Co-degree Graphs and Order Elements

Abstract

Let G be a finite group. For an irreducible character $\chi$ of G, the number $\chi ^c(1)=\frac{[G:\mathrm{ker}\chi ]}{\chi (1)}$ is called the co-degree of $\chi$. Let $\mathrm{Codeg}(G)$ denote the set of co-degrees of the irreducible (complex) characters of G. The co-degree graph of G is defined as the simple undirected graph whose vertices are prime divisors of the numbers in $\mathrm{Codeg}(G)$ and two distinct vertices p and q are joined by an edge if and only if pq divides some number in $\mathrm{Codeg}(G)$. In this paper, we first study the finite non-solvable groups whose co-degree graphs have no complete vertices. Then, we show that if the co-degree graph of G has no complete vertices, then for every $x \in G$, G admits an irreducible character whose co-degree is divisible by o(x), the order of x. Finally, we inspect the finite groups whose co-degree graphs are m-regular.

1 Introduction and Preliminaries

In this paper, G is a finite group and p is a prime number. Let $\mathrm{Irr}(G)$ denote the set of irreducible (complex) characters of G, $\mathrm{Fit}(G)$ be the Fitting subgroup of G, and let Z(G) be the center of G. The solvable radical of G is the largest solvable normal subgroup of G. For a normal subgroup N of G and $\theta \in \mathrm{Irr}(N)$, let $I_G(\theta )$ denote the inertia group of $\theta$ in G and let $\mathrm{Irr}(\theta ^G)$ be the set of the irreducible constituents of the induced character $\theta ^G$. Also, if n is a positive integer, we use $\pi (n)$ to show the set of the prime divisors of n and by $\pi (G)$, we mean the set of the prime divisors of |G|. For an irreducible character $\chi$ of G, the number $\chi ^c(1)=\frac{[G:\mathrm{ker}\chi ]}{\chi (1)}$ is called the co-degree of $\chi$ (see [20]). Set $\mathrm{Codeg}(G)=\{\chi ^c(1):\chi \in \mathrm{Irr}(G)\}$. In [1,2,3, 5, 6], some properties of the co-degrees of irreducible characters of finite groups have been studied.

For a set of (positive) integers $\mathcal {A}$, the prime graph associated with $\mathcal {A}$ is defined as a simple undirected graph whose vertices are prime divisors of the numbers in $\mathcal {A}$ and two distinct vertices p and q are joined by an edge if and only if pq divides some number in $\mathcal {A}$. If $\mathcal {A}$ is the set of the irreducible character degrees, conjugacy class sizes or order elements, then many interesting results about the prime graph associated with $\mathcal {A}$ have been obtained, for instance see [8, 9, 16].

In this paper, by the co-degree graph $\Delta (G)$, we mean the prime graph associated with $\mathrm{Codeg}(G)$. By [20], the vertex set of $\Delta (G)$ is $\pi (G)$ and if $\Delta (G)$ is disconnected, then G is a Frobenius or a 2-Frobenius group. We note that the vertex $p \in \pi (G)$ is an incomplete vertex in $\Delta (G)$ when p is not adjacent to at least one other vertex; otherwise, p is called a complete vertex. By [4, Theorem A], if G has an incomplete vertex, then $\mathrm{Fit}(G)\ne \{1\}$. As mentioned in [4], there exists a non-solvable group G of order $2^3.3.5^2.11^2$ such that every element of $\mathrm{Codeg}(G)$ is either $5.11^2$ or it is co-prime to 11. This example shows the existence of non-solvable groups such that their co-degrees graphs have no complete vertices. In Theorems A and B, we are going to study such groups. Note that in GAP library of primitive groups, the group G stated in the example given above can be accessed via G=PrimitiveGroup(11$\hat{}$ 2,57).

Theorem A

Let G be non-solvable and N be the solvable radical of G. If 2 is an incomplete vertex in $\Delta (G)$, then G/N has the unique minimal normal subgroup S/N and $S/N \cong \mathrm{Alt}_7$ or $\mathrm{PSL}_2(q)$, where q is an odd prime power. In particular, a Sylow 2-subgroup of G is a generalized quaternion group and $2 \mid |N|$.

Theorem B

Let G be non-solvable and N be the solvable radical of G. If every vertex of $\Delta (G)$ is incomplete in $\Delta (G)$, then $G/N \cong \mathrm{PSL}_2(p)$ or $\mathrm{PGL}_2(p)$, where p is an odd prime. In particular, G has the normal subgroups L and M such that $L \le M \le N$, $N/M \le Z(G/M)$, M is the Hall subgroup of G, $\Delta (L)$ is complete and G/N has the unique minimal normal subgroup S/N such that $[G:S] \mid 2$, $S/M \cong \mathrm{SL}_2(p)$ and $S/L \cong M/L \times S/M$.

Note that in Theorem B, $\pi (M)=\pi (N)-\{2\}$ and $[N:M]=2$.

By [14, 21], if G has an element of order m, then G admits an irreducible character whose co-degree is divisible by $\prod _{r \in \pi (m)}r$. Recently, G. Qian in [22] has proven that if G is solvable, then for every $x \in G$, G admits an irreducible character $\chi$ such that o(x), the order of x, divides $\chi ^c(1)$. Also, he has conjectured that the same result holds true for every finite group. In this paper, we prove that:

Theorem C

If every vertex of $\Delta (G)$ is incomplete, then for every $x \in G$, G admits an irreducible character whose co-degree is divisible by o(x).

We note that $\Delta (G)$ is m-regular, if every vertex of G is adjacent to exactly m other vertices of $\Delta (G)$. For a positive integer n, let $C_n$ denote the cyclic group of order n, $G_1$ be a Frobenius group $C_{91} \rtimes C_6$ and $G_2$ be a Frobenius group whose kernel is a direct product of an elementary abelian group of order $11^2$, an elementary abelian group of order $41^2$ and an elementary abelian group of order $59^2$, and its complement is isomorphic to $SL_2(5)$. Also, assume that $G_3$ is a Frobenius group whose kernel is a direct product of an elementary abelian group of order $41^2$, an elementary abelian group of order $211^2$, an elementary abelian group of order $491^2$ and an elementary abelian group of order $631^2$, and its complement is isomorphic to $SL_2(5) \times C_7$. Then, $\Delta (C_5)$ and $\Delta (D_{10})$ are 0-regular, $\Delta (C_{15})$ and $\Delta (G_1)$ are 1-regular, $\Delta (C_{30})$ and $\Delta (G_2)$ are 2-regular and also, $\Delta (G_3)$ is 3-regular. These examples motivate us to study some finite groups with m-regular co-degree graphs in the following theorem.

