Finite groups for which all proper subgroups have consecutive character degrees

1.   Introduction

Throughout this paper, all considered groups are finite. Denote the set of all complex irreducible characters of a group $G$ by $\mathrm{Irr}(G)$, denote the linear characters of $G$ by $\mathrm{Lin}(G)$, and write $\mathrm{cd}(G)$ for the set of all complex irreducible character degrees of $G$, i.e.,

We use usually the degrees of $G$ instead of the more precise irreducible character degrees.

Huppert determined the structure of finite groups with consecutive character degrees; see Section 32 of ^[14]. In order to argue in short, we give a conception.

Definition 1.1. Let $G$ be a finite group. If the degrees from $\mathrm{cd}(G)-\{1\}$ are consecutive, then we call $G$ a $\mathrm{CCD}$-group (or a non-$\mathrm{CCD}$-group otherwise). We also assume that an abelian group is a $\mathrm{CCD}$-group.

Qian in ^[26] improved the result of Huppert ^[14] and considered $\mathrm{CCD}$-groups.

Let $d_i(G)$ or $d_i$ without confusion be the smallest $i$-th degree in $\mathrm{cd}(G)-\{1\}$. Tiep and Zalesskii in ^[32] determined the first three smallest character degrees of classical groups. Recently, Liu, Lei and Li in ^[21] considered non-solvable $\mathrm{CCD}$-groups by using the result of ^[32].

On the other hand, some authors considered the influence of subgroups with certain properties on the structure of finite groups. In particular, the structures of finite groups are determined if the proper subgroups are all abelian ^[1] or solvable ^[27,31] or supersolvable ^[6] or nilpotent ^[18] or minimal non-nilpotent ^[2].

Recently some scholars combined the representation theory of finite groups with subgroup properties. In particular, the structures of finite groups are determined if all their subgroups have prime power degrees or degrees divisible by two primes; see ^[20,22].

In this paper, we consider the influence of proper (maximal) subgroups whose degrees are consecutive on the structure of finite groups. Let $A*B$ be the central product of two groups $A$ and $B$. To argue in short, we begin with the following definition.

Definition 1.2. Let $\sum(G)$ be the set of all proper subgroups of a group $G$. If for each $M\in\sum(G)$, $M$ is a $\mathrm{CCD}$-group, then we call $G$ an $\mathrm{SCCD}$-group (or a non-$\mathrm{SCCD}$-group otherwise). We always assume that an abelian group is an $\mathrm{SCCD}$-group.

Theorem 1.3. Let $G$ be a non-solvable $\mathrm{SCCD}$-group. Then, $G$ is isomorphic to one of the following groups:

(1) $\mathrm{PSL}_2(q)$ where $q\geq4$ is even or $q\geq5$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

(2) $S\times H$ where $S\in\{A_5, S_5\}$ and $H$ is abelian.

(3) $\mathrm{SL}_2(q)$ where $q\geq7$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

(4) $\mathrm{SL}_2(5)*H$ with $H/C_2$ abelian.

We call a group $G$ a $\mathrm{CC}$-group if $\mathrm{cd}(G) = \{1, 2, 3, \cdots, k\}$. Huppert investigated the structure of finite $\mathrm{CC}$-groups $G$; see Sec. 32 of ^[14]. Let $m$ be a positive integer. Then, what can we see if for each $H\in\sum G$, $\mathrm{cd}(H) = \{1, 2, 3, \cdots, m\}$? We also call a group $G$ an $\mathrm{SCC}$-group if for each $H\in\sum G$, $H$ is a $\mathrm{CC}$-group. As an application of Theorem 1.3, we have the following result.

Theorem 1.4. Let $G$ be an $\mathrm{SCC}$-group. Then, $G$ is solvable. In particular, one of the following statements holds:

(i) $\mathrm{cd}(G)\subseteq \{1, 2\}$.

(ii) $G$ is nilpotent, and $G$ is isomorphic to a metabelian $p$-group or a direct product of a 2-group $P$ and $2'$-group $H$ where $\mathrm{cd}(P) = \{1, 2\}$ and $\mathrm{cd}(H) = \{1\}$.

(iii) $G$ is a Frobenius group with kernel $F$ and complement $H$, respectively, such that

(iii.a) $\mathrm{cd}(F) = \{1, 2\}$ and $H$ is of a prime order such that for each non-identity $W\in\sum G$, $|W|\not\equiv1\pmod{|H|}$.

(iii.b) $H$ is of prime order or 4; $F$ is abelian. Furthermore, if $H$ is of order 4, then $F$ is isomorphic to $C_p\times C_q$ with $q\equiv p\equiv-1\pmod{4}$ or $C_p$ with $p\equiv1\pmod{4}$; if $H$ is of prime order $> 2$, then $|H|\|(|F|-1)$.

(iv) $G$ is a direct product of two groups $H_1$ and $H_2$ where $H_1$ and $H_2$ are metabelian with $\mathrm{cd}(H_i) = \{1, 2\}$, $i = 1, 2$; or only one of $H_1$ and $H_2$ is metabelian, say, $H$ with $\mathrm{cd}(H) = \{1, 2\}$.

The rest of the paper is organized as follows. In Section 2, we assemble some basic results which will be used throughout this paper. In Section 3, we determine the structure of the simple $\mathrm{CCD}$-groups. In Section 4, we classify the non-solvable $\mathrm{SCCD}$-groups. In Section 5, we give an application of Theorem 1.3 and also determine the structure of an $\mathrm{SCC}$-group.

Let $A:B$ or $A.B$ be an extension of a group $A$ by a group $B$. Let $\mathrm{GL}_n(q)$ be the general linear group of dimension $n$ over a finite field of order $q$. Let $m, n$ be positive integers, denote by $m\|(n-1)$ if $m|(n-1)$ but $m\nmid (\frac{n}{p}-1)$ for all nontrivial divisors $p$ of $n$. The other notation is standard; please see ^[7,15], for instance.

2.   Basic results

In this section, some needed results are shown. First, some information about Frobenius groups is given.

Lemma 2.1 (Theorems 13.3 and 13.8 of ^[9]). Let $G = K:H$ be a Frobenius group with kernel $K$ and complement $H$. Then,

(i) $|H|\bigl|(|K|-1)$.

(ii) Any subgroup of $H$ of order $p^2$ or $pq$ is cyclic where $p$ and $q$ are primes.

(iii) If $|H|$ is even, $K$ is abelian.

(iv) In any case, $K$ is nilpotent.

(v) Assume that $K$ has $h(K)$ conjugacy classes, $H$ has $h(H)$ conjugacy classes, and $G$ has characters with the following forms:

(a) $h(H)$ irreducible characters $\chi_1$, $\cdots$, $\chi_{h(H)}$ with $K$ in their kernel. If $\mu_1$, $\cdots$, $\mu_{h(H)}$ are the irreducible characters of $H$, then these satisfy

(b) Whenever $\tau\neq1_K$ is an irreducible character of $K$, then $\tau^G$ is an irreducible character of $G$. This gives $[h(K)-1]/|H|$ irreducible characters of $G$ with $K$ not in their kernel. Such $\tau^G$ satisfies

where $\rho_H$ is the regular character of $H$.

Related to Frobenius groups, we need some character degrees for special cases to determine the group structure. From [19,Table 1] and [24,Theorem] and using the notation of ^[3], we get that $^2B_2(q)$ with $q = 2^{2m+1}$ and $^2G_2(q)$ with $q = 3^{2m+1}$ have maximal subgroups of the form $K:C_{q-1}$, where $K$ is a Sylow 2 or 3-subgroup of $^2B_2(q)$ or $^2G_2(q)$, respectively. Now, we have the following result by using the above Lemma.

Lemma 2.2. Let $G\in\{^2B_2(q)$ with $q = 2^{2m+1}\geq2^3$, $^2G_2(q)$ with $q = 3^{2m+1}\geq3^3\}$ and let $K$ be a Sylow $\pi(q)$-subgroup of $G$. Then, $G$ has a maximal subgroup of the form $K:C_{q-1}$ that is a non-$\mathrm{CCD}$-group.

Proof. The first claim follows from above arguments.

