Non-solvable groups whose character degree graph has a cut-vertex. III

Abstract

Let G be a finite group. Denoting by $\textrm{cd}(G)$ the set of the degrees of the irreducible complex characters of G, we consider the character degree graph of G: this, is the (simple, undirected) graph whose vertices are the prime divisors of the numbers in $\textrm{cd}(G)$, and two distinct vertices p, q are adjacent if and only if pq divides some number in $\textrm{cd}(G)$. This paper completes the classification, started in Dolfi et al. (Non-solvable groups whose character degree graph has a cut-vertex. II, 2022. https://doi.org/10.1007/s10231-022-01299-3) and Dolfi et al. (Non-solvable groups whose character degree graph has a cut-vertex. I, 2022. https://doi.org/10.48550/arXiv.2207.10119), of the finite non-solvable groups whose character degree graph has a cut-vertex, i.e., a vertex whose removal increases the number of connected components of the graph. More specifically, it was proved in Dolfi et al. (Non-solvable groups whose character degree graph has a cut-vertex. I, 2022. https://doi.org/10.48550/arXiv.2207.10119 that these groups have a unique non-solvable composition factor S, and that S is isomorphic to a group belonging to a restricted list of non-abelian simple groups. In Dolfi et al. (Non-solvable groups whose character degree graph has a cut-vertex. II, 2022. https://doi.org/10.1007/s10231-022-01299-3) and Dolfi et al. (Non-solvable groups whose character degree graph has a cut-vertex. I, 2022. https://doi.org/10.48550/arXiv.2207.10119) all isomorphism types for S were treated, except the case $S\cong \textrm{PSL}_{2}(2^a)$ for some integer $a\ge 2$; the remaining case is addressed in the present paper.

1 Introduction

The character degree graph $\Delta (G)$ of a finite group G is a very useful tool for studying the arithmetical structure of the set $\textrm{cd}(G)=\{\chi (1):\chi \in \textrm{Irr}(G)\}$, i.e., the set of the irreducible (complex) character degrees of G. As many results in the literature show, there is a profound interaction between the group structure of G and certain graph-theoretical properties (in particular, connectivity properties) of $\Delta (G)$.

In the papers [5, 6] we considered the problem of classifying the finite non-solvable groups G such that $\Delta (G)$ has a cut-vertex, which is a vertex whose removal (together with all the edges incident to it) produces a graph having more connected components than the original. Among the various properties of such a group G, it is proved in [6] that G has a unique non-solvable composition factor S, and that S is isomorphic to one of the simple groups in the following list: the projective special linear group $\textrm{PSL}_{2}(t^a)$ (where $t^a$ is a prime power greater than 3), the Suzuki group $\textrm{Sz}(2^a)$ (where $2^a-1$ is a prime number), $\textrm{PSL}_3(4)$, the Mathieu group $\textrm{M}_{11}$, and the first Janko group $\textrm{J}_1$. The aforementioned papers carry out an analysis (and provide a complete classification) of all the possibilities, except for the case $S\cong \textrm{PSL}_{2}(2^a)$ when $\Delta (G)$ is connected; the present work addresses the remaining case, thus completing the classification of these groups. We refer the reader to [5, 6] for a thorough description of the problem and, in particular, for the full statements of the relevant theorems (see the introductions of [5, 6], and Section 2 of [6]).

The situation that remains to be studied is treated in the following Theorems 1 and 2, which deal with the cases $2^a>4$ and $2^a=4$, respectively (see [6, Theorem A, Case (f)], and [6, Theorem B]), and which are the main results of this paper.

In order to clarify the statements we mention that, for $H = \textrm{SL}_2(t^a)$ (where $t^a$ is a prime power), an H-module V over the field ${\mathbb {F}}_t$ of order t is called the natural module for H if V is isomorphic to the standard module for $\textrm{SL}_2(t^a)$, or any of its Galois conjugates, seen as an ${\mathbb {F}}_t[H]$-module. We will freely use this terminology also referred to the conjugation action of a group on a suitable elementary abelian normal subgroup. For our purposes, it is important to recall that the standard module for $\textrm{SL}_2(t^a)$ is self-dual.

Also, given a finite group G, we denote by $R = R(G)$ the solvable radical (i.e., the largest solvable normal subgroup), and by $K = K(G)$ the solvable residual (i.e., the smallest normal subgroup with a solvable factor group) of G. Equivalently, K(G) is the last term of the derived series of G.

Theorem 1

Let R and K be, respectively, the solvable radical and the solvable residual of the finite group G and assume that G has a composition factor $S \cong \textrm{SL}_{2}(2^a)$, with $a \ge 3$. Then, $\Delta (G)$ is a connected graph and has a cut-vertex p if and only if G/R is an almost simple group with socle isomorphic to S, ${V}(G) = \pi (G/R) \cup \{ p\}$ and one of the following holds.

• (a) $K\cong S$ is a minimal normal subgroup of G; also, either $p=2$ and ${V}(G/K)\cup \pi (G/KR)=\{2\}$, or $p\ne 2$, ${V}(G/K)=\{p\}$, and G/KR has odd order.