Theorem D

Let $\Delta (G)$ be m-regular.

• (i) If $m=0$, then G is an r-group, for some prime r, or $|\pi (G)|=2$ and G is either a Frobenius or a 2-Frobenius group. In particular, G is solvable.

• (ii) If $m=1$, then G is solvable and $|\pi (G)| \in \{2,4\}$. If $|\pi (G)|=4$, then G is a Frobenius or a 2-Frobenius group.

• (iii) If $m=2\ne |\pi (G)|-1$, then $|\pi (G)| \le 10$. If $|\pi (G)| \ge 6$, then G is a Frobenius or a 2-Frobenius group. Also, if G is non-solvable, then $|\pi (G)|=6$ and G is a Frobenius group whose Frobenius kernel is abelian and its Frobenius complement has a normal subgroup of index at most 2 which is isomorphic to $\mathrm{SL}_2(5)$.

• (iv) If $m=3 \ne |\pi (G)|-1$ and G is non-solvable, then G has the normal subgroups K and S such $[G:S] \mid 2$ and either $|\pi (G)|=8$ and G is a Frobenius group with Frobenius kernel K such that $S/K \cong L \times \mathrm{SL}_2(5)$, where L is an r-group, for some prime $r \not \in \{2,3,5\}$ or $|\pi (G)|=6$ and $S/K \cong \mathrm{SL}_2(t)$, where $t\in \{5,7,17\}$.

We note that for the proof of Theorem D-(i), our argument is similar to the argument given in [24, Theorem 1.1]. However, for convenience, we have brought the proof of Theorem D-(i) here.

Let $\sigma ^c(G)=\mathrm{max}\{|\pi (\chi ^c(1))|: \chi \in \mathrm{Irr}(G)\}$ and $\sigma ^e(G)=\mathrm{max}\{|\pi (o(g))|: g \in G\}$. Using [15, P. 463], there exists a function C defined on the positive integers such that for every finite solvable group G, $|\pi (G)| \le C(\sigma ^e(G))\sigma ^e(G)$. It follows from [15, P. 464] that there exists an upper bound K for C. By [17, Corollary 1.4], $|\pi (G)|\le (K+3/2)\sigma ^c(G)$. A. Moreto in [18, Theorem B] proved that if G is solvable such that every prime p divides at most k members of $\mathrm{Codeg}(G)$, then $|\pi (G)| \le 24k \mathrm{log}_2k+360k$. Here, considering Theorem D leads us to prove Theorem E and propose the question after Theorem E:

Theorem E

Let G be non-solvable, and every vertex in $\Delta (G)$ be adjacent to at most m other vertices. Then, $m \ge 2$ and $|\pi (G)| \le 2(m+1)$. In particular, if $|\pi (G)|=2(m+1)$, then G is a Frobenius group and has the normal subgroups L and S such that $L \le S$, $|\pi (L)|=|\pi (G/L)|=m+1$, $[G:S] \mid 2$ and $S/L \cong U\times \mathrm{SL}_2(5)$, where $\pi (U) \cap \{2,3,5\}=\emptyset$ and $|\pi (U)|=m-2$.

Question

Let G be solvable and every vertex in $\Delta (G)$ be adjacent to at most m other vertices. Is there an upper bound for $|\pi (G)|$ in terms of a function respect to m (independent of G)?

2 Proof of the Main Results

In this section, we first present a lemma that will be used frequently in this paper.

Lemma 2.1

[20, Lemma 2.1] Let N be a normal subgroup of G. Then, $\mathrm{Codeg}(G/N) \subseteq \mathrm{Codeg}(G)$. Also, if $\psi \in \mathrm{Irr}(N)$, then for every $\chi \in \mathrm{Irr}(\psi ^G)$, $\psi ^c(1) \mid \chi ^c(1)$.

Corollary 2.2

If N is a normal subgroup of G, then $\Delta (N)$ and $\Delta (G/N)$ are sub-graphs of $\Delta (G)$.

Proof

The proof follows immediately from Lemma 2.1. $\square$

Lemma 2.3

Let $p,q \in \pi (G)$.

• (i) [20, Theorem A] There exists a $\chi \in \mathrm{Irr}(G)$ such that $p \mid \chi ^c(1)$. In particular, the vertex set of $\Delta (G)$ is $\pi (G)$.

• (ii) [20, Theorem E] If p and q are not adjacent in $\Delta (G)$, then G has no elements of order pq.

• (iii) [4, Theorem A] If p and q are not adjacent in $\Delta (G)$, then $\mathrm{Fit}(G) \ne \{1\}$. In particular, if $\mathrm{Fit}(G)= \{1\}$, then $\Delta (G)$ is complete.

• (iv) [20, Theorem C] If ${\Delta }(G)$ is disconnected, then $\Delta (G)$ has exactly two connected components and G is either a Frobenius group or a 2-Frobenius group.

• (v) [20, Theorem E] The diameter of ${\Delta }(G)$ is at most 3, i.e., for every $\Delta \subseteq \pi (G)$ with $|\Delta | \ge 3$, there exist $p,q \in \Delta$ such that p and q are adjacent in $\Delta (G)$.

2.1 Proof of Theorem A

For $r \in \pi (G)$, let $\pi _r$ denote the set of vertices of $\Delta (G)$ that are not adjacent to r. In order to prove Theorem A, we first prove the following lemmas:

Lemma 2.4

Let L be a solvable normal subgroup of G. If $p,q \in \pi (L)$ are distinct, then there exists a normal subgroup N of G such that $N \le L$, $r \mid |N|$ and $s \not \mid |N|$, where $\{p,q\}=\{r,s\}$.