Now, we prove that $K:C_{q-1}$ is a non-$\mathrm{CCD}$-group. Let $H: = C_{q-1}$. Now, $K:H$ is a Frobenius group. So, we need to prove that $K:H$ is not a $\mathrm{CCD}$-group. Assume the contrary, and then $K:H$ is a $\mathrm{CCD}$-group.

To get the desired result, we consider two cases in what follows.

Case 1: $G = ^2B_2(q)$.

Let $\tau\in \mathrm{Irr}(K)- \mathrm{Lin}(K)$. Note that $K$ is a Sylow 2-subgroup of $^2B_2(q)$ since $|K| = |^2B_2(q)|_2 = q^2$. We see that $Z(K)$ is of order $q$, so $K/Z(K)$ of order $q$ is abelian.

For any $1\neq d\in \mathrm{cd}(K)$, we have that $d\mid2^{m}$. In fact, $|K/Z(K)| = q$, so for each $d\in \mathrm{cd}(K)$, one has $d\bigl|\sqrt{q} = 2^{\frac{2m+1}{2}}$, i.e., $d\bigl| 2^m$.

On the other hand, $G/K\cong H$ is cyclic, so $(q-1)\in \mathrm{cd}(G)$, as the cyclic group $H$ acts faithfully on the nilpotent group $K$.

Lemma 2.1(v)(b) implies that $\mathrm{cd}(K:H) = \{1, d, q-1, d(q-1)\}$ with $d\neq1$. Hypothesis and $d\bigl|2^m$ show that $d+2 = d(q-1)$, i.e., $d = \frac{2}{q-2}$, so $1\leq q-2\leq 2$ implies $q = 4\ngeqq8$, a contradiction to the hypothesis.

From now on, we have proved that $K:H$ is a non-$\mathrm{CCD}$-group when $G = ^2B_2(q)$.

Case 2: $G = ^2G_2(q)$.

Similarly, we can get that $K$ is a Sylow 3-subgroup of $^2G_2(q)$) since $|K| = |^2G_2(q)|_3 = q^3$. By a Theorem of ^[34], one has that the center $Z(K)$ of $K$ is of order $q$.

Let $d\in \mathrm{cd}(K)- \mathrm{Lin}(K)$. Then, $d\bigl|q$, and in particular, $d\leq q$. Note that $(K:H)/K\cong H$ is a cyclic group, so $(q-1)\in \mathrm{cd}(G)$, as the cyclic group $H$ acts faithfully on the nilpotent group $K$. Now, Lemma 2.1(v)(b) implies that $\mathrm{cd}(K:H) = \{1, q-1, d, d(q-1)\}$. Hypothesis forces that $q(q-1)-1 = q$ for $d = q$ or $d+2 = d(q-1)$ for $d\bigl|q$, so $q = 3\ngeqq3^3$, a contradiction. Therefore, $K:H$ is a non-$\mathrm{CCD}$-group when $G = ^2G_2(q)$.

Now, the lemma is proved.

Let $\max G$ or $\max (G)$ be the set of all proper maximal subgroups with respect to their subgroup-order divisibility. Then, $\max G\subseteq \sum G$.

Lemma 2.3. (1) $G$ is an $\mathrm{SCCD}$-group if and only if every proper subgroup $H$ of $G$ is a $\mathrm{CCD}$-group.

(2) Let $N$ be a non-trivial normal subgroup of an $\mathrm{SCCD}$-group $G$. Then, $G/N$ is a $\mathrm{CCD}$-group.

Let $\Phi(G)$ be the Frattini subgroup of a group $G$.

Proof. (1) By Definition of an $\mathrm{SCCD}$-group, we can get this.

(2) Since $N$ is non-trivial and normal in $G$, we assume that $N$ is minimal. Assume that $G/N$ is a non-$\mathrm{CCD}$-group. If $N\nleq\Phi(G)$, then choose a maximal subgroup $M\in\max G$ with $N\nleq M$, so $G = MN$. Note that $M\cap N = N$ or $M\cap N = 1$. If the former, then $N\leq M$, a contradiction to the choice of $M$. Thus, $G/N$ is isomorphic to $M$, which is a $\mathrm{CCD}$-group, a contradiction also. If $N\leq \Phi(G)$, then for $M\in\max G$, $N\leq M$. Since $G/N$ is a non-$\mathrm{CCD}$-group, we have that $G/N$ is a non-$\mathrm{SCCD}$-group, and so in $G/N$ there is a non-$\mathrm{CCD}$-subgroup, say, $M/N$. It follows that $M$ is a non-$\mathrm{CCD}$-group, a contradiction to $M\in\max G$ (note that for each $M\in\max G$, $M$ is a $\mathrm{CCD}$-group).

In general, the product of two $\mathrm{CCD}$-groups is not a an $\mathrm{CCD}$-group.

Example 2.4. Let $A_1$ be any maximal proper subgroup of a non-abelian $p$-group $A$ of order $p^4$ with $\mathrm{cd}(A_1) = \{1, p\}$, and let $B$ be a $q$-group with $\mathrm{cd}(B) = \{1, q\}$, where $q, p$ are different primes. Let $G = A\times B$. Then, $G$ is not an $\mathrm{SCCD}$-group. We will show this by contradiction. Assume that $G$ is an $\mathrm{SCCD}$-group. Then, $A_1\times B$ is a maximal subgroup of $G$, and $\mathrm{cd}(A_1\times B) = \{1, p, q, pq\}$ by Theorem 4.21 of ^[15]. Say $p < q$. Then, $p+1 = q$ by assumption, so $p = 2$, and $q = 3$. Thus, $\mathrm{cd}(A_1\times B) = \{1, 2, 3, 6\}$, and $A_1\times B$ is not a $\mathrm{CCD}$-group, a contradiction.

There is a $\mathrm{CCD}$-group that is not an $\mathrm{SCCD}$-group.

Example 2.5. Let $G = \mathrm{PGL}_2(q)$ with $q > 5$ odd. Then, $\mathrm{cd}(G) = \{1, q-1, q, q+1\}$ by ^[30], so $G$ is a $\mathrm{CCD}$-group. Obviously, $\mathrm{PSL}_2(q)$ is a normal subgroup of $\mathrm{PGL}_2(q)$ with index $\gcd (2, q-1)$, and

If $G$ is an $\mathrm{SCCD}$-group, then $\mathrm{PSL}_2(q)$ is a $\mathrm{CCD}$-group. On the other hand, hypothesis forces $\frac{q+(-1)^{(q-1)/2}}{2}+1 = q-1$ or $1+1 = \frac{q+(-1)^{(q-1)/2}}{2}$, so the two equations have solutions $q = 3, 5$, a contradiction to the hypothesis. It follows that $\mathrm{PSL}_2(q)$ is not a $\mathrm{CCD}$-group, so $G$ is not an $\mathrm{SCCD}$-group.

An extension of an $\mathrm{SCCD}$-group by an $\mathrm{SCCD}$-group is not an $\mathrm{SCCD}$-group usually.

Example 2.6. Let $S_5$ be the symmetric group of degree 5. Then, $S_5$ is an $\mathrm{SCCD}$-group (we can check it by ^[7]). Also, $2^4$, an elementary group of order $2^4$, is an $\mathrm{SCCD}$-group. Now, let $G = 2^4:S_5$. It is easy to see that $2^4:A_5$, say, $H$, is a proper subgroup of $G$. If $G$ is an $\mathrm{SCCD}$-group, then $H$ is a $\mathrm{CCD}$-group by Lemma 2.3. On the other hand, by ^[4] $\mathrm{cd}(H) = \{1, 3, 4, 5, 15\}$, which means that $H$ is not a $\mathrm{CCD}$-group. It follows that $G$ is not an $\mathrm{SCCD}$-group.

Let $\mathrm{PSL}_n^+(q) = \mathrm{PSL}_n(q)$ and $\mathrm{PSL}_n^-(q) = \mathrm{PSU}_n(q)$. Let $\mathrm{SL}_n^+(q) = \mathrm{SL}_n(q)$ and $\mathrm{SL}_n^-(q) = \mathrm{SU}_n(q)$. In order to prove our main result, we need some information about the structure of some special subgroup of a non-abelian simple group due to Liu ^[20].