• (b) K contains a minimal normal subgroup L of G such that $K/L\cong S$ and L is the natural module for K/L; also, $p\ne 2$, ${V}(G/K)=\{p\}$, G/KR has odd order and, for a Sylow 2-subgroup T of G, we have $T'=(T\cap K)'$.

In all cases, p is a complete vertex and the unique cut-vertex of $\Delta (G)$.

Theorem 2

Let R and K be, respectively, the solvable radical and the solvable residual of the finite group G and assume that G has a composition factor $S \cong \textrm{SL}_{2}(4)$. Then, $\Delta (G)$ is a connected graph and has a cut-vertex p if and only if G/R is an almost simple group with socle isomorphic to S, ${V}(G) = \{2,3,5\} \cup \{ p\}$ and one of the following holds.

• (a) K is isomorphic either to $\textrm{SL}_{2}(4)$ or to $\textrm{SL}_{2}(5)$, and ${V}(G/K) =\{p\}$; if $p=5$, then $K\cong \textrm{SL}_{2}(4)$ and $G=K\times R$.

• (b) K contains a minimal normal subgroup L of G with $|L| = 2^4$. Moreover, $G = KR$ and

  • (i) either L is the natural module for K/L, $p \ne 2$ and ${V}(G/K)=\{p\}$,

  • (ii) or L is isomorphic to the restriction to K/L, embedded as $\Omega _4^-(2)$ into $\textrm{SL}_4(2)$, of the standard module of $\textrm{SL}_4(2)$. Moreover $p = 5$, $G=K\times R_0$ where $R_0=\textbf{C}_{G}(K)$, and ${V}(R_0)={V}(G/K) \subseteq \{ 5\}$.

• (c) K contains a minimal normal subgroup L of G such that K/L is isomorphic to $\textrm{SL}_{2}(5)$, and

  • (i) either L is the natural module for K/L, $p\not =5$ and ${V}(G/K)=\{p\}$,

  • (ii) or L is isomorphic to the restriction to K/L, embedded in $\mathrm{{SL}}_4(3)$, of the standard module of $\textrm{SL}_4(3)$, $p=2$ and ${V}(G/K)\subseteq \{2\}$.

In all cases, p is a complete vertex and the unique cut-vertex of $\Delta (G)$.

To conclude this introduction, we display in Table 1 the graphs related to the groups as in Theorems 1 and 2, so, all the possible connected graphs having a cut-vertex p, of the form $\Delta (G)$ where G is a finite group with a composition factor isomorphic to $\textrm{SL}_{2}(2^a)$, $a\ge 2$. The first row of the table shows the graphs arising from Theorem 1, whereas the second row shows the graphs arising from Theorem 2in the case when p is larger than 5. As regards the remaining graphs coming from Theorem 2, they are displayed in the third row of the table, and they are all the paths of length 2 with vertex set $\{2,3,5\}$. Each of them actually occurs for groups as in Theorem 2(a) (it is enough to consider the direct product $\textrm{SL}_{2}(4)\times R$ where R is a non-abelian q-group, for $q\in \{2,3,5\}$). Also, case (b)(ii) is associated to the path $2-5-3$, and case (c)(ii) to the path $3-2-5$.

All the groups considered in the following discussion will be tacitly assumed to be finite groups.

Table 1 The graphs related to the groups of Theorem 1 and Theorem 2

2 Preliminaries

Given a group G, we denote by $\Delta (G)$ the character degree graph (or degree graph for short) of G as defined in the Introduction. Our notation concerning character theory is standard, and we will freely use basic facts and concepts such as Ito-Michler’s theorem, Clifford’s theory, Gallagher’s theorem, character triples and results about extension of characters (see [8]).

For a positive integer n, the set of prime divisors of n will be denoted by $\pi (n)$, and we simply write $\pi (G)$ for $\pi (|G|)$. If q is a prime power, then the symbol ${\mathbb {F}}_q$ will denote the field of order q.

We start by recalling some structural properties of the groups $\textrm{SL}_{2}(2^a)$.

Remark 2.1

The group $\textrm{SL}_{2}(2^a)=\textrm{PSL}_{2}(2^a)$ has order $2^a(2^a-1)(2^a+1)$, and the proper subgroups of this group are of the following types ([7, II.8.27]):

• (i$_+$) dihedral groups of order $2(2^a+ 1)$ and their subgroups;

• (i$_-$) dihedral groups of order $2(2^a- 1)$ and their subgroups;

• (ii) Frobenius groups with elementary abelian kernel of order $2^a$ and cyclic complements of order $2^a -1$, and their subgroups;

• (iii) $A_4$ when a is even or $A_5$ when 5 divides $|\textrm{SL}_{2}(2^a)|$;

• (iv) $\textrm{SL}_{2}(2^b)$, where b is a proper divisor of a.

When dealing with subgroups of $\textrm{SL}_{2}(2^a)$, we will refer to the above labels to identify the type of these subgroups. By a subgroup of type (i) we will mean a subgroup that is either of type (i$_-$) or of type (i$_+$).