Proof

We complete the proof by induction on |G|. Let M be a minimal normal subgroup of G such that $M \le L$. Since M is an elementary abelian t-group, for some prime t and $p,q \in \pi (L)$, $M \ne L$. If $t \in \{p,q\}$, then the lemma follows. Otherwise, $p,q \in \pi (L/M)$. By induction, there exists a normal subgroup N/M of G/M such that $N/M \le L/M$, $r \mid |N/M|$ and $s \not \mid |N/M|$, where $\{p,q\}=\{r,s\}$. Therefore, $N \unlhd G$, $N \le L$, $r \mid |N|$ and $s \not \mid |N|$, as wanted. $\square$

Lemma 2.5

Let N be the solvable radical of G and S/N be a minimal normal subgroup of G/N. If $p\in \pi (S/N)$ such that $\pi _p\ne \emptyset$, then a Sylow p-subgroup of G is cyclic or $p=2$ and a Sylow 2-subgroup of G is a generalized quaternion group.

Proof

Let $q \in \pi _p$. Since $\mathrm{Fit}(G/N)=\{N\}$, Lemma 2.3(iii) shows that $\Delta (G/N)$ is complete. So, Corollary 2.2 forces $q \in \pi (N)-\pi (G/N)$. First, let $p \in \pi (N)$. By Lemma 2.4, there exists a normal subgroup K of G such that $K \le N$, $r \mid |K|$ and $t \not \mid |K|$, where $\{r,t\}=\{p,q\}$. Let $R \in \mathrm{Syl}_r(K)$. By the Frattini argument, $G/K \cong N_G(R)/N_K(R)$. As $t \not \mid |K|$, there exists $T \in \mathrm{Syl}_t(G)$ such that $T \le N_G(R)$. By Lemma 2.3(ii), G has no elements of order rt. Hence, $\langle R,T\rangle$ is a Frobenius group. Therefore, either T is cyclic or $t=2$ and T is a generalized quaternion group. So, if $t=p$, then the lemma follows. Otherwise, $t=q \ne 2$, because $2 \not \in \pi _p$ and $q \in \pi _p$. Thus, T is cyclic. Also, $q \in \pi (N)-\pi (G/N)$. Therefore, $T \in \mathrm{Syl}_t(N)$. By the Frattini argument, $G/N \cong N_G(T)/N_N(T)$. Thus, $N_G(T)$ has a chief factor which is isomorphic to S/N. Obviously, S/N is a direct product of some isomorphic non-abelian simple groups. On the other hand, $N_G(T)/C_G(T) \lesssim \mathrm{Aut}(T)$ and $\mathrm{Aut}(T)$ is abelian. This forces $C_G(T)$ to have a chief factor isomorphic to S/N. Thus, $p \in \pi (S/N) \subseteq \pi (C_G(T))$. This implies that G has an element of order pq. This is a contradiction with Lemma 2.3(ii). Next, assume that $p \not \in \pi (N)$ and $Q \in \mathrm{Syl}_q(N)$. By the Frattini argument, $G/N \cong N_G(Q)/N_N(Q)$. So, there exists a $P \in \mathrm{Syl}_p(G)$ such that $P \le N_G(Q)$. By Lemma 2.3(ii), G has no elements of order pq. Hence, $\langle P,Q\rangle$ is a Frobenius group. Therefore, either P is cyclic or $p=2$ and P is a generalized quaternion group. $\square$

Proof of Theorem A

Since $\pi _2 \ne \emptyset$, Lemma 2.5 forces a Sylow 2-subgroup of G to be cyclic or a generalized quaternion group. However, G is non-solvable. So, a Sylow 2-subgroup of G is a generalized quaternion group. It follows from Brauer–Suzuki theorem [7] that a Sylow 2-subgroup of G/N is a dihedral group. Hence, $2 \mid |N|$ and G/N has the unique minimal normal subgroup S/N. By [10] and [11], $S/N \cong \mathrm{Alt}_7$ or $\mathrm{PSL}_2(q)$, where q is an odd prime power. Now, the proof is complete. $\square$

2.2 Proof of Theorem B

Here, we first prove the following interesting and useful lemmas.

Lemma 2.6

Let N be the solvable radical of G and S/N be a minimal normal subgroup of G/N. If a Sylow p-subgroup of G is cyclic, for some $p \in \pi (N) \cap \pi (S/N)$, then S/N is a simple group and p divides the order of the Schur multiplier of S/N.

Proof

Since $p \mid |S/N|$ and a Sylow p-subgroup of G is cyclic, the Sylow p-subgroups of G/N are cyclic. However, S/N is a minimal normal subgroup G/N. Hence, S/N is a direct product of some isomorphic non-abelian simple groups. Nevertheless, S/N is a simple group. Let $K_1$ be the largest normal subgroup of G such that $K_1 \le N$ and $p \mid |N/K_1|$. Let $K_2/K_1$ be a minimal normal subgroup of $G/K_1$ such that $K_2 \le N$. By our assumption, $p \not \mid |N/K_2|$. Hence, $K_2/K_1$ is an elementary abelian p-group. Since a Sylow p-subgroup of G is cyclic, $K_2/K_1$ is cyclic. Consequently, $|K_2/K_1|=p$. We claim that $C_{N/K_1}(K_2/K_1)=K_2/K_1$. If not, then since $p \not \mid |N/K_2|$, $N/K_1$ has a $p'$-subgroup $K_3/K_1$ such that $C_{N/K_1}(K_2/K_1)=(K_2/K_1)(K_3/K_1)$, by Schur–Zassenhaus theorem. However, $K_2/K_1 \le Z(C_{N/K_1}(K_2/K_1))$. Thus, $C_{N/K_1}(K_2/K_1)=(K_2/K_1)\times (K_3/K_1)$. So, $K_3/K_1$ is normal in $G/K_1$, $K_1 < K_3 \le N$ and $p \mid |N/K_3|$. This is a contradiction with our assumption on $K_1$. Therefore,

$$\begin{aligned} C_{N/K_1}(K_2/K_1)=K_2/K_1. \end{aligned}$$

(1)