Lemma 2.7 (Lemma 2 of ^[20]).Let $q$ be a prime power, and let $\epsilon = \pm$.

(1) Let $n\geq8$. Then, $A_n$ has a subgroup $A_{n-1}$.

(2) Let $n\geq4$. Then, $\mathrm{PSL}_n^\epsilon(q)$ has a subgroup isomorphic to $\mathrm{SL}_{n-1}^\pm(q)$ or $\mathrm{PSL}_{n-1}^\pm(q)$, and $\mathrm{SL}_n^\epsilon(q)$ has a subgroup of the form $\mathrm{SL}_{n-1}^\epsilon(q)$.

(3) Let $n\geq2$. Then, $\mathrm{PSp}_{2n}(q)$ has a subgroup $\mathrm{PSp}_{2(n-1)}(q)$.

(4) Let $n\geq3$, with $q$ odd. Then, $\Omega_{2n+1}(q)$ contains a subgroup $\Omega_{2n-1}(q)$.

(5) Let $n\geq4$. Then, $\mathrm{P}\Omega_{2n}^\epsilon(q)$ has a subgroup $\mathrm{P}\Omega_{2n-2}^{\epsilon}(q)$ with $q$ odd or $\mathrm{PSp}_{2n-2}(q)$ with $q$ even.

3.   Simple $\mathrm{SCCD}$-groups

By the classification theorem of finite simple groups, a non-abelian simple group is isomorphic to an alternating group $A_n$, a Sporadic simple group or Tits group, an exceptional group of Lie type or a classical group of Lie type. Now, we will determine the structure of the simple $\mathrm{SCCD}$-groups by the following Lemmas.

Lemma 3.1. Let $G$ be an alternating $A_n$ of degree $n\geq5$. Assume that $G$ is an $\mathrm{SCCD}$-group, and then $G$ is isomorphic to $A_5$ or $A_6$.

Proof. We assume that $G$ is an $\mathrm{SCCD}$-group, so we need to check whether its subgroups are $\mathrm{CCD}$-groups or not.

Let $n = 5$. Then, by [7,pp. 2], $\max A_5 = \{A_4, D_{10}, S_3\}$. Note that $\mathrm{cd}(A_4) = \{1, 3\}$, $\mathrm{cd}(D_{10}) = \mathrm{cd}(S_3) = \{1, 2\}$. It follows that all the subgroups of $A_5$ are $\mathrm{CCD}$-groups and $A_5$ is an $\mathrm{SCCD}$-group, so $G\cong A_5$.

Let $n = 6$, and then by [7,pp. 4], $\max A_6 = \{A_5, 3^2:4, S_4\}$. As $\mathrm{cd}(A_5) = \{1, 3, 4, 5\}$, $\mathrm{cd}(3^2:4) = \{1, 4\}$ and $\mathrm{cd}(S_4) = \{1, 2, 3\}$, one has that the subgroups $A_5, 3^2:4, S_4$ of $A_6$ are $\mathrm{CCD}$-groups, and so $A_6$ is an $\mathrm{SCCD}$-group. Thus, $G\cong A_6$.

Let $n = 7$, and then by [7,pp. 10], $A_7$ contains a subgroup $A_6$, but by [7,pp. 4] $\mathrm{cd}(A_6) = \{1, 5, 8, 9, 10\}$. Now, $A_6$ is not a $\mathrm{CCD}$-group. It follows that $A_7$ is a non-$\mathrm{SCCD}$-group.

If $n\geq 8$, then we get that $A_n$ has a subgroup $A_{n-1}$ by Lemma 2.7, so one has the subgroup series

It follows that $A_n$ with $n\geq8$ is not an $\mathrm{SCCD}$-group since $A_7$ is a non-$\mathrm{CCD}$-group.

Lemma 3.2. There does not exist an $\mathrm{SCCD}$-group for a Sporadic simple group.

Proof. By ^[7], there is a subgroup $H\in\max G$ which is a non-$\mathrm{CCD}$-group.

Lemma 3.3. There does not exist a simple exceptional $\mathrm{SCCD}$-group of Lie type.

Proof. It is well-known in this case that, for a simple group $G$ of exceptional Lie type, $G$ is isomorphic to $^2B_2(q)$ with $q = 2^{2m+1}\geq8$, $^2G_2(q)$ with $q = 3^{2m+1}$, $m\geq1$, $G_2(q)$, $^3D_4(q)$, $F_4(q)$, $^2F_4(q^2)$, $E_6^\epsilon(q)$, $E_7(q)$, or $E_8(q)$. We deal with these case by case.

The following two subgroup series are from Table 1:

and

However, by ^[23], one has

and hypothesis $d_2+1 = d_3$ implies $q = 2$ or $4$, a contradiction. It follows that $G_2(q)$ and $^2F_2(q^2)$ are non-$\mathrm{SCCD}$-groups. Note that $^2F_4(2)'$ is a non-$\mathrm{SCCD}$-group by ^[7]. Also, we can show from Table 1 that $^3D_4(q)$, $F_4(q)$, and $E_6^\epsilon(q)$ are non-$\mathrm{SCCD}$-groups. Now, if $\mathrm{P\Omega}_{12}^+(q)$ is a non-$\mathrm{CCD}$-group, then $E_7(q)$ and $E_8(q)$ are non-$\mathrm{SCCD}$-groups. In fact,

Now, by Theorem 4.21 of ^[15], $\mathrm{PSL}_2(q)\times \mathrm{PSL}_2(q)$ is not a $\mathrm{CCD}$-group, so $\mathrm{P\Omega}_{12}^+(q)$ is a non-$\mathrm{CCD}$-group. Thus, $E_7(q)$ and $E_8(q)$ are non-$\mathrm{SCCD}$-groups.

Now, by Lemma 2.2, $^2B_2(q)$ with $q = 2^{2m+1}\geq8$ and $^2G_2(q)$ with $q = 3^{2m+1}$, $m\geq1$, are non-$\mathrm{SCCD}$-groups.

Proposition 3.4. Let $G$ be a simple classical $\mathrm{SCCD}$-group of Lie type. Then, $G$ is isomorphic to $\mathrm{PSL}_2(q)$ where $q$ is even, or $q$ is an odd prime, or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

Proof. The Classification Theorem of Finite Simple Groups shows that $G$ may be isomorphic to one of the following groups: $\mathrm{PSL}_{l+1}(q)$, $\mathrm{PSU}_{l+1}(q)$, $\Omega_{2l+1}(q)$, $\mathrm{PSp}_{2l}(q)$, $\mathrm{P\Omega}_{2l}^+(q)$ or $\mathrm{P\Omega}_{2l}^-(q)$, where $q$ is a power of a prime. Now, we need to show whether these groups are $\mathrm{SCCD}$-groups or not.

Case 1: $\mathrm{PSL}_{l+1}(q)$ with $l\geq1$.

Subcase 1: $l = 1$.

Note from Table 2 that $\mathrm{cd}(E_q:C_{(q-1)/k}) = \{1, \frac{q-1}{k}\}$, $\mathrm{cd}(D_{2(q\pm1)/k}) = \{1, 2\}$, $\mathrm{cd}(S_4) = \{1, 2, 3\}$, $\mathrm{cd}(A_4) = \{1, 3\}$, and $\mathrm{cd}(A_5) = \{1, 3, 4, 5\}$. Note that $E_q:C_{(q-1)/k}$, $D_{2(q\pm1)/k}$, $S_4$, $A_4$ and $A_5$ are $\mathrm{CCD}$-groups, so in the following, we only need to consider maximal subgroup $H: =$ $\mathrm{PSL}_2(q_0).\gcd(\gcd(2, q-1), b)$ with $q = q_0^b$, $q_0\neq2$ and $b$ a prime.

If $q = 2^n$, then $\gcd(\gcd(2, q-1), p) = 1$, $H$ is isomorphic $\mathrm{PSL}_2(q_0)$, and $\mathrm{PSL}_2(q_0)$ is a $\mathrm{CCD}$-group because $\mathrm{cd}(\mathrm{PSL}_2(q_0)) = \{1, q_0-1, q_0, q_0+1\}$. Thus, $\mathrm{PSL}_2(q)$ is an $\mathrm{SCCD}$-group for $q$ even.