Lemma 2.2

Let $G\cong \textrm{SL}_{2}(2^a)$, where $a\ge 2$. Let u be a prime divisor of $2^a-1$, and let U be a subgroup of G with $|U|=u^b$ for a suitable $b\in {{\mathbb {N}}}-\{0\}$. Then U lies in the normalizer in G of precisely two Sylow 2-subgroups of G.

Proof

See [5, Lemma  2.2].$\square$

Next, some properties of the degree graph of simple and almost-simple groups.

Theorem 2.3

([15, Theorem 5.2]) Let $S \cong \textrm{PSL}_{2}(t^a)$ or $S \cong \textrm{SL}_{2}(t^a)$, with t prime and $a \ge 1$. Let $\rho _{+} = \pi (t^a+1)$ and $\rho _{-} = \pi (t^a-1)$. For a subset $\rho$ of vertices of $\Delta (S)$, we denote by $\Delta _{\rho }$ the subgraph of $\Delta = \Delta (S)$ induced by the subset $\rho$. Then

• (a) if $t=2$ and $a \ge 2$, then $\Delta (S)$ has three connected components, $\{t\}$, $\Delta _{\rho _{+}}$ and $\Delta _{\rho _{-}}$, and each of them is a complete graph.

• (b) if $t > 2$ and $t^a > 5$, then $\Delta (S)$ has two connected components, $\{t\}$ and $\Delta _{\rho _{+} \cup \rho _{-}}$; moreover, both $\Delta _{\rho _{+}}$ and $\Delta _{\rho _{-}}$ are complete graphs, no vertex in $\rho _{+} - \{ 2 \}$ is adjacent to any vertex in $\rho _{-} - \{ 2\}$ and 2 is a complete vertex of $\Delta _{\rho _{+} \cup \rho _{-}}$.

Theorem 2.4

Let G be an almost-simple group with socle S, and let $\delta = \pi (G) - \pi (S)$. If $\delta \ne \emptyset$, then S is a simple group of Lie type, and every vertex in $\delta$ is adjacent to every other vertex of $\Delta (G)$ that is not the characteristic of S. Moreover, if $S\cong \textrm{SL}_{2}(2^a)$ and $a \ge 3$, then any prime in $\pi (G/S)$ is adjacent to every other vertex of $\Delta (G)$, except possibly to 2.

Proof

The first claim is Theorem 3.9 of [6]. As for the second claim, by Theorem A of [16] we see that both $(2^a-1)|G/S|$ and $(2^a+1)|G/S|$ are irreducible character degrees of G. $\square$

Lemma 2.5

Let G be a group and let R be its solvable radical. Assume that G/R is an almost-simple group with socle isomorphic to $\textrm{PSL}_{2}(t^a)$, for a prime t with $t^a> 4$ and $t^a\ne 9$. Then, denoting by K the solvable residual of G, one of the following conclusions holds.

• (a) K is isomorphic to $\textrm{PSL}_{2}(t^a)$ or to $\textrm{SL}_{2}(t^a)$;

• (b) K has a non-trivial normal subgroup L such that K/L is isomorphic to $\textrm{PSL}_{2}(t^a)$ or to $\textrm{SL}_{2}(t^a)$, and every non-principal irreducible character of $L/L'$ is not invariant in K.

Proof

See [5, Lemma  2.5].$\square$

Lemma 2.6

Let G be a group, let R be its solvable radical and K its solvable residual. Assume that L is a normal subgroup of G, contained in K, such that $K/L\cong \textrm{SL}_{2}(2^a)$ with $a \ge 2$, and L is isomorphic to the natural module for K/L. Let T be a Sylow 2-subgroup of KR, let $T_0 = T \cap R$ and $T_1 = T \cap K$. Then $L \le \textbf{Z}(T_0)$. Furthermore, every non-principal T-invariant character $\lambda \in \textrm{Irr}(L)$ extends to $T_1$ and, assuming that $T_0/L$ is abelian, $\lambda$ extends to T if and only if $T'= T_1'$.

Proof

Observe that $L = K \cap R$ is an elementary abelian 2-group of order $2^{2a}$, $T_0$ is normalized by K and $T = T_0T_1$. As $\textbf{Z}(T_0) \cap L$ is non-trivial and normal in K, by the irreducibility of L as a K-module it follows that $L \le \textbf{Z}(T_0)$.

It is well known that $\textbf{N}_{K}(T_1)/L=\textbf{N}_{K/L}(T_1/L)$ is a subgroup of type (ii) of K/L whose order is $2^a\cdot (2^a-1)$ (in fact, $\textbf{N}_{K/L}(T_1/L)$ can be identified with the subgroup of lower-triangular matrices of $\textrm{SL}_{2}(2^a)$); thus we write $\textbf{N}_{K}(T_1) = T_1D$, where D is cyclic of order $2^a-1$. By looking at the action of $T_1$ on the natural module L, we see that $Z = \textbf{Z}(T_1) = (T_1)'$ is a normal subgroup of order $2^a$ of $T_1D$. Since L is a self-dual K-module, we have $|\textbf{C}_{{\widehat{L}}}(T_1)| = |\textbf{C}_{L}(T_1)| = 2^a = |\widehat{L/Z}|$ and hence, as certainly the characters of L/Z are $T_1$-invariant, we conclude that the $T_1$-invariant characters of L are precisely the elements of $\widehat{L/Z}$. They are clearly T-invariant and they extend to $T_1$, because $T_1/Z$ is abelian.