On the other hand, $(S/K_1)/C_{S/K_1}(K_2/K_1) \lesssim \mathrm{Aut}(K_2/K_1)$ and $\mathrm{Aut}(K_2/K_1)$ is a cyclic group. Hence, $C_{S/K_1}(K_2/K_1)$ has a chief factor isomorphic to S/N. By (1), $\frac{C_{S/K_1}(K_2/K_1)}{K_2/K_1} = \frac{C_{S/K_1}(K_2/K_1)}{C_{N/K_1}(K_2/K_1)} \cong \frac{C_{S/K_1}(K_2/K_1)N/K_1}{N/K_1} \unlhd \frac{S/K_1}{N/K_1} \cong S/N$. Since S/N is simple, $\frac{C_{S/K_1}(K_2/K_1)}{K_2/K_1} \cong S/N$. Also, $K_2/K_1 \le Z(C_{S/K_1}(K_2/K_1))$ and $|K_2/K_1|=p$. So, $(C_{S/K_1}(K_2/K_1))' \cap K_2/K_1=\{K_1\}$ or $K_2/K_1$. Therefore, either $(C_{S/K_1}(K_2/K_1))'\cong S/N$ and $C_{S/K_1}(K_2/K_1)=K_2/K_1 \times (C_{S/K_1}(K_2/K_1))'$ or $(C_{S/K_1}(K_2/K_1))'=C_{S/K_1}(K_2/K_1)$. The former case is impossible, because $p \mid |S/N|,|K_2/K_1|$ and a Sylow p-subgroup of G is cyclic. In the latter case, $C_{S/K_1}(K_2/K_1)$ is the Schur cover of S/N. Hence, $|K_2/K_1|=p$ divides the order of the Schur multiplier of S/N, as desired. $\square$

Lemma 2.7

Let N be the solvable radical of G and S/N be a minimal normal subgroup of G/N. If a Sylow 2-subgroup of G is a generalized quaternion group, then S/N is a simple group and G has a normal subgroup M such that $M \le N$, $2 \not \mid |M|$ and S/M is a Schur cover of S/N.

Proof

Since a Sylow 2-subgroup of G is a generalized quaternion group, Brauer–Suzuki theorem [7] shows that a Sylow 2-subgroup of G/N is a dihedral group. So, $2 \mid |N|$ and S/N is simple. Suppose that $N_1$ is the largest normal subgroup of G such that $N_1 \le N$ and $2 \not \mid |N_1|$. Then, $G/N_1$ has the largest normal subgroup $L/N_1$ which is a 2-group and $L \le N$. Since G/N is non-solvable, $4 \mid |G/N|$. Hence, $4 \mid [G/N_1:L/N_1]$. Also, a Sylow 2-subgroup of $G/N_1$ is a generalized quaternion group. So, $L/N_1$ is cyclic. Thus, $\mathrm{Aut}(L/N_1)$ is an abelian group of order $|L/N_1|/2$. On the other hand, $\frac{N/N_1}{C_{N/N_1}(L/N_1)} \lesssim \mathrm{Aut}(L/N_1)$. We claim that $L=N$. If not, G/L has a normal minimal subgroup K/L such that $K \le N$. Obviously, $2 \not \mid |K/L|$. Therefore, $K/N_1 \le C_{N/N_1}(L/N_1)$. By Schur–Zassenhaus theorem, there exists a $2'$-subgroup $K_1/N_1$ of $K/N_1$ such that $(K/N_1)=(L/N_1)(K_1/N_1)$. Since $L/N_1 \le Z(C_{N/N_1}(L/N_1))$ and $K/N_1 \le C_{N/N_1}(L/N_1)$, $(K/N_1)=(L/N_1) \times (K_1/N_1)$. Therefore, $K_1/N_1$ is normal in $G/N_1$, $N_1 <K_1 \le N$ and $2 \not \mid |K_1|$. This is a contradiction with our assumption on $N_1$. This forces $L=N$. However, $\frac{S/N_1}{C_{S/N_1}(N/N_1)} \lesssim \mathrm{Aut}(N/N_1)$ and $\mathrm{Aut}(N/N_1)$ is abelian. Thus, $C_{S/N_1}(N/N_1)$ is non-solvable. Also, $N/N_1 < C_{S/N_1}(N/N_1)$ and $\frac{C_{S/N_1}(N/N_1)}{N/N_1} \unlhd \frac{S/N_1}{N/N_1} \cong S/N$. Therefore, $C_{S/N_1}(N/N_1)=S/N_1$. Set $N_2/N_1=(N/N_1) \cap (S/N_1)'$. Then, $S/N_2 =(S'N_2/N_2) \times (N/N_2)$. Consequently,

$$\begin{aligned} (P\cap S')N_2/N_2 \times N/N_2=PN_2/N_2, \end{aligned}$$

(2)

where $P \in \mathrm{Syl}_2(S)$. Since $P \cap S'$ is not cyclic, we get that P is a generalized quaternion group. It follows from (2) that $N/N_2=\{N_2\}$. So, $N/N_1 \le (S/N_1)'$. This forces $(S/N_1)'=S/N_1$. Note that $N/N_1 \le Z(C_{S/N_1}(N/N_1))=Z(S/N_1)$, considering $C_{S/N_1}(N/N_1)=S/N_1$. Therefore, $S/N_1$ is a Schur cover of S/N, as desired. $\square$

Proof of Theorem B

Since 2 is incomplete in $\Delta (G)$, Theorem A forces G/N to have the unique minimal normal subgroup S/N and $S/N \cong \mathrm{Alt}_7$ or $\mathrm{PSL}_2(q)$, where q is a power of an odd prime p. Let $t \in \pi (S/N)-\{2\}$. By Lemma 2.5, a Sylow t-subgroup of G is cyclic. Hence, a Sylow t-subgroup of S/N is cyclic. This shows that the Sylow subgroups of S/N of odd orders are cyclic. However, a Sylow 3-subgroup of $\mathrm{Alt}_7$ is not cyclic and a Sylow p-subgroup of $\mathrm{PSL}_2(q)$ is cyclic if and only if $q=p$. Therefore, $S/N \cong \mathrm{PSL}_2(p)$. Since S/N is the unique minimal normal subgroup of G/N, $C_{G/N}(S/N)=\{N\}$. Consequently, $G/N \lesssim \mathrm{Aut}(S/N)=\mathrm{Aut}(\mathrm{PSL}_2(p)) \cong \mathrm{PGL}_2(p)$. So, $G/N \cong \mathrm{PSL}_2(p)$ or $\mathrm{PGL}_2(p)$. Therefore, $[G:S]=[G/N:S/N]$ divides 2.