Let $q = r^n$ be odd, where $r$ is a prime. If $n = 1$, then $G\cong \mathrm{PSL}_2(q)$ with $q$ a prime, as in this case, and $\mathrm{PSL}_2(q)$ possibly contains $E_q:C_{(q-1)/k}$, $D_{2(q-1)/k}$, $D_{2(q+1)/k}$, $S_4$, $A_4$, $A_5$ as its maximal subgroups. Note that all these possible maximal subgroups are $\mathrm{CCD}$-groups. If $n > 1$, then $\gcd(2, q-1) = 2$, so $G$ has a maximal subgroup $\mathrm{PSL}_2(q_0).(2, b)$ with $b\bigl|n$.

● If $b = 2$, then by Theorem A of ^[35], we can assume that $\mathrm{PGL}_2(q_0)\cong \mathrm{PSL}_2(q_0).2$ or $\mathrm{PGL}_2(q_0)\nleqslant \mathrm{PSL}_2(q_0).2$. Observe that $\mathrm{PSL}_2(q_0)$ is a normal subgroup of $\mathrm{PGL}_2(q_0)$ or $\mathrm{PSL}_2(q_0).2$. Then, we have by ^[8] that $\mathrm{cd}(\mathrm{PSL}_2(q_0)) = \{1, \frac{q_0+\varepsilon}{2}, q_0-1, q_0, q_0+1\}$ with $\varepsilon = (-1)^{(q-1)/2}$. It follows from hypothesis that $1 = \frac{q_0+\varepsilon}{2}$, or $\frac{q_0+\varepsilon}{2}+1 = q_0-1$, and so $q_0 = 3$ or $5$.

● If $b\geq3$ is a prime, then $\gcd(2, b) = 1$, and $\mathrm{PSL}_2(q)$ contains a maximal subgroup $\mathrm{PSL}_2(q_0)$ by Table 2. As $\mathrm{cd}(\mathrm{PSL}_2(q_0)) = \{1, \frac{q_0+\varepsilon}{2}, q_0-1, q_0, q_0+1\}$ with $q = q_0^b$ and $\varepsilon = (-1)^{(q-1)/2}$, one gets from hypothesis that $q_0 = 3$ or 5, since the possible maximal subgroups $E_q:Z_{(q-1)/k}$, $D_{(q-1)/k}$, $D_{2(q+1)/k}$, $S_4$, $A_4$, $A_5$ for $\mathrm{PSL}_2(q)$ are $\mathrm{CCD}$-groups.

It follows that $\mathrm{PSL}_2(r^b)$ is an $\mathrm{SCCD}$-group where $r\in\{3, 5\}$ and $b\geq2$ is a prime.

Subcase 2: $l = 2$.

If $q = q_0^b$ with $b$ a prime, then by Table 3, $G$ contains a maximal subgroup $\mathrm{PSL}_3(q_0)$. Note from ^[29] that $\mathrm{cd}(\mathrm{PSL}_3(q_0)) = \{1$, $q_0(q_0+1)$, $q_0^2+q_0+1$, $(q_0+1)(q_0-1)^2$, $q_0^3-1$, $q_0^3$, $q_0(q_0^2+q_0+1)$, $(q_0+1)(q_0^2+q_0+1)\}$. It is easy to see that the degrees of $\mathrm{PSL}_3(q_0)$ are not consecutive. Thus, $q$ is a prime.

$k = \gcd(3, q-1)$, and $l = \gcd(3, q+1)$. Here, $m\equiv r(n)$ denotes $m\equiv r\pmod{n}$.

Since $q$ is an odd prime, we can get by Table 3 that $G$ contains a maximal subgroup $\mathrm{SO}_3(q)\cong \mathrm{SL}_2(q)$. We know from ^[16] that $\mathrm{cd}(\mathrm{SL}_2(q)) = \{1, \frac{q-1}{2}, \frac{q+1}{2}, q-1, q, q+1\}$, so by hypothesis, $q = 5$. By [7,pp. 38], $\mathrm{PSL}_3(5)$ contains a maximal subgroup $4^2:S_3$, and by ^[4], $\mathrm{cd}(4^2:S_3) = \{1, 2, 3, 6\}$, a contradiction to hypothesis. It follows that $q = 2$, and so $G\cong \mathrm{PSL}_3(2)\cong \mathrm{PSL}_2(7)$ is considered above.

Subcase 3: $l = 3$.

By Lemma 2.7, $\mathrm{PSL}_4(q)$ contains a subgroup isomorphic to either $\mathrm{PSL}_3(q)$ or $\mathrm{SL}_3(q)$. By ^[29], we obtain that either

or

so hypothesis forces $d_2+1 = d_3$, but the equation has no solution in $\mathbb{N}$, the nonnegative integer set. Now, $\mathrm{PSL}_3(q)$ is not a $\mathrm{CCD}$-group, and so $\mathrm{PSL}_4(q)$ is not an $\mathrm{SCCD}$-group.

Subcase 4: $l\geq4$.

Applying Lemma 2.7 repeatedly, we have that

Note that $\mathrm{PSL}_3(q)$ and $\mathrm{SL}_3(q)$ are not $\mathrm{CCD}$-groups by Subcase 3 of Case 1. Thus, $\mathrm{PSL}_{l+1}(q)$ is not an $\mathrm{SCCD}$-group.

Case 2: $\mathrm{PSU}_{l+1}(q)$ with $l\geq2$.

Note that $\mathrm{PSU}_2(q)\cong \mathrm{PSL}_2(q)$. We deal with this by two cases: $l = 2$ and $l\geq3$.

$\bullet$ Let $l = 2$.

We know that $\mathrm{PSU}_3(2)\cong 3^2.Q_8$ is solvable, so $q\geq3$. $\mathrm{PSU}_3(q)$ has a subgroup of the form: $E_{q}^{1+2}:C_{q^2-1}$. By the Main theorem of ^[28], $\mathrm{cd}(E_{q}^{1+2}) = \{1, q\}$. By Lemma 2.1, we get that either

or

so hypothesis shows that $q+1 = q^2-1$ or $q+1 = \frac{q^2-1}{3}$. We have $q = 2\ngtr3$ or $q = 4$. Thus, by ATLAS ^[7], $\mathrm{PSU}_3(4)$ contains a maximal subgroup $5^2:S_3$, and by ^[4], $\mathrm{cd}(5^2:S_3) = \{1, 2, 3, 6\}$, a contradiction to hypothesis.

Now, we have that, for $q\geq3$, $\mathrm{PSU}_3(q)$ is not an $\mathrm{SCCD}$-group; in particular, $\mathrm{PSU}_3(q)$ is not a $\mathrm{CCD}$-group.

$\bullet$ Let $l\geq3$.

Applying Lemma 2.7 repeatedly, we get that

and so $\mathrm{PSU}_{l+1}(q)$ is not an $\mathrm{SCCD}$-group.

Case 3: $\mathrm{PSp}_{2l}(q)$ with $l\geq2$.

Let $l = 2$. Since $\mathrm{PSp}_4(2)\cong S_6$, where $S_n$ is a symmetric group of degree $n$, is not simple, we assume that $q\geq3$. By [17,pp. 209], $\mathrm{PSp}_4(q)$ has a maximal subgroup $(\mathrm{PSp}_2(q)\times \mathrm{PSp}_2(q)).2 > \mathrm{PSp}_2(q)\times \mathrm{PSp}_2(q)$. Note that $\mathrm{PSp}_2(q)\cong \mathrm{PSL}_2(q)$ and that $S_6$ is a non-$\mathrm{CCD}$-group.

If $q$ is odd or $q\geq4$ is even, then $q-1, q, q+1\in \mathrm{cd}(\mathrm{PSL}_2(q))$, so by Theorem 4.21 of ^[15], one has that $(q+1)^2$, $q(q+1)$ $\in \mathrm{cd}(\mathrm{SL}_2(q)\times \mathrm{SL}_2(q))$. By hypothesis, $(q+1)q+1 = (q+1)^2$, and so $q = 2$, a contradiction.

Now, we get that $\mathrm{PSp}_4(q)$ is not an $\mathrm{SCCD}$-group.