Let $\lambda \in \textrm{Irr}(L)$ be a non-principal T-invariant character and assume that $\lambda$ has an extension $\tau \in \textrm{Irr}(T)$. Since $\tau (1) = \lambda (1) = 1$, we have $T' \le \ker {\tau }$. Assuming that $T_0/L$ is abelian, then $T/L = T_0/L \times T_1/L$ is abelian and $T' \le L$. So, $T' = T' \cap L \le \ker {\tau _L} = \ker {\lambda }$ and, as $\lambda \ne 1_L$, hence $T' < L$. Observing that $T'$ is normalized by D and that D acts irreducibly on L/Z, we conclude that $T' \le Z$ and, since $Z = T_0' \le T'$, that $T' = T_0'$.

Conversely, if $\lambda \in \textrm{Irr}(L)$ is T-invariant, then as observed above $Z \le \ker {\lambda }$ and, assuming $Z = T'$, clearly $\lambda$, seen as a character of L/Z, extends to the abelian group T/Z. $\square$

Remark 2.7

Let K be a group having a normal subgroup L with $K/L\cong \textrm{SL}_{2}(2^a)$ (for $a\ge 2$), and such that L is isomorphic to the natural K/L-module. Then, as a consequence of the previous lemma, we can see that the graph $\Delta (K)$ is disconnected with two connected components, whose vertex sets are $\{2\}$ and $\pi (K)-\{2\}$ respectively, and which are both complete subgraphs of $\Delta (K)$.

In fact, by Theorem 2.3, $\Delta (K/L)$ has three connected components with vertex sets $\pi (2^a-1)$, $\pi (2^a+1)$ and $\{2\}$ respectively, which are all complete subgraphs of $\Delta (K/L)$. On the other hand, if $\lambda$ is any non-principal character in $\textrm{Irr}(L)$, then $I_K(\lambda )$ is a Sylow 2-subgroup of K, and Lemma 2.6 guarantees that $\lambda$ extends to $I_K(\lambda )$; our claim then easily follows by Clifford’s theory.

Theorem 2.8

Let G be a non-solvable group such that $\Delta (G)$ is connected and it has a cut-vertex p. Then, denoting by R the solvable radical of G, we have that G/R is an almost-simple group such that ${V}(G)=\pi ({G/R})\cup \{p\}$.

Proof

See [6, Theorem 3.15].$\square$

To conclude this preliminary section, we recall the statements of three crucial results proved in [5, 6], concerning certain module actions of $\textrm{SL}_{2}(t^a))$.

Let H and V be finite groups, and assume that H acts by automorphisms on V. Given a prime number q, we say that the pair (H, V) satisfies the condition ${\mathcal {N}}_q$ if q divides $|H: \textbf{C}_{H}(V)|$ and, for every non-trivial $v\in V$, there exists a Sylow q-subgroup Q of H such that $Q\trianglelefteq \textbf{C}_{H}(v)$ (see [2]).

If (H, V) satisfies ${\mathcal {N}}_q$ then V turns out to be an elementary abelian r-group for a suitable prime r, and V is in fact an irreducible module for H over the field ${\mathbb {F}}_r$ (see Lemma 4 of [17]).

Lemma 2.9

([6, Lemma 3.10]) Let t, q, r be prime numbers, let $H = \textrm{SL}_2(t^a)$ (with $t^a \ge 4$) and let V be an ${\mathbb {F}}_r[H]$-module. Then (H, V) satisfies ${\mathcal {N}}_q$ if and only if either $t^a = 5$ and V is the natural module for $H/\textbf{C}_{H}(V) \cong \textrm{SL}_2(4)$ or V is faithful and one of the following holds.

• (1) $t = q = r$ and V is the natural ${\mathbb {F}}_r[H]$-module (so $|V| = t^{2a}$);

• (2) $q = r = 3$ and $(t^a, |V|) \in \{(5, 3^4), (13, 3^6)\}$.

Theorem 2.10

([5, Theorem 3.3]) Let V be a non-trivial irreducible module for $G=\textrm{SL}_{2}(t^a)$ over the field ${{\mathbb {F}}}_q$, where $t^a\ge 4$ and q is a prime number, $q\ne t$. For odd primes $r\in \pi (t^a-1)$ and $s\in \pi (t^a+1)$ (possibly $r=q$ or $s=q$) let R, S be respectively a Sylow r-subgroup and a Sylow s-subgroup of G, and let T be a Sylow t-subgroup of G. Then, considering the sets

$$\begin{aligned}V_{I_-}=\{v\in V\,|\, {\text { there exists }} z\in G {\text { such that }}R ^z\trianglelefteq \textbf{C}_{G}(v)\},\\V_{I_+}=\{v\in V\,|\, {\text { there exists }} z\in G {\text { such that }}S ^z\trianglelefteq \textbf{C}_{G}(v)\},\\V_{II}=\{v\in V\,|\, {\text { there exists }} z \in G {\text { such that }}T^z\trianglelefteq \textbf{C}_{G}(v)\},\end{aligned}$$

we have that $V-\{0\}$ strictly contains $V_{I_-}\cup V_{II}$, $V_{I_+}\cup V_{II}$, and $V_{I_-}\cup V_{I_+}$, unless one of the following holds.