By Lemma 2.5, a Sylow 2-subgroup of G is a generalized quaternion group and for every $t \in \pi (G/N)-\{2\}$, a Sylow t-subgroup of G is cyclic. Also, the order of the Schur multiplier of $S/N \cong \mathrm{PSL}_2(p)$ is 2. Thus, Lemmas 2.6 and 2.7 guarantee the existence of a normal Hall subgroup M of G such that $S/M \cong \mathrm{SL}_2(p)$ and $N/M \le Z(S/M)$ is of order 2. In particular,

$$\begin{aligned} \pi (M)=\pi (G) -\pi (G/N),~ N/M \le Z(G/M)~\mathrm{and}~[N:M]=2. \end{aligned}$$

(3)

Now, suppose that K is the smallest non-solvable normal subgroup of G. Obviously, $K \le S$ and KM is normal in S. So, $K/(K\cap M) \cong KM/M$ is normal in $S/M \cong \mathrm{SL}_2(p)$. Set $L =K \cap M$. As K is non-solvable, $KM=S$ and $K/L \cong \mathrm{SL}_2(p)$. Thus, $S/L=KM/L =K/L \times M/L \cong \mathrm{SL}_2(p) \times M/L$. We claim that $\Delta (L)$ is complete. Working towards a contradiction, suppose that there are $r,s \in \pi (L)$ which are not adjacent in $\Delta (L)$. Then, $r,s \ne 2$. Without loss of generality, we can apply the same argument given in the proof of Lemma 2.5 to conclude that a Sylow r-subgroup of G is cyclic. Let T be the largest normal subgroup of G such that $T \le L$ and $r \mid |L/T|$. Assume that $L_1/T$ is a minimal normal subgroup of G/T such that $L_1 \le L$. By our assumption on T, $L_1/T$ is an elementary abelian r-group and $r \not \mid |L/L_1|$. Since a Sylow r-subgroup of G is cyclic, $|L_1/T|=r$. Also, $L_1/T$ is normal in K/T. Hence, $\frac{K/T}{C_{K/T}(L_1/T)} \lesssim \mathrm{Aut}(L_1/T)$ and $\mathrm{Aut}(L_1/T)$ is a cyclic group. Therefore, $C_{K/T}(L_1/T)$ is non-solvable. So, our assumption on K forces

$$\begin{aligned} C_{K/T}(L_1/T)=K/T. \end{aligned}$$

(4)

On the other hand, the assumption on T forces $r \not \mid |L/L_1|$. By (3), $r \not \mid |K/L|$. Therefore, $r \not \mid |K /L_1|$. It follows from Schur–Zassenhaus theorem that there exists a subgroup $K_1/T$ of K/T such that $r\not \mid |K_1/T|$ and $K/T=(K_1/T)(L_1/T)$. By (4), $L_1/T \le Z(K/T)$. This shows that $K/T=(K_1/T) \times (L_1/T)$. So, $K' \le K_1$ is a non-solvable normal subgroup of G whose order is strictly less than |K|. This is a contradiction with our assumption on K. This guarantees that $\Delta (L)$ is complete, as desired. $\square$

2.3 Proof of Theorem C

In this sub-section, let p be an odd prime. In the following, we first bring some lemmas and a proposition that will be used in the proof of Theorem C.

Lemma 2.8

[22, Proposition 2.2] Let $Z \unlhd G$, $\lambda \in \mathrm{Irr}(Z)$ and G/Z be solvable. If $x \in I_G(\lambda )$, then there exists a $\chi \in \mathrm{Irr}(\lambda ^G)$ such that $o(xZ) \chi (1) \mid [G:Z] \lambda (1)$, where o(xZ) denotes the order of xZ in G/Z.

Lemma 2.9

Let S be a normal subgroup of G such that $|Z(G)|=2$, $[G:S]=l \mid 2$, $S \cong \mathrm{SL}_2(p)$ and $G/Z(G) \cong \mathrm{PSL}_2(p)$ or $\mathrm{PGL}_2(p)$. Then:

• (i) for every $x \in G$, $o(x) \in \{p,lp\}$ or $o(x) \mid l(p\pm 1)$;

• (ii) G admits the faithful irreducible characters whose degrees are $p+1$ and $p-1$.

Proof

The proof follows by checking the character table of $\mathrm{SL}_2(p)$ (see [23]). $\square$

Now, we fix the following hypothesis:

Hypothesis $*$. Let N, U and S be normal subgroups of G such that $N \le U \le S$, $U/N \le Z(G/N)$ is of order 2, $\mathrm{gcd}(|N|,|G/N|)=1$, [G : S] divides 2, $S/N \cong \mathrm{SL}_2(p)$ and $G/U \cong \mathrm{PSL}_2(p)$ or $\mathrm{PGL}_2(p)$, where p is an odd prime.

Proposition 2.10

Let N, U and S be the normal subgroups of G satisfying Hypothesis $*$ and let $2 \mid |C_G(N)|$. Then:

• (i) G has the unique involution z. In particular, $z \in Z(G)$.

• (ii) If $\psi _1,\psi _2 \in \mathrm{Irr}(G)$ such that $2 \mid |\mathrm{ker}\psi _1|$ and $2 \not \mid |\mathrm{ker}\psi _2|$, then $2 \not \mid |\mathrm{ker}\psi _1\psi _2|$.

• (iii) If $H \le G$ and $\psi \in \mathrm{Irr}(H)$ such that $2 \not \mid |\mathrm{ker}\psi |$, then $2 \not \mid |\mathrm{ker} \psi ^G|$.