Let $l\geq3$. Then, by Lemma 2.7, we get a subgroup series

and $\mathrm{PSp}_{2l}(q)$ is not an $\mathrm{SCCD}$-group.

Case 4: $\Omega_{2l+1}(q)$ with $l\geq2$.

If $l = 2$, then $\Omega_5(q)\cong \mathrm{PSp}_4(q)$ is a non-$\mathrm{SCCD}$-group by Case 3.

If $l = 3$, then $\Omega_5(q) < \Omega_7(q)$ by Lemma 2.7. We know from Case 3 that $\Omega_5(q)$ is a non-$\mathrm{SCCD}$-group, and so is $\Omega_7(q)$.

Let $l\geq4$. Then, Lemma 2.7 shows that $\Omega_{2l+1}(q)$ has a subgroup series

so $\Omega_{2l+1}(q)$ is a non-$\mathrm{SCCD}$-group.

Case 5: $\mathrm{P\Omega}_{2l}^\varepsilon(q)$ with $l\geq4$, $\varepsilon = \pm$.

We first consider $\mathrm{P\Omega}_{2l}^+(q)$. If $l = 4$, then by [17,pp. 214], $\mathrm{P\Omega}_8^+(q)$ contains a subgroup $\mathrm{P\Omega}_7(q)$. It follows from Case 4 that $\mathrm{P\Omega}_8^+(q)$ is a non-$\mathrm{SCCD}$-group.

If $l\geq5$, then using Lemma 2.7, $\mathrm{P\Omega}_{2l}^+(q)$ contains a subgroup isomorphic to

so Cases 3 and 4 imply that $\mathrm{P\Omega}_{2l}^+(q)$ is a non-$\mathrm{SCCD}$-group.

Similarly we can see that $\mathrm{P\Omega}_{2l}^-(q)$ with $l\geq4$ is a non-$\mathrm{SCCD}$-group.

Now, we can prove the following result.

Theorem 3.5. Let $G$ be a non-abelian simple $\mathrm{SCCD}$-group. Then, $G$ is isomorphic to $\mathrm{PSL}_2(q)$ where $q\geq4$ is even or $q\geq5$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

Note $A_5\cong \mathrm{PSL}_2(5)\cong \mathrm{PSL}_2(4)$, and $A_6\cong \mathrm{PSL}_2(3^2)$.

Proof. Lemmas 3.1– 3.3 and Proposition 3.4 imply the result.

4.   Non-solvable $\mathrm{SCCD}$-groups

In this section, we first show the structure of almost simple $\mathrm{SCCD}$-groups and then give that of non-solvable $\mathrm{SCCD}$-groups.

Note that $\mathrm{PGL}_2(5)\cong S_5$, and let $\mathrm{Out}(A)$ be the outer-automorphism group of a group $A$.

Theorem 4.1. Let $G$ be an $\mathrm{SCCD}$-group, and let $S$ be a non-abelian simple group such that $S\leq G\leq \mathrm{Aut}(S)$. Then, $G$ is isomorphic to one of the following groups:

(1) $\mathrm{PSL}_2(q)$ where $q\geq4$ is even or $q\geq5$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime,

(2) $S_5$.

Proof. Let $G$ be an almost simple $\mathrm{SCCD}$-group with socle $S$. Thus, $S$ is also a simple $\mathrm{SCCD}$-group, and by Theorem 3.5, $S\cong \mathrm{PSL}_2(q)$ where $q\geq4$ is even or $q$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

If $q = 2^f$ for some $f\geq2$, then $G$ is isomorphic to $\mathrm{PSL}_2(2^f).d$ with $d\bigl|f$. If $G$ is a simple group, then $G$ is isomorphic to $\mathrm{PSL}_2(q)$ as $\mathrm{PSL}_2(q)\cong \mathrm{PGL}_2(q)\cong \mathrm{SL}_2(q)$. If $f = 2$, then $G\cong S_5$ is an $\mathrm{SCCD}$-group. If $f\geq3$, then $E_{2^f}:C_{2^f-1}:C_f$ is a subgroup of $\mathrm{PSL}_2(2^f).f$ (see ^[11]). By Lemma 2.1, one has $\mathrm{cd}(E_{2^f}:C_{2^f-1}:f) = \{1, f, 2^f-1\}$. By hypothesis, we get $f+1 = 2^f-1$, so $f = 2\ngeqq3$, a contradiction.

Now, $q$ is odd, and by Theorem 3.5, these cases will be considered: $q\geq5$ is an odd prime, or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

Case 1: $q\geq5$ is an odd prime.

Note from [7,pp. xvi] that $| \mathrm{Out}(\mathrm{PSL}_2(q))| = 2$ for $q$ an odd prime. It follows that $G$ is possibly isomorphic to $\mathrm{PGL}_2(q)$. Note that $\mathrm{PSL}_2(q)\in\max \mathrm{PGL}_2(q)$, and

Lemma 2.3 gives that $\frac{q+(-1)^{\frac{q-1}{2}}}{2}+1 = q-2$, so $q = 5$. As $\mathrm{PSL}_2(5)\cong \mathrm{PSL}_2(4)\cong A_5$, $\mathrm{PGL}_2(5)\cong S_5$ is considered above.

Case 2: $q = 3^p$ or $5^p$ with $p$ a prime.

Let $\mathfrak{S} = \{\mathrm{PGL}_2(q), \mathrm{PSL}_2(q).2, \mathrm{PSL}_2(q).p, \mathrm{PGL}_2(q).p, \mathrm{PSL}_2(q).(2p)\}$.

In this case, we obtain that $| \mathrm{Out}(\mathrm{PSL}_2(q))| = 2\cdot p$, so $G$ is possibly isomorphic to $S\in \mathfrak{S}$ by Corollary 6.2 of ^[35]. Observe that for each $S\in \mathfrak{S}$, $S$ has a subgroup isomorphic to $\mathrm{PSL}_2(q)$. Note that

so $\mathrm{PSL}_2(q)$ is not a $\mathrm{CCD}$-group, as $q > 5$ is a power of an odd prime. It follows that for every $S\in \mathfrak{S}$, $S$ is a non-$\mathrm{SCCD}$-group.

Now, we will give the proof of Theorem 1.3.

Proof of Theorem 1.3.

Proof. As $G$ is non-solvable, we have a normal subgroup series $1\leq H\leq K\leq G$ such that $K/H$ is isomorphic to a direct product of isomorphic non-abelian simple groups and that $|G/K|$ divides the order of the outer-automorphism group $\mathrm{Out}(K/H)$; (see ^[36]).

Assume that

where $S$ is isomorphic to one of the groups as listed in Theorem 3.5. If $m\geq2$, then by Theorem 4.21 of ^[15], $L\times S\times S\cdots\times S$ is a non-$\mathrm{CCD}$-group where $L$ is a non-abelian subgroup of $S$. Now, $m = 1$, and so $K/H$ is a simple group isomorphic to $S\in \mathfrak{S}$, where $\mathfrak{S}$ is a set which consists of the groups $\mathrm{PSL}_2(q)$ where $q\geq4$ is even or $q\geq5$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

Case 1: $K/H$ is isomorphic to $\mathrm{PSL}_2(q)$ with $q = 2^f$, $f\geq2$.

We know that when $q$ is even, $\mathrm{PGL}_2(q)\cong \mathrm{PSL}_2(q) \cong \mathrm{SL}_2(q)$. Thus, by Theorem 4.1, $G/H$ is possibly isomorphic to $\mathrm{PSL}_2(q).f$. However, $E_q:C_{q-1}:C_f$ is a subgroup of $\mathrm{PSL}_2(q).f$, so $q = 4$, and $G/H$ is isomorphic to $A_5$ or $S_5$, as the proof of Theorem 4.1. We see from Chap 2, Theorem 6.10 of ^[12] that $G'/H\cong A_5$. We know that the order of the Schur multiplier of $A_5$ is 2, and $|\mathrm{Out}(A_5)| = 2$.