• (a) $G\cong \textrm{SL}_{2}(5)$, $s=3$, $|V|=3^4$ and $V{\setminus }\{0\}=V_{I_+}$,

• (b) $G\cong \textrm{SL}_{2}(13)$, $r=3$, $|V|=3^6$ and $V{\setminus }\{0\}=V_{I_-}$.

Theorem 2.11

([5, Theorem 3.4]) Let T be a Sylow t-subgroup of $G\cong \textrm{SL}_{2}(t^a)$ (where $t^a\ge 4$) and, for a given odd prime divisor r of $t^{2a}-1$, let R be a Sylow r-subgroup of G. Assuming that V is a t-group such that G acts by automorphisms (not necessarily faithfully) on V and $\textbf{C}_{V}(G)=1$, consider the sets

$$\begin{aligned}{} & {} V_I=\{v\in V\,|\, {\text { there exists }} x\in G {\text { such that }}R^x\trianglelefteq \textbf{C}_{G}(v)\},{\text { { and}}}\\{} & {} V_{II}=\{v\in V\,|\, {\text { there exists }} x \in G {\text { such that }}T^x\trianglelefteq \textbf{C}_{G}(v)\}. \end{aligned}$$

Then, the following conditions are equivalent.

• (a) $V_I$ and $V_{II}$ are both non-empty and $V-\{1\}=V_I\cup V_{II}$.

• (b) $G\cong \textrm{SL}_{2}(4)$, and V is an irreducible G-module of dimension 4 over ${\mathbb {F}}_2$. More precisely, V is the restriction to G, embedded as $\Omega _4^-(2)$ into $\mathrm{{SL}}_4(2)$, of the standard module of $\mathrm{{SL}}_4(2)$.

3 The structure of the solvable residual

Let G be a group having a composition factor isomorphic to $\textrm{SL}_{2}(2^a)$ (with $a\ge 2$), such that $\Delta (G)$ is connected and has a cut-vertex: as the first step in our analysis, our purpose is to describe the structure of the solvable residual K of G. In particular we will see that, except for two sporadic cases, either we have $K\cong \textrm{SL}_{2}(2^a)$, or $K\cong \textrm{SL}_{2}(5)$, or K contains a minimal normal subgroup L of G such that either $K/L\cong \textrm{SL}_{2}(2^a)$ or $K/L\cong \textrm{SL}_{2}(5)$ and L is the natural module for K/L.

We collect the main results of this section in the following single statement (which is the counterpart in characteristic 2 of [5, Theorem 4.1]). This will be proved by treating separately the case $a>2$ and the case $a=2$, in Theorems 3.2 and 3.4, respectively.

Theorem 3.1

Assume that the group G has a composition factor isomorphic to $\textrm{SL}_{2}(2^a)$ with $a\ge 2$, and let p be a prime number. Assume also that $\Delta (G)$ is connected with cut-vertex p. Then, denoting by K the solvable residual of G, one of the following conclusions holds.

• (a) K is isomorphic to $\textrm{SL}_{2}(2^a)$ or to $\textrm{SL}_{2}(5)$;

• (b) K contains a minimal normal subgroup L of G such that K/L is isomorphic either to $\textrm{SL}_{2}(2^a)$ or to $\textrm{SL}_{2}(5)$ and L is the natural module for K/L.

• (c) $a=2$, and K contains a minimal normal subgroup L of G such that K/L is isomorphic to $\textrm{SL}_{2}(4)$. Moreover, L is isomorphic to the restriction to K/L, embedded as $\Omega _4^-(2)$ into $\textrm{SL}_4(2)$, of the standard module of $\textrm{SL}_4(2)$.

• (d) $a=2$, and K contains a minimal normal subgroup L of G such that K/L is isomorphic to $\textrm{SL}_{2}(5)$. Moreover, L is isomorphic to the restriction to K/L, embedded in $\mathrm{{SL}}_4(3)$, of the standard module of $\textrm{SL}_4(3)$.

We will then start by treating the case $a>2$. Before stating the next theorem we recall that, for m and n integers larger than 1, a prime divisor q of $m^n-1$ is called a primitive prime divisor if q does not divide $m^b -1$ for all $1 \le b <n$. In this case, n is the order of m modulo q, so n divides $q-1$. In view of [10, Theorem 6.2], $m^n - 1$ always has primitive prime divisors except when $n = 2$ and $m= 2^c -1$ for some integer c (i.e., m is a Mersenne number), or when $n=6$ and $m= 2$.