Proof

Since |N| is odd and $2 \mid |C_G(N)|$, $2 \mid |C_G(N)N/N|$. However, $C_G(N)N/N$ is a normal subgroup of G/N. Thus, $C_G(N)N/N\in \{G/N,S/N\}$ or $C_G(N)N/N$ is a cyclic group of order 2. This shows that $C_G(N)N/N$ has the unique central involution. On the other hand, $\mathrm{gcd}(|N|,|G/N|)=1$ and $C_G(N)N/N \cong C_G(N)/Z(N)$. Hence, $\mathrm{gcd}(|Z(N)|,|C_G(N)/Z(N)|)=1$. By Schur–Zassenhaus theorem, $C_G(N)$ has a subgroup H such that $H \cong C_G(N)/Z(N) \cong C_G(N)N/N$, $H \cap Z(N) =\{1\}$ and $C_G(N)=H Z(N)$. So, $C_G(N)=H \times Z(N)$. We note that $H \cap N=\{1\}$, because $\mathrm{gcd}(|N|,|H|)=1$. As $H \cong HN/N=HZ(N)N/N=C_G(N)N/N$ and $C_G(N)N/N$ has the unique central involution, we get that H has the unique central involution. Thus, $C_G(N)$ has the unique central involution z. Since $C_G(N)$ is a normal subgroup of G, $\langle z\rangle$ is normal in G. Therefore, $z \in Z(G)$. Also, a Sylow 2-subgroup of G is a generalized quaternion group. Hence, z is the unique involution of G, as desired in (i). Now, let $\psi _1,\psi _2 \in \mathrm{Irr}(G)$ such that $2 \mid |\mathrm{ker}\psi _1|$ and $2 \not \mid |\mathrm{ker}\psi _2|$. Thus, $z \in \mathrm{ker}\psi _1$ and $z \not \in \mathrm{ker}\psi _2$. So, $z \not \in \mathrm{ker}\psi _1\psi _2$. Since z is the unique involution of G, we get that $2 \not \mid |\mathrm{ker}\psi _1\psi _2|$, as wanted in (ii). Finally, let $H \le G$ and $\psi \in \mathrm{Irr}(H)$ such that $2 \not \mid |\mathrm{ker}\psi |$. Since $z \in Z(G)$, $\psi ^G(z)=[G:H] \dot{\psi }(z)$. However, $z \not \in \mathrm{ker}\psi$. Hence, $\dot{\psi }(z) \ne \psi (1)$, so $\psi ^G(1) \ne \psi ^G(z)$. Consequently, $z \not \in \mathrm{ker}\psi ^G$. As z is the unique involution of G, we have $2 \not \mid |\mathrm{ker}\psi ^G|$. Now, the proof is complete. $\square$

Proposition 2.11

Let N, U and S be normal subgroups of G satisfying Hypothesis $*$ and let M be a normal subgroup of G such that $M \le N$. If $\lambda \in \mathrm{Irr}(M)$ and $x \in I_G(\lambda )$, then there exists a $\chi \in \mathrm{Irr}(\lambda ^G)$ such that $o(xM) \chi (1) \mid [G:M] \lambda (1)$ and if $2 \mid |C_G(N)|$, then $2 \not \mid |\mathrm{ker}\chi |$.

Proof

First let $I_G(\lambda ) \ne G$. If $I_G(\lambda )$ is non-solvable, then since $I_G(\lambda )N/N \le G/N$, we get that $I_S(\lambda )/(N \cap I_S(\lambda )) \cong \mathrm{SL}_2(r)$ for some prime r. So, the inductive hypothesis and Lemma 2.8 guarantee the existence of $\psi \in \mathrm{Irr}(I_G(\lambda ))$ such that

$$\begin{aligned} o(xM) \psi (1) \mid [I_G(\lambda ):M] \lambda (1). \end{aligned}$$

(5)

Also, if $I_G(\lambda )$ is non-solvable and $2 \mid |C_{I_G(\lambda )}(I_N(\lambda ))|$, then $2 \not \mid |\mathrm{ker} \psi |$. Recall that if $2 \mid |C_G(N)|$, then G has a central involution z and hence, $z \in C_{I_G(\lambda )}(I_N(\lambda ))$. Thus, if $I_G(\lambda )$ is non-solvable and $2 \mid |C_G(N)|$, then $2 \not \mid |\mathrm{ker}\psi |$. Next, let $I_G(\lambda )$ is solvable and $2 \mid |C_G(N)|$. By Proposition 2.10(i), G has the unique involution z which is central in G. Thus, $M \times \langle z\rangle \trianglelefteq G$. Let $\mathrm{Irr}(\langle z\rangle )=\{1_{\langle z\rangle },\lambda _2\}$. Since $(\lambda \lambda _2)(1)=\lambda (1)$, $I_G(\lambda \lambda _2)=I_G(\lambda )$ and $\mathrm{Irr}((\lambda \lambda _2)^{I_G(\lambda )})\subseteq \mathrm{Irr}(\lambda ^{I_G(\lambda )})$, we can replace $\lambda$ by $\lambda \lambda _2$ in the above argument. Since $\psi \in \mathrm{Irr}((\lambda \lambda _2)^{I_G(\lambda )})$, $z \not \in \mathrm{ker}\psi$. Also, z is the unique involution of G. Thus, $2 \not \mid |\mathrm{ker}\psi |$. Clifford correspondence [13, Theorem 6.11] forces $\chi =\psi ^G \in \mathrm{Irr}(\lambda ^G)$ and $\chi (1)=[G:I_G(\lambda )]\psi (1)$. It follows from (5) that

$$\begin{aligned} o(xM)\chi (1)= & {} o(xM)[G:I_G(\lambda )] \psi (1) \mid ([I_G(\lambda ):M] \lambda (1))[G:I_G(\lambda )] \!=\! [G:M] \lambda (1). \end{aligned}$$

Further, Proposition 2.10(iii) shows that if $2 \mid |C_G(N)|$, then $2\not \mid |\mathrm{ker}\chi |$, as wanted. Now, assume that $I_G(\lambda )=G$. If $N=M$, then since $\mathrm{gcd}(|G/N|,|N|)=1$, we get from [13, Exercise 6.18] that there exists a $\chi \in \mathrm{Irr}(\lambda ^G)$ such that $\chi _N=\lambda$. By Lemma 2.9, there exists a faithful character $\psi _2 \in \mathrm{Irr}(G/N)$ such that $o(xN)\psi _2(1) \mid |G/N|$. It follows from [13, Corollary 6.17] and Proposition 2.10(ii) that $\chi \psi _2 \in \mathrm{Irr}(\lambda ^G)$ and $2 \not \mid |\mathrm{ker}\psi \psi _2|$. Therefore, $o(xN)(\chi \psi _2)(1)=o(xN)\psi _2(1)\lambda (1)$ divides $[G:N]\lambda (1)$, as needed. Finally, assume that $N \ne M$ and L/M is a chief factor of G such that $L\le N$. Then, L/M is an elementary abelian p-group for some prime p. Using [13, Exercise 6.12], $\lambda ^L=e\theta$, where $e^2=[L:M]$ and $\theta \in \mathrm{Irr}(L)$ or $\lambda ^L=\theta _1+\cdots +\theta _t$, where $\theta _1, \ldots ,\theta _t \in \mathrm{Irr}(L)$, $\theta _1(1)=\cdots =\theta _t(1)=\lambda (1)$ and $t=[L:M]$. Now, rewriting the argument given in the proof of [22, Proposition 2.2] guarantees that there exists $\psi \in \mathrm{Irr}(\lambda ^L)$ and a divisor $t_0$ of o(xL) such that $x^{t_0} \in I_G(\psi )$, where either $t_0=1$ or $\lambda ^L=\theta _1+\cdots +\theta _t$, $t_0$ is a power of p and $t_0 o(xM)/o(xL) \mid t=[L:M]$. By the inductive hypothesis, there exists $\chi \in \mathrm{Irr}(\psi ^G)$ such that

