Lemma 2.1, it is easy to see that $G$ is not a SNPD-group, a contradiction.

Table 4. Degrees of the sporadic simple groups and the Tits group

Group	Chars.	Degrees			Group	Chars.	Degrees
$M_{11}$	$\chi_{5}$	$11$	$=$	$11$	$Fi_{22}$	$\chi_{2}$	78	$=$	$2\cdot 3\cdot 13$
	$\chi_{10}$	$55$	$=$	$5\cdot 11$		$\chi_{6}$	$3,003$	$=$	$3\cdot 7\cdot 11\cdot 13$
$M_{12}$	$\chi_{2}$	11	$=$	11	$Fi_{23}$	$\chi_{2}$	782	$=$	$2\cdot 17\cdot 23$
	$\chi_{11}$	66	$=$	$2\cdot 3\cdot 11$		$\chi_{5}$	$25,806$	$=$	$2\cdot 3\cdot 11\cdot 17\cdot 23$
$M_{22}$	$\chi_{2}$	$21$	$=$	$3\cdot 7$	$Fi_{24}^{\prime}$	$\chi_{2}$	$8,671$	$=$	$13\cdot 23\cdot 29$
	$\chi_{8}$	$210$	$=$	$2\cdot 3\cdot 5\cdot 7$		$\chi_{4}$	$249,458$	$=$	$2\cdot 11\cdot 17\cdot 23\cdot 29$
$M_{23}$	$\chi_{2}$	$22$	$=$	$2\cdot 11$	$McL$	$\chi_{2}$	22	$=$	$2\cdot 11$
	$\chi_{10}$	$770$	$=$	$2\cdot 5\cdot 7\cdot 11$		$\chi_{5}$	770	$=$	$2\cdot 5\cdot 7\cdot 11$
$M_{24}$	$\chi_{2}$	$23$	$=$	$23$	$He$	$\chi_{2}$	51	$=$	$3\cdot 17$
	$\chi_{9}$	$483$	$=$	$3\cdot 7\cdot 23$		$\chi_{1}3$	$4,080$	$=$	$2^{4}\cdot 3\cdot 5\cdot 17$
$J_{1}$	$\chi_{6}$	77	$=$	$7\cdot 11$	$Ru$	$\chi_{2}$	378	$=$	$2\cdot 3^{3}\cdot 7$
	$\chi_{9}$	120	$=$	$2^{3}\cdot 3\cdot 5$		$\chi_{6}$	$3,276$	$=$	$2^{2}\cdot 3^{2}\cdot 7\cdot 13$
$J_{2}$	$\chi_{2}$	14	$=$	$2\cdot 7$	$Suz$	$\chi_{2}$	143	$=$	$11\cdot 13$
	$\chi_{8}$	70	$=$	$2\cdot 5\cdot 7$		$\chi_{4}$	780	$=$	$2^{2}\cdot 3\cdot 5\cdot 13$
$J_{3}$	$\chi_{6}$	324	$=$	$2^{2}\cdot 3^{4}$	$O^{\prime}N$	$\chi_{3}$	$13,376$	$=$	$2^{6}\cdot 11\cdot 19$
	$\chi_{10}$	1,140	$=$	$2^{2}\cdot 3\cdot 5\cdot 19$		$\chi_{11}$	$52,668$	$=$	$2^{2}\cdot 3^{2}\cdot 7\cdot 11\cdot 19$
$J_{4}$	$\chi_{2}$	1,333	$=$	$31\cdot 43$	$HN$	$\chi_{2}$	133	$=$	$7\cdot 19$
	$\chi_{4}$	$299,367$	$=$	$3^{2}\cdot 29\cdot 31\cdot 37$		$\chi_{6}$	$8,778$	$=$	$2\cdot 3\cdot 7\cdot 11\cdot 19$
$Co_{1}$	$\chi_{3}$	299	$=$	$13\cdot 23$	$Ly$	$\chi_{2}$	$2,480$	$=$	$2^{4}\cdot 5\cdot 31$
	$\chi_{5}$	8,855	$=$	$5\cdot 7\cdot 11\cdot 23$		$\chi_{5}$	$48,174$	$=$	$2\cdot 3\cdot 7\cdot 31\cdot 37$
$Co_{2}$	$\chi_{2}$	23	$=$	23	$Th$	$\chi_{2}$	248	$=$	$2^{3}\cdot 31$
	$\chi_{5}$	$1,771$	$=$	$7\cdot 11\cdot 23$		$\chi_{6}$	$30,628$	$=$	$2^{2}\cdot 13\cdot 19\cdot 31$
$Co_{3}$	$\chi_{2}$	23	$=$	23	$B$	$\chi_{2}$	$4,371$	$=$	$3\cdot 31\cdot 47$
	$\chi_{8}$	1,771	$=$	$7\cdot 11\cdot 23$		$\chi_{4}$	$1,139,374$	$=$	$2\cdot 17\cdot 23\cdot 31\cdot 47$
$HS$	$\chi_{2}$	22	$=$	$2\cdot 11$	$M$	$\chi_{2}$	$196,883$	$=$	$47\cdot 59\cdot 71$
	$\chi_{10}$	770	$=$	$2\cdot 5\cdot 7\cdot 11$		$\chi_{3}$	$21,296,876$	$=$	$2^{2}\cdot 31\cdot 41\cdot 59\cdot 71$
${}^{2}F_{4}(2)^{\prime}$	$\chi_{7}$	$300$	$=$	$3\cdot 5^{2}\cdot 2^{2}$
	$\chi_{15}$	$675$	$=$	$3^{3}\cdot 5^{2}$

We now suppose that $S$ is a simple group of Lie type. The Steinberg character $St$ of $S$ has prime power degree and extends to ${\rm Aut}(S)$. By Lemma 2.1, $St(1)^{t}\in{\rm cd}(G)$. So, if $G$ is a SNPD-group then all complex irreducible characters of $G$ have prime power degrees. By [15, Proposition B], $G$ is isomorphic to $L_{2}(4)$ or $L_{2}(8)$ up to an abelian direct factor.

Finally, we suppose that $S=A_{n}$ with $n\geq 5$. Since $A_{5}\cong L_{2}(4)$ and $A_{6}\cong L_{2}(9)$, we may assume that $n\geq 7$. With the notation in the proof of Theorem 1.2, we have that the chosen characters are all extendible to ${\rm Aut}(A_{n})=S_{n}$ for $n>7$. By Lemma 2.1, we conclude that $\{(d_{1})^{t},(d_{2})^{t},(d_{3})^{t},(d_{4})^{t}\}\subset{\rm cd(G)}$. Therefore, for $n>7$, the proof of Theorem 1.2 shows that $G$ is not a SNPD-group, a contradiction. Let $n=7$. If $t>1$, then $M/N$ has an irreducible character of degree divisible by $2\cdot 3\cdot 5\cdot 7$. By Clifford’s theory, $G$ has an irreducible character of degree $\chi(1)$ divisible by $2\cdot 3\cdot 5\cdot 7$. Clearly, $2=|\pi((d_{1})^{t})|<|\pi(\chi(1))|$, and so $G$ is not a SNPD-group, a contradiction. Hence $t=1$, and so $G/N\cong A_{7}$ or $S_{7}$. Now Lemma 2.4 applies. ∎

Declaration of competing interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.