In the following, for a normal subgroup N of a group G, and a character $\theta \in \textrm{Irr}(N)$, we denote by $\textrm{Irr}(G| \theta )$ the set of all irreducible characters of G that lie over $\theta$.

Theorem 3.2

Assume that the group G has a composition factor isomorphic to $\textrm{SL}_{2}(2^a)$ with $a>2$, and let p be a prime number. Assume also that $\Delta (G)$ is connected with cut-vertex p. Then, denoting by K the solvable residual of G, one of the following conclusions holds.

• (a) K is isomorphic to $\textrm{SL}_{2}(2^a)$;

• (b) K contains a minimal normal subgroup L of G such that K/L is isomorphic to $\textrm{SL}_{2}(2^a)$ and L is the natural module for K/L.

Proof

Let R be the solvable radical of G. By Theorem 2.8, we have that G/R is an almost-simple group with socle isomorphic to $\textrm{SL}_{2}(2^a)$, and ${V}(G)=\pi ({G/R})\cup \{p\}$. Note that, since $a>2$, Lemma 2.5 applies here; so either we get conclusion (a), or K has a non-trivial normal subgroup L such that K/L is isomorphic to $\textrm{SL}_{2}(2^a)$, and every non-principal irreducible character of $L/L'$ is not invariant in K. Therefore, we can assume that the latter condition holds.

Consider then a non-principal $\xi$ in $\textrm{Irr}(L/L')$: as $I_K(\xi )/L$ is a proper subgroup of $K/L\cong \textrm{SL}_{2}(2^a)$, its possible structures are described in Remark 2.1. In particular, if 2 is not a divisor of $|K:I_K(\xi )|$, then $I_K(\xi )/L$ contains a Sylow 2-subgroup of K/L as a normal subgroup. Assuming for the moment that this happens for every non-principal $\xi \in \textrm{Irr}(L/L')$, Lemma 2.9 (together with the paragraph preceding it) yields that the dual group $\widehat{L/L'}$ is the natural module for K/L, and the same holds for $L/L'$ by self-duality; so, in order to get the desired conclusion, we only have to show that $L'$ is trivial (note that, once this is proved, $L=\textbf{O}_{2}(K)$ is a minimal normal subgroup of G), and this is what we do next.

For a proof by contradiction assume $L'\ne 1$, and consider a chief factor $L'/Z$ of K. As observed in Remark 2.7, the graph $\Delta (K/L')$ has two connected components having vertex sets $\{2\}$ and $\pi (K/L')-\{2\}$, respectively; since the vertex set of $\Delta (G)$ is $\pi (G/R)\cup \{p\}$ and, also in view of Theorem 2.4, $\pi (G/R)-\{2\}$ is now a clique of $\Delta (G)$, we see that the cut-vertex p of $\Delta (G)$ cannot be 2, and that p is the unique vertex adjacent to 2 in $\Delta (G)$.

Now, let $\lambda$ be a non-principal irreducible character of $L'/Z$, and let $\chi \in \textrm{Irr}(K/Z\, |\, \lambda )$. If $\psi$ is an irreducible constituent of $\chi _{L/Z}$ lying over $\lambda$, then clearly $\psi (1)\ne 1$, and since $L'/Z$ is an abelian normal subgroup of L/Z whose index is a 2-power, we conclude that $\psi (1)$ (whence $\chi (1)$) is a multiple of 2. As a consequence, we get $\pi (|K:I_K(\lambda )|)\subseteq \{2,p\}$. Observe that $I_K(\psi )$ is a proper subgroup of K, as otherwise (the Schur multiplier of K/L being trivial) $\psi$ would extend to K yielding a contradiction via Gallagher’s theorem; of course $I_K(\lambda )$ is a proper subgroup of K as well, unless $L'/Z$ lies in $\textbf{Z}(K/Z)$.

We conclude this part of the proof by considering three situations that are exhaustive, and that all lead to a contradiction.

(i) $L'/Z\not \subseteq \textbf{Z}(L/Z).$

Consider the normal subgroup $\textbf{C}_{L'/Z}(L/L')$ of K/Z; since $L'/Z$ is a chief factor of K and it is not centralized by $L/L'$, we deduce that $\textbf{C}_{L'/Z}(L/L')$ is trivial. Thus we can apply the proposition appearing in the Introduction of [3], which ensures that the second cohomology group $\mathrm{{H}}^2(K/L',L'/Z)$ is trivial, and therefore K/Z is a split extension of $L'/Z$; in particular, every irreducible character of $L'/Z$ extends to its inertia subgroup in K/Z. Now, let $\lambda$ be any non-principal character in $\textrm{Irr}(L'/Z)$: since $\pi (|K:I_K(\lambda )|)\subseteq \{2,p\}$, Gallagher’s theorem implies that $I_K(\lambda )/L'$ contains a unique Sylow q-subgroup of $K/L'$ for every prime $q\in \pi (2^{2a}-1)-\{p\}$. But this yields a contradiction via, for example, Proposition 3.13 of [6]; in fact, according to that result, $K/L'$ should have a cyclic solvable radical (whereas $L/L'=\textbf{O}_{2}(K/L')$ is non-cyclic).