If $[G', H] = 1$, then $G\cong H\times A_5$ or $G\cong H\times S_5$. If $H$ is non-abelian, then we can assume that $\mathrm{cd}(H) = \{1, m\}$. Now, by ^[7], $H\times A_4$ and $H\times S_4$ are subgroups of $H\times A_5$ and $H\times S_5$ respectively, so $\mathrm{cd}(H\times A_4) = \{1, m, 3, 3m\}$ and $\mathrm{cd}(H\times S_4) = \{1, 2, 3, m, 2m, 3m\}$. We see that $H\times A_4$ and $S_4\times H$ are $\mathrm{CCD}$-subgroups, so $3+1 = 3m$ or $m+1 = 3m$, but the two equations have no solutions in $\mathrm{N}$. So, $H$ is abelian.

If $|[G', H]|\neq1$, then $G\cong \mathrm{SL}_2(5)*H$, and $H\geq C_2$. We similarly get that $H/C_2$ is abelian as the case $[G', H] = 1$.

Case 2: $K/H$ is isomorphic to $\mathrm{PSL}_2(q)$ where $q\geq5$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

Where $\mathrm{N}1$ maximal under $< \delta >$ with $|\delta| = (q-1, 2)$; $\mathrm{N}2$ maximal under subgroups not contained in $< \varphi >$ with $|\varphi| = e$, $q = p^e$, $p$ a prime.

From Case 1, we see that $q\geq7$. By Theorem 4.1, $G/H$ is isomorphic to $\mathrm{PSL}_2(q)$ where $q\geq7$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime. Now, by [12,Chap 2,Theorem 6.10], $G'/H\cong \mathrm{PSL}_2(q)$.

If $[G', H] = 1$, then $G\cong H\times \mathrm{PSL}_2(q)$. We can show similarly that $H$ is abelian in this case. If $H\neq1$, then $\mathrm{PSL}_2(q)$ is a proper subgroup of $G$, and so $\mathrm{PSL}_2(q)$ is a $\mathrm{CCD}$-group. We see that $\mathrm{cd}(\mathrm{PSL}_2(q)) = \{1, \frac{q+(-1)^{(q-1)/2}}{2}, q-1, q, q+1\}$, so $q-1 = \frac{q+(-1)^{(q-1)/2}}{2}+1$. Now, $q = 5$, a contradiction. So, $G\cong \mathrm{PSL}_2(q)$, as desired.

If $[G', H]\neq1$, then $H\geq C_2( = Z(\mathrm{SL}_2(q)))$, and $G$ is isomorphic to $\mathrm{SL}_2(q)*H$. If $H > C_2$, then $\mathrm{SL}_2(q)$ is a proper subgroup of $G$, so it is a $\mathrm{CCD}$-subgroup. We see that

so $q = 5$, a contradiction. Thus, $H = C_2$, and $\mathrm{SL}_2(q)C_2\cong\mathrm{SL}_2(q)$. Furthermore, if $q$ is prime, then by Table 4, $\mathrm{SL}_2(q)$ possibly has subgroups $E_q:C_{q-1}$, $D_{2(q\pm1)}$, $2.S_4$, $2.A_4$ and $2.A_5$, which are $\mathrm{CCD}$-groups. If $q\in\{3^p$, $5^p\}$ with $p$ a prime, then Table 4 shows that $\mathrm{SL}_2(q)$ possibly subgroups $E_q:C_{q-1}$, $D_{2(q\pm1)}$, $\mathrm{SL}_2(q_0)$, $\mathrm{PSL}_2(q_0)$ and $2.A_5$, which are $\mathrm{CCD}$-groups, too. Thus, $G$ is isomorphic to $\mathrm{SL}_2(q)$, as wanted.

5.   Applications

In this section, we will prove that an $\mathrm{SCC}$-group is solvable (see Theorem 5.1) and then determine the structure of an $\mathrm{SCC}$-group.

Theorem 5.1. Let $G$ be a finite $\mathrm{SCC}$-group. Then, $G$ is solvable.

Proof. Assume that the result is not true, and then $G$ is a non-solvable group. It is easy to get that a $\mathrm{CC}$-group must be a $\mathrm{CCD}$-group, so by hypothesis, for each $H\in \sum G$, $H$ is a $\mathrm{CCD}$-group. Thus, we have that $G$ is an $\mathrm{SCCD}$-group. Now, Theorem 1.3 shows that $G$ has one of the following structure:

$(1)$ $\mathrm{PSL}_2(q)$ where $q\geq4$ is even or $q\geq5$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

$(2)$ $S\times H$ where $S\in\{A_5, S_5\}$ and $H$ is abelian.

$(3)$ $\mathrm{SL}_2(q)$ where $q\geq7$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

$(4)$ $\mathrm{SL}_2(5)*H$ with $H/C_2$ abelian.

So, four cases are dealt with in what follows.

Case 1: $\mathrm{PSL}_2(q)$ where $q\geq4$ is even or $q\geq5$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

By Table 2, $E_q:C_{(q-1)/k}\in\max \mathrm{PSL}_2(q)$, so $\mathrm{cd}(E_q:C_{(q-1)/k}) = \{1, (q-1)/k\} = \{1, 2\}$. It follows that $q = 5$. If $q = 5$, then by [7,pp. 2], $A_4\in\max A_5$. We know that $\mathrm{cd}(A_4) = \{1, 3\}$, a contradiction.

Case 2: $S\times H$ where $S\in\{A_5, S_5\}$ and $H$ is abelian.

Obviously, $A_4$ is a subgroup of $S$. The transitivity of subgroups shows that $A_4$ is a subgroup of $G$, so $\mathrm{cd}(A_4) = \{1, 3\}$ gives a contradiction.

Case 3: $\mathrm{SL}_2(q)$ where $q\geq7$ is an odd prime or $q\in\{3^p$, $5^p\}$ with $p$ a prime.

Now, Table 4 means that $E_q:C_{q-1}\in\max \mathrm{SL}_2(q)$, and hypothesis shows that $\mathrm{cd}(E_q:C_{q-1}) = \{1, q-1\} = \{1, 2\}$, so $q = 3\ngeqslant7$, a contradiction.

Case 4: $\mathrm{SL}_2(5)*H$ with $H/C_2$ abelian.

We see that $\mathrm{SL}_2(5)$ has a subgroup of the form $E_5:C_4$, and $\mathrm{cd}(E_5:C_4) = \{1, 4\}$, a contradiction.

It follows from the four cases that $G$ is solvable, as desired.

Let $\rho(G)$ be the set of prime divisors of degrees of the irreducible characters of a group $G$, that is,

Lemma 5.2. Let $G$ be a non-abelian $\mathrm{SCC}$-group. Then, $|\rho(G)|\leq 2$.

Proof. We know that for each non-abelian $A\in \sum G$, $A$ is a $\mathrm{CC}$-group, so $2\in \mathrm{cd}(A)$. It follows that $2\in\rho(G)$. We know from Theorem 5.1 that $G$ is solvable, so the character degree graph $\mathrm{T}(G)$ of $G$ has at most two connected components; see ^[25]. Let $\rho(G) = \{2, p_1, p_2, \cdots, p_s\}$ with $p_1 < p_2 < \cdots < p_s$. Let $s\geq2$, and let $H$ be a Hall $\{p_1, p_2, \cdots, p_s\}$-subgroup of $G$. Then $H$ is a $\mathrm{CC}$-group, so $\mathrm{cd}(H)\subseteq \{1, p_1, p_2, p_1p_2, \cdots\}$. We get a contradiction as all $p_i$s are $\geq3$. So, $H$ is abelian and non-normal in $G$ by Theorem 7.1 of ^[10]. As $s\geq2$, Hall $\{2, p_2\}$-subgroup $H_{2, p_2}$ of $G$ is a $\mathrm{CC}$-group, so $\mathrm{cd}(H_{2, p_2}) = \{1, 2, p_2, \cdots\}$. It follows that $p_2 = 3$, a contradiction, as $p_2 > p_1\geq3$. It follows that the Sylow $p_2$-subgroup $H_{p_2}$ of $H_{2, p_2}$ is normal and abelian in $H_{2, p_2}$. We also can get that the Sylow $p_i$-subgroup $H_{p_i}$ of the Hall $\{2, p_i\}$-subgroup $H_{2, p_i}$ of $G$ is abelian and normal in $H_{2, p_i}$. It follows that the Hall $\{p_2, \cdots, p_s\}$-subgroups $H_{ p_2, p_3, \cdots, p_s}$ of $H_{2, p_2, \cdots, p_s}$ are abelian and normal in $H_{2, p_2, \cdots, p_s}$. We see that $G = H_{p_1, p_2, \cdots, p_s}H_{2, p_2, \cdots, p_s}$ and $H_{p_1, p_2, \cdots, p_s}\cap H_{2, p_2, \cdots, p_s} = H_{p_2, \cdots, p_s}$, so $H_{p_2, \cdots, p_s}$ is abelian and normal in $G$, a contradiction to Theorem 7.1 of ^[10]. It follows that $s = 1$, the desired result.