$$\begin{aligned} o(x^{t_0}L)\chi (1) \mid [G:L]\psi (1) \end{aligned}$$

(6)

and if $2 \mid |C_G(N)|$, then $2 \not \mid |\mathrm{ker}\chi |$. If $t_0=1$, then since $p\psi (1)/\lambda (1) \mid |L/M|$ and $o(xM) \mid po(xL)$, we have $o(xM) \chi (1) \mid [G:M] \lambda (1)$, regarding (6). Otherwise, $\psi (1)=\lambda (1)$ and $o(x^{t_0}L)=o(xL)/t_0$. Thus, $o(xM)=t_0o(x^{t_0}L)o(xM)/o(xL) \mid [L:M]o(x^{t_0}L)$. It follows from (6) that $o(xM) \chi (1) \mid [G:M] \lambda (1)$, as desired. $\square$

Proposition 2.12

Let N, U and S be normal subgroups of G satisfying Hypothesis $*$. Then, for every $x \in G$, G admits an irreducible character whose co-degree is divisible by o(x).

Proof

We complete the proof by induction on |G|. Let $\{N_1,\ldots ,N_t\}$ be the set of all minimal normal subgroups of G which are subgroups of N. Then, for every $1 \le i \le t$, $N_i$ is an elementary abelian $p_i$-group, for some prime $p_i$. If there exists an $1 \le i \le t$ such that $o(xN_i)=o(x)$, then we get from induction that there exists a $\chi \in \mathrm{Irr}(G/N_i)$ such that $o(x)=o(xN_i) \mid \chi ^c(1)$. So, Lemma 2.1 completes the proof. Now, suppose that for every $1 \le i \le t$, $o(xN_i)\ne o(x)$. Set $x_i=x^{o(x)/p_i}$. If $x_i =1$, then $o(x)=p_i$ and hence, Lemma 2.3(i) completes the proof. Next, we assume that $x_i \ne 1$, for every $1 \le i \le t$. Then, $x_i \in N_i$ and hence, $\{p_1,\ldots ,p_t\}$ are pair-wise distinct. Since $x \in C_G(x_i)$, we get from [13, Theorem 6.32] that there exists a non-principal character $\lambda _i \in \mathrm{Irr}(N_i)$ such that $x \in I_G(\lambda _i)$. Set $\lambda =\lambda _1 \times \cdots \times \lambda _t$ and $M=N_1 \times \cdots \times N_t$. Then, $M \unlhd G$ and $\lambda \in \mathrm{Irr}(M)$. By Proposition 2.11, there exists a $\chi \in \mathrm{Irr}(\lambda ^G)$ such that

$$\begin{aligned} o(xM)\chi (1) \mid [G:M] \lambda (1). \end{aligned}$$

(7)

Also, if $2 \mid |C_G(N)|$, then $2 \not \mid |\mathrm{ker}\chi |$. If $\mathrm{ker} \chi \ne \{1\}$, then since $\mathrm{ker} \chi \unlhd G$, there exists a minimal normal subgroup L of G such that $L \le \mathrm{ker} \chi$. If $L \le N$, then $L \in \{N_1,\ldots ,N_t\}$. So, $L=N_i$, for some $1 \le i \le t$. Thus, $\chi _{N_i}=\chi (1)1_{N_i}$. This is a contradiction, because $\chi \in \mathrm{Irr}(\lambda _i^G)$. This shows that $L \not \le N$. However, $L \ne L \cap N \unlhd G$ and L is a minimal normal subgroup of G. Therefore, $L \cap N=\{1\}$. So, $L \times N \unlhd G$ and $L \cong LN/N \unlhd G/N$. Hence, considering the normal subgroups of G/N forces $2 \mid |Z(L)|$. Consequently, $2 \mid |C_G(N)|$ and $2 \mid |\mathrm{ker} \chi |$, which is a contradiction. This shows that $\mathrm{ker} \chi = \{1\}$. On the other hand, $(p_1 \ldots p_t)o(xM)=o(x)$ and $(p_1 \ldots p_t)[G:M]=(p_1 \ldots p_t)|G|/(|N_1|\ldots |N_t|) \mid |G|$. It follows from (7) that $o(x) \chi (1) \mid |G| \lambda (1)=|G|$. Consequently, o(x) divides $\chi ^c(1)$, as wanted. $\square$

We note that in some parts of the proof of Proposition 2.12, we follow the ideas given in the proof of [22, Theorem B].

Proof of Theorem C

If G is solvable, then [22, Theorem B] completes the proof. Now, let G be non-solvable. By Theorem C, G has the normal subgroups N, U and S satisfying Hypothesis $*$. So, the theorem follows from Proposition 2.12. $\square$

2.4 Proof of Theorem D

First let $m=0$. Then, every connected component of $\Delta (G)$ is a single point. However, $\Delta (G)$ has at most two connected components, by Lemma 2.3(iv). It follows that $|\pi (G)| =1$ or $\Delta (G)$ is disconnected and $|\pi (G)| =2$. Thus, G is solvable. Also, if $\Delta (G)$ is disconnected, then Lemma 2.3(iv) shows that G is a Frobenius or a 2-Frobenius group.