(ii) $L'/Z\subseteq \textbf{Z}(L/Z)$, but $L'/Z\not \subseteq \textbf{Z}(K/Z).$

First, we note that $L'/Z$ is a 2-group in this case, as otherwise L/Z would be isomorphic to the direct product $(L'/Z)\times (L/L')$ and it would then be abelian, a clear contradiction. Also, for a non-principal $\lambda$ in $\textrm{Irr}(L'/Z)$, we already observed that $I_K(\lambda )$ is a proper subgroup of K such that $\pi (|K:I_K(\lambda )|)\subseteq \{2,p\}$.

We claim that $I_K(\lambda )/L$ cannot be a subgroup of type (iv) of K/L unless it is also of type (iii). In fact, assume $I_K(\lambda )/L\cong \textrm{SL}_{2}(2^b)$ where $b>2$ and $a=bc$ for some $c>1$. If c is an odd number, then $2^b+1$ is a divisor of $2^a+1$ and it is easy to see that $\pi (|K:I_K(\lambda )|)$ contains at least two odd primes, not our case. On the other hand, if c is even, then $2^{2b}-1$ divides $2^a-1$ and again (recalling [10, Proposition 3.1]) we reach a contradiction unless $c=2$ and $p=2^a+1$ (note that p is neither 3 nor 5). Now we look at $I_K(\psi )$, where $\psi$ lies in $\textrm{Irr}(L/Z\,|\,\lambda )$ (recall that $\psi (1)$ is a multiple of 2, and that $I_K(\psi )$ is contained in $I_K(\lambda )$ because $L'/Z$ is central in L/Z): we have $\pi (|I_K(\lambda ):I_K(\psi )|)\subseteq \{2\}$, and therefore $I_K(\psi )/L$ is either the whole $I_K(\lambda )/L$ or it is necessarily isomorphic to $A_5$. In any case we get the adjacency of 2 with odd primes different from p, a contradiction.

So, assume that $I_K(\lambda )/L$ is of type (iii) isomorphic to $A_4$: then there must be a prime in $\pi (|K:I_K(\lambda )|)-\{2,3\}$, and this prime is necessarily p. This forces the 3-part of |K/L| to be 3, yielding the contradiction that either $2^a-1=3$ or $2^a+1=3$. On the other hand, let $I_K(\lambda )/L$ be of type (iii) isomorphic to $A_5$. If $\pi (|K:I_K(\lambda )|)-\{3,5\}\subseteq \{2\}$, then either the 3-part or the 5-part of |K/L| is forced to be 3 or 5 respectively, and we get a contradiction from the fact that one among 3 and 5 is $2^a-1$ or $2^a+1$; if $\pi (|K:I_K(\lambda )|)-\{3,5\}$ contains an odd prime (which is p), then the 3-part and the 5-part of |K/L| are 3 and 5, respectively, and we get the same contradiction as before unless $3\cdot 5=2^a-1$, i.e., $K/L\cong \textrm{SL}_{2}(2^4)$ and $p=17$ is the only vertex adjacent to 2 in $\Delta (G)$. But in the latter case, taking $\psi \in \textrm{Irr}(L/Z\,|\,\lambda )$, we see that $I_K(\psi )$ cannot be a proper subgroup of $I_K(\lambda )$ (otherwise $|I_K(\lambda ):I_K(\psi )|$ would be divisible by 3 or 5 and we would get the adjacency between one of these primes and 2); thus, recalling that $I_K(\psi )\subseteq I_K(\lambda )$, we get $I_K(\psi )/L\cong A_5$. Working with character triples we now get the adjacency between 2 and 3, again a contradiction. Our conclusion so far is that, for every non-principal $\lambda \in \textrm{Irr}(L'/Z)$, the subgroup $I_K(\lambda )/L$ of K/L is either of type (i) or of type (ii).

Next, assume that $I_K(\lambda )/L$ is a subgroup of type (i$_+$). Then we get $p=2^a-1$ and, since 2 cannot be adjacent in $\Delta (G)$ to any prime in $\pi (2^a+1)$, for every non-principal $\nu \in \textrm{Irr}(L'/Z)$ the subgroup $I_K(\nu )/L$ must be either of type (i$_+$) containing a unique Hall $\pi (2^a+1)$-subgroup of K/L, or of type (ii) containing a unique Sylow 2-subgroup of K/L. Now, the former situation cannot occur for every $\nu$, by Lemma 2.9; on the other hand, if the latter situation occurs for some non-principal $\nu \in \textrm{Irr}(L'/Z)$, then we reach a contradiction via Theorem 2.11 (recall that $L'/Z$ is a 2-group).