Lemma 5.3. Let $G$ be a nilpotent non-abelian $\mathrm{SCC}$-group. Then, $G$ is isomorphic to a metabelian $p$-group or a direct product of a 2-group $P$ and $2'$-group $H$ where $\mathrm{cd}(P) = \{1, 2\}$ and $\mathrm{cd}(H) = \{1\}$.

Proof. Let $P_i$ be Sylow $p_i$-subgroups of $G$ with primes $p_i$, and $p_1 < p_2 < \cdots < p_s$. As $G$ is nilpotent, write $G = P_1\times P_2\times \cdots\times P_s$. By Lemma 5.2, $\rho(G) = \{p\}$, or $\rho(G) = \{p, q\}$ with different primes $p, q$.

$\bullet$ $\rho(G) = \{p\}$.

If $s = 1$ and $p = 2$, then $G$ is a 2-group. We know that for each $H\in\sum (G)$, $H$ is a $\mathrm{CC}$-group, so $\mathrm{cd}(H) = \{1\}$ or $\mathrm{cd}(H) = \{1, 2\}$. Thus, $G$ is a 2-group whose proper subgroups all have degrees 1 or 2. If $s\geq2$ and $p = 2$, then $G$ is a direct product of a non-abelian 2-subgroup $P$ and a Hall $2'$-subgroup $H$. As $H\neq 1$, we have $\mathrm{cd}(P) = \{1, 2\}$. If $H$ is non-abelian, then $Z(P)\times H$ is a non-abelian $\mathrm{CC}$-group, so $\mathrm{cd}(Z(P)\times H) = \{1, p', \cdots \}$. Hypothesis shows that $p' = 2$, a contradiction. Hence, $H$ is abelian. We now draw a conclusion that $G = P\times H$ with $\mathrm{cd}(P) = \{1, 2\}$, and $H$ abelian.

Now, we consider $p > 2$. If $s = 1$, then assumption gives that $G$ is a metabelian $p$-group. If $s\geq2$, then $G = P\times H$ where $P$ is a $p$-subgroup and $H$ is a $p'$-subgroup. If $P$ is non-abelian, then $P$ is a $\mathrm{CC}$-subgroup of $G$, and so $\mathrm{cd}(P) = \{1, p, \cdots\}$. Now, hypothesis gives that $p = 2\ngeqq3$, a contradiction. Hence, $P$ is abelian, and so is $G$. Otherwise, there is a subgroup which is not a $\mathrm{CC}$-subgroup.

$\bullet$ $\rho(G) = \{p, q\}$.

In this case, $G = P\times H$ with $\rho(P) = \{p\}$ and $\rho(H) = \{q\}$. By the above arguments, we can assume that $p = 2$ and $\mathrm{cd}(P) = \{1, 2\}$. Thus, $C_2\times H$ is a $\mathrm{CC}$-subgroup of $G$, and $\mathrm{cd}(C_2\times H) = \{1, q, \cdots\}$. Hence, $q = 2$, too, a contradiction. So, this case can not occur.

Lemma 5.4. Let $G = F:H$ be a Frobenius group with kernel $F$ and complement $H$, respectively. If $G$ is an $\mathrm{SCC}$-group, then one of the following holds:

(i) $\mathrm{cd}(F) = \{1, 2\}$, and $H$ is a prime order such that for each non-identity $W\in\sum F$, $|W|\not\equiv1\pmod{|H|}$.

(ii) $F$ is abelian, and $H$ is of order a prime or 4. Furthermore, if $H$ is of order 4, then $F$ is isomorphic to $C_p\times C_q$ with $q\equiv p\equiv-1\pmod{4}$ or $C_p$ with $p\equiv1\pmod{4}$. If $H$ is of prime order $> 2$, then $|H|\|(|F|-1)$.

Proof. We divide the proofs into two cases: $F$ is non-abelian, and $F$ is abelian.

Case 1: $F$ is non-abelian

Then, $F$ is a $\mathrm{CC}$-subgroup of $G$, and so $2\in \mathrm{cd}(F)$. We know that $F$ is nilpotent, so by Lemma 5.3, $\mathrm{cd}(F) = \{1, 2\}$. Thus, $F$ is nilpotent with $\mathrm{cd}(F) = \{1, 2\}$.

If $H$ is non-abelian, then Lemma 2.1 gives that $H$ is of odd order, and so by Lemma 5.3 $H$ is metabelian. Let $H_p$ with $|H_p| = p$ be a proper subgroup of $H$ with certain prime $p$. Then, $F:H_p$ is also a Frobenius subgroup of $G$. By assumption, $F:H_p$ is a $\mathrm{CC}$-group, and so by Lemma 2.1, $\mathrm{cd}(F:H_p) = \{1, 2, p, 2p\}$, a contradiction. Thus, $H$ is abelian. We claim that $H$ is of prime order. In fact, if $|H| = pq$ with primes $p, q$, then $F:H_q$ and $F:H_p$ are Frobenius subgroups of $G$, so they are $\mathrm{CC}$-groups. By Lemma 2.1, we have that $\mathrm{cd}(F:H_p) = \{1, 2, p, 2p\}$, and $\mathrm{cd}(F:H_q) = \{1, 2, q, 2q\}$, a contradiction. This means that $H$ is of prime order, say, $q$.

By Lemma 2.1, $|F|\equiv 1\pmod{q}$. Let $F_{p_i}, i = 1, 2, \cdots, n$, be the Sylow $p_i$-subgroup of $F$. If $|F_1F_2\cdots F_s|\equiv 1\pmod{q}$, then $|F_{s+1}\cdots F_n|\equiv 1\pmod{q}$, so $W: = F_{s+1}\cdots F_n:H$ is a Frobenius group. Now, $\mathrm{cd}(W) = \{1, q\}$, a contradiction, as $W$ is a $\mathrm{CC}$-subgroup of $G$. Thus, $|K|\not\equiv 1\pmod{q}$ for all $K\in\sum F$, but $|F|\equiv 1\pmod{q}$.

Case 2: $F$ is abelian.

We claim that

In fact, if $H$ is non-abelian, and $H$ has two subgroups $C_p$ and $C_q$ with primes $p, q$, then $\mathrm{cd}(F:C_p) = \{1, p\}$, and $\mathrm{cd}(F:C_q) = \{1, q\}$. By hypothesis, $F:C_{p}$ and $F:C_{q}$ are $\mathrm{CC}$-groups, so we have that $p = q = 2$ is possible. It follows that either $p = 2$ and $H$ is a 2-group of order $\leq 4$ or $H$ is of odd prime order.

If $H$ is of order 2, then $F$ is of odd order, and for all $W\in\sum G$, $W$ is a $\mathrm{CC}$-group, since $\mathrm{cd}(W) = \{1\}$ or $\{1, 2\}$.

If $H$ is of order 4, then $F$ is of odd order, too.

● If $F$ is a $p$-group, and then say $|F| = p^s$. Let $s\geq3$, then $G$ has a subgroup $C_{p^2}:C_4$ which is a $\mathrm{CC}$-group. We know that $\mathrm{cd}(C_{p^2} :C_4) = \{1, 4\}$, a contradiction. Thus, $s\leq 2$: namely, $|F| = p$ or $p^2$. If $|F| = p$, then $p\equiv 1\pmod{4}$; if $|F| = p^2$, then $p^2\equiv 1\pmod {4}$. If $|F| = p^2$ and $p\equiv 1\pmod{4}$, then $G$ has a Frobenius subgroup $C_p:C_4$ which is not a $\mathrm{CC}$-group. Thus, $|F| = p^2$ with $p\equiv-1\pmod{4}$, or $|F| = p$ with $p\equiv1\pmod{4}$.