Now, let $m=1$. Then, every connected component of $\Delta (G)$ is a path of length 1. However, $\Delta (G)$ has at most two connected components, by Lemma 2.3(iv). It follows that $|\pi (G)| =2$ or $\Delta (G)$ is disconnected and $|\pi (G)| =4$. Thus, if $\Delta (G)$ is connected, then G is solvable. If $\Delta (G)$ is disconnected, then Lemma 2.3(iv) shows that G is a Frobenius or a 2-Frobenius group. If G is a Frobenius group with the Frobenius complement H, then $\pi (H)$ is a connected component of $\Delta (G)$. Thus, $|\pi (H)|=2$. Hence, H is solvable. Since the kernel of a Frobenius group is nilpotent, G is solvable. Also, the 2-Frobenius groups are solvable. Nevertheless, $|\pi (G)| \in \{2,4\}$ and G is solvable, as wanted.

Next, assume that $m=2$. Thus, every connected component of $\Delta (G)$ is a cycle. By Lemma 2.3(v), diameter of $\Delta (G)$ is at most 3. Thus, every connected component of $\Delta (G)$ contains at most 5 vertices. However, $\Delta (G)$ has at most two connected components, by Lemma 2.3(iv). It follows that $|\pi (G)| \le 5$ or $\Delta (G)$ is disconnected and $|\pi (G)| \le 10$. In the latter case, Lemma 2.3(iv) shows that G is either a Frobenius or a 2-Frobenius group. Note that $m < |\pi (G)|-1$ and the co-degree graph of a simple group is complete. So, 2-regularity of $\Delta (G)$ and [19, Theorem 18.6] shows that G is solvable or G is a Frobenius group with the abelian kernel N such that G/N has a normal subgroup of index at most 2 which is isomorphic to $\mathrm{SL}_2(5)$.

Assume that $m = 3$ and G is non-solvable. Since $m < |\pi (G)|-1$, every vertex of $\Delta (G)$ is incomplete. It follows that from Theorem B that G has the normal subgroups N and S such that $\mathrm{gcd}(|N|,|G/N|)=1$, $[G:S] \mid 2$ and $S/N \cong \mathrm{SL}_2(p)$. However, $\Delta (\mathrm{SL}_2(p))$ is complete, $|\pi (\mathrm{SL}_2(p))| \ge 3$ and by Lemma 2.2, $\Delta (S/N)$ is a sub-graph of $\Delta (G)$. So, $|\pi (\mathrm{SL}_2(p))| \le 4$.

If $|\pi (\mathrm{SL}_2(p))| =4$, then for every $r \in \pi (N)$ and $s \in \pi (S/N)$, rs divides none of the elements of $\mathrm{Codeg}(G)$. It follows from Lemma 2.3(ii) that $\mathrm{SL}_2(p)$ acts fixed point freely on N. Therefore, S is a non-solvable Frobenius group with the kernel N and its complement is isomorphic to S/N. Thus, N is abelian and it follows from [19, Theorem 18.6] that $S/N \cong \mathrm{SL}_2(5) \times L$, where L is a group whose order is a prime power. This is a contradiction, because $S/N \cong \mathrm{SL}_2(p)$ and $|\pi (\mathrm{SL}_2(p))|=4$.

Next, let $|\pi (\mathrm{SL}_2(p))| = 3$. Applying [12], $p \in \{5,7,17\}$. By Theorem B, there exists a normal subgroup K of G such that $\Delta (K)$ is complete and $S/K \cong N/K \times S/N$. If $N/K \ne \{K\}$, then since $\Delta (S/N)$ is complete, we get that $|\pi (N/K)|=1$, $|\pi (K)|=4$ and S/K acts fixed point freely on K. Therefore, $S/K \cong \mathrm{SL}_2(5) \times N/K$, by [19, Theorem 18.6]. Nevertheless, $|\pi (G)|=8$. If $N/K=\{K\}$, then since $\Delta (K)$ is complete and $\Delta (G/K)$ is 2-regular, we get that $|\pi (K)|=3$ and for every $t\in \pi (K)$, there exists exactly one element $r \in \pi (G/K)$ such that t and r are adjacent in $\Delta (G)$. Therefore, $|\pi (G)|=6$, as wanted. $\square$

2.5 Proof of Theorem E

If $\Delta (G)$ has a complete vertex, then $m=|\pi (G)|-1$ and hence, $|\pi (G)| =(m+1) < 2(m+1)$, as wanted. Now, suppose that every vertex of $\Delta (G)$ is incomplete. By Theorem B, G has the normal subgroups L, M and S such that $L\le M \le S$, $[G:S] \mid 2$, $\Delta (L)$ is complete and $S/L \cong \mathrm{SL}_2(p) \times M/L$, for some odd prime p. So,

$$\begin{aligned} \pi (G) = \pi (S/L) \cup \pi (L). \end{aligned}$$

(8)

Let $r \in \pi (\mathrm{SL}_2(p))$ and $t \in \pi (S/L)$. If $t \in \pi (\mathrm{SL}_2(p))$, then Lemma 2.3(iii) and Corollary 2.2 force r and t to be adjacent in $\Delta (G)$. If $t \in \pi (M/L)$, then G/L contains an element of order tr, so does G. By Lemma 2.3(ii), r and t are adjacent in $\Delta (G)$. So, r is adjacent to at least $|\pi (S/L)|-1$ other vertices. Therefore, $|\pi (S/L)|-1 \le m$. Also, $\Delta (L)$ is complete. Hence, every vertex of $\Delta (L)$ is adjacent to at least $|\pi (L)|-1$ other vertices. Consequently, $|\pi (L)|-1 \le m$. It follows from (8) that $|\pi (G)| \le |\pi (S/L)|+|\pi (L)| \le 2(m+1)$, as desired.

Now, suppose that $|\pi (G)| =2(m+1)$. Thus, the above argument shows that $\pi (L) \cap \pi (G/L)=\pi (L) \cap \pi (S/L) =\emptyset$ and $|\pi (L)|=|\pi (S/L)|= m+1$. Therefore, G has a subgroup H such that $H \cong G/L$ and $G=HL$. However, $\Delta (L)$ is complete. This signifies that no element of $\pi (L)$ is adjacent to the elements of $\pi (S/L)=\pi (H)$. Hence, Lemma 2.3(ii) shows that H acts fixed freely on L. So, [19, Theorem 18.6] completes the proof. $\square$