If $I_K(\lambda )/L$ is a subgroup of type (i$_-$) then, as above, for every non-principal $\nu \in \textrm{Irr}(L'/Z)$, the subgroup $I_K(\nu )/L$ must be either of type (i$_-$) containing a unique Hall $\pi (2^a-1)$-subgroup of K/L or of type (ii). Observe that if, in the latter case, $|K:I_K(\nu )|$ is divisible by 2, then $I_K(\nu )/L$ must contain a Hall $\pi (2^a-1)$-subgroup of K/L; hence, by the structure of the subgroups of type (ii), $I_K(\nu )/L$ should contain a full Sylow 2-subgroup of K/L as well, against the fact that $|K:I_K(\lambda )|$ is even. Therefore $I_K(\nu )/L$ actually contains a (unique) Sylow 2-subgroup of K/L whenever it is a subgroup of type (ii), and now we reach a contradiction as in the previous paragraph.

We conclude that, for every non-principal $\lambda \in \textrm{Irr}(L'/Z)$, the subgroup $I_K(\lambda )/L$ of K/L is of type (ii), and the same argument as in the paragraph above shows that it must contain a full Sylow 2-subgroup of K/L. This yields (via Lemma 2.9) that $L'/Z$ is the natural module for K/L, so that $I_K(\lambda )/L$ is a Sylow 2-subgroup of K/L for every non-principal $\lambda \in \textrm{Irr}(L'/Z)$. Considering $\psi \in \textrm{Irr}(L/Z)$ lying over such a $\lambda$, and recalling once again that $\psi (1)$ is even and $I_K(\psi )\subseteq I_K(\lambda )$, Clifford’s theory yields that the primes in $\pi (K/L)$ are pairwise adjacent in $\Delta (G)$ and, also in view of Theorem 2.4, every odd prime divisor of |K/L| is a complete vertex of $\Delta (G)$. This is clearly not compatible with the existence of a cut-vertex of $\Delta (G)$.

(iii) $L'/Z\subseteq \textbf{Z}(K/Z).$

As in case (ii), we have that $L'/Z$ is a 2-group. If $\lambda$ is a non-principal irreducible character of $L'/Z$, then $\lambda$ is fully ramified with respect to the K/Z-chief factor $L/L'$ (see Exercise 6.12 of [8]); therefore, the unique $\psi$ in $\textrm{Irr}(L/Z\,|\,\lambda )$ is such that $I_K(\psi )=I_K(\lambda )=K$. The fact that the Schur multiplier of K/L is trivial implies that $\psi$ extends to K, yielding a clear contradiction via Gallagher’s theorem.

To conclude the proof, we will show that $I_K(\xi )/L$ contains a unique Sylow 2-subgroup of K/L for every non-principal $\xi$ in $\textrm{Irr}(L/L')$. To this end, we will proceed through a number of steps.

(a) For every non-principal $\xi \in \textrm{Irr}(L/L')$, the subgroup $I_K(\xi )/L$ of K/L cannot be of type (iv), unless it is also of type (iii).

For a proof by contradiction, let $\xi \in \textrm{Irr}(L/L')$ be such that $I_K(\xi )/L\cong \textrm{SL}_{2}(2^b)$ for some $b> 2$ properly dividing a. Thus, 2 is a divisor of $|K:I_K(\xi )|$. Since the Schur multiplier of $I_K(\xi )/L$ is trivial, $\xi$ extends to $I_K(\xi )$ and this yields (via Clifford’s correspondence and Gallagher’s theorem) that 2 is adjacent in $\Delta (G)$ to every prime in $\pi (K/L)-\{2\}$. Moreover, taking into account Theorem 2.4 (which, together with Theorem 2.3, will be freely used from now on and should be kept in mind), also each prime in $\pi (G/R)-\pi (K/L)$ is adjacent to every prime in $\pi (K/L)-\{2\}$. Finally, $2^{2a}-1$ has a primitive prime divisor q because $a\ne 3$; this prime q, which clearly belongs to $\pi (2^a+1)$, is a divisor of $|K:I_K(\xi )|$, so every prime in $\pi (2^b-1)$ is adjacent to q in $\Delta (G)$. As easily seen, this setting is not compatible with the existence of a cut-vertex of $\Delta (G)$.

(b) For every non-principal $\xi \in \textrm{Irr}(L/L')$, the subgroup $I_K(\xi )/L$ of K/L cannot be isomorphic to $A_5$.

Assume the contrary, and take $\xi \in \textrm{Irr}(L/L')$ such that $I_K(\xi )/L\cong A_5$. Working with character triples, we observe that $\textrm{Irr}(K\,|\, \xi )$ contains characters whose degrees are divisible by every prime in $\pi (|K:I_K(\xi )|)\cup \{3\}$, which contains $\pi (K/L)-\{5\}$ (note that 2 divides $|K:I_K(\xi )|$ because $2^a>4$); thus the 5-part of |K/L| is 5, otherwise the primes in $\pi (K/L)$ would be pairwise adjacent in $\Delta (G)$, easily contradicting the existence of a cut-vertex of $\Delta (G)$. Observe also that, since neither $2^a-1$ nor $2^a+1$ can be 5, there exists an odd prime q in $\pi (K/L)-\{5\}$ that is adjacent to 5 in $\Delta (K/L)$; as q is now a complete vertex in the subgraph of $\Delta (