● If $F$ is a non-$p$-group, then $F$ has two subgroups $C_p$ and $C_q$. Note that $p, q$ are odd, so $p\equiv\pm 1\pmod{4}$ and $q\equiv\pm1\pmod{4}$. If $p\equiv q\equiv 1\pmod{4}$, then $G$ has a subgroup of the form $C_p:C_4$ which is not a $\mathrm{CC}$-group, a contradiction. If $q\equiv-1\pmod{4}$ and $p\equiv1\pmod{4}$, or $q\equiv1\pmod{4}$ and $p\equiv-1\pmod{4}$, then $q^2p\equiv1\pmod{4}$. However, $C_p:C_4$ is a Frobenius group, so it is a $\mathrm{CC}$-group, a contradiction (note in this case that $pq\equiv1\pmod{2}$, but $pq\equiv-1\pmod{4}$). Now, we draw a conclusion that $p\equiv q\equiv -1\pmod{4}$. If $|F| = (pq)^{2s}q^t$ or $(pq)^{2s}p^t$ with $s\geq1, t\geq0$, then a $\mathrm{CC}$-subgroup $C_{pq}:C_4$ is a Frobenius group, a contradiction. Thus, $|F| = pq$, and $F$ is isomorphic to $C_p\times C_q$, as $F$ is nilpotent.

Let $H$ be of prime order $r > 2$. Note that the nilpotence of $F$ gives that

As with the above arguments, two cases are considered: $s = 1$ and $s\geq2$.

● $s = 1$. Then, $F$ is a $p$-group for certain prime $p$, so $r\equiv1\pmod{|F|}$. If $F$ has a subgroup $K$ such that $K:H$ is a Frobenius group, then $K:H$ is not a $\mathrm{CC}$-group. Thus, $r\|(|F|-1)$.

● $s\geq2$. Then, similarly, we can show that $r\|(|F|-1)$.

Now, the lemma is complete.

Lemma 5.5. Let $G$ be a non-abelian $\mathrm{SCC}$-group. Assume that $G$ is a direct product of at least two $\mathrm{CC}$-groups. Then, $G$ is isomorphic to $H_1\times H_2$ where at least one of $H_1$ and $H_2$ is metabelian with $\mathrm{cd}(H_i)\subseteq \{1, 2\}$.

Proof. Let $G = H_1\times H_2\times \cdots H_s$ where all $H_i$ are $\mathrm{CC}$-groups. If $H_i$ are all abelian, then $G$ is abelian, so we assume that for some $i$, $H_i$ is non-abelian, say, $H_1$. If the other $H_i$ for $i = 2, \cdots s$ are abelian, then $G$ is an $\mathrm{SCC}$-group when $\mathrm{cd}(H_1) = \{1, 2\}$. If $\mathrm{cd}(H_1) = \{1, p\}$ for some prime $p > 2$, then $G$ is not an $\mathrm{SCC}$-group when $H_i\neq 1$, $i = 2, 3, \cdots, s$. It follows that $G = H_1$, a contradiction. Thus, we assume that $G = H_1\times H_2\times \cdots \times H_s$ with $\mathrm{cd}(H_1) = \mathrm{cd}(H_2) = \{1, 2\}$. If $s\geq3$, then $H_1\times H_2$ is a $\mathrm{CC}$-subgroup of $G$, but by Theorem 4.21 of ^[15], $\mathrm{cd}(H_1\times H_2) = \{1, 2, 4\}$, a contradiction. Thus, $s = 2$, and $G = H_1\times H_2$, where $\mathrm{cd}(H_1) = \mathrm{cd}(H_2) = \{1, 2\}$, and $H_1$ and $H_2$ are metabelian. We similarly can get that $G = H_1\times H_2\cdots H_k\times H_{k+1}\times \cdots H_s$ is not an $\mathrm{SCC}$-group if $\mathrm{cd}(H_i) = \{1, 2\}$, $i = 1, 2, \cdots, k$, and $k\geq3$.

Proof of Theorem 1.4.

Proof. If $G$ is an $\mathrm{SCC}$-group, then $G$ is solvable by Theorem 5.1. If $G$ is a nilpotent group or a Frobenius group or a direct product of two $\mathrm{CC}$-groups, then we can get (ii), (iii) and (iv) by Lemmas 5.3, 5.4 and 5.5. Thus, in what follows, we assume that $G$ is neither a nilpotent group, nor a Frobenius group nor a direct product of two $\mathrm{CC}$-groups. If $G$ is abelian or $\mathrm{cd}(G) = \{1, 2\}$, then for all $H\in\sum G$, $H$ is always a $\mathrm{CC}$-group, so we get (i). Thus, we assume that $\mathrm{cd}(G)\neq\{1\}$ and $\mathrm{cd}(G)\neq\{1, 2\}$. Since $G$ is a solvable $\mathrm{SCC}$-group, we get that $G$ has a factor; say that $N/M$ is such a factor. Note that $N\in\sum G$, so $N$ is a $\mathrm{CC}$-group. Now,

see ^[13] or [14,Sec. 32]. By pp. 168 of ^[13],

It is easy to see that $M$ is abelian. (In fact, let $\theta\in \mathrm{Irr}(M)\setminus \mathrm{Lin}(M)$. Then, $I: = \mathrm{I}_N(\theta)$, the inertia subgroup of $\theta$ in $N$, is equal to $N$ or $KM$ where $K\in\sum N/M$. If $I = KM$, then $\theta^N(1) = |N:I|\theta(1)$, so $\theta(1)$, $\theta^N(1)\in \mathrm{cd}(N)$. Note that $N$ is a $\mathrm{CC}$-group, so $\mathrm{cd}(N) = \{1, \theta(1), 2, \theta^N(1)\}$ or $\mathrm{cd}(N) = \{1, \theta(1), 3, \theta^N(1)\}$ or $\mathrm{cd}(N) = \{1, \theta(1), 4, \theta^N(1)\}$, contradicting Theorem of ^[13].)

Let $\overline{G} = G/M$ and let $\overline{N} = N/M$. Then,

Set $L = \{xM:xM\in C_{\overline{G}}(\overline{N}\}$, and then $G/L\cong\overline{G}/C_{\overline{G}}(\overline{N})$. Notice that $L$ is also abelian as the proof as $M$ is abelian. So, we have that $\mathrm{Aut}(\overline{N}) = S_3$, $S_4$ and $3^2:D_8$ when $\overline{N}$ is isomorphic to $S_3$, $A_4$ and $3^2:2^2$, respectively; see ^[33]. Observe that the Schur multipliers of $\overline{N}$ are trivial, so $G$ is possibly isomorphic to $S_3\times L$, $A_4\times L$, $S_4\times L$, $(3^2 :C_4)\times L$ or $(3^2:D_8)\times L$. If $G$ is isomorphic to $S_3\times L$, then it is contained in (i); if $G$ is isomorphic to $S_4\times L$ or $3^2:D_8$, then $A_4\in \sum S_4$ and $3^3:2^2\in\sum (3^2:D_8)\times L$ are $\mathrm{CC}$-groups, a contradiction. Thus, $A_4$ and $3^2:2^2$ are not proper subgroups of $G$, and so $L = 1$. Now, $G$ is isomorphic to $A_4$ or $3^2:C_4$, as desired.

6.   Conclusions

From Theorem 1.3, we know that a group whose non-linear character degrees are consecutive is possibly non-solvable, but a group whose irreducible character degrees is consecutive are solvable. Chen in ^[5] gave the information of groups whose irreducible character degrees are arithmetic numbers. In comparison with Theorem 5.1, is the group solvable if proper subgroups all have this property?

Acknowledgments

The author was supported by the NSF of China (Grant No: 11871360), and also the first author was supported by the Opening Project of Sichuan Province University Key Laboratory of Bridge Non-destruction Detecting and Engineering Computing (Grant Nos: 2022QYJ04) and by the Project of High-Level Talent of Sichuan University of Arts and Science (Grant No: 2021RC001Z).

Conflict of interest

The authors declare that he has no conflict of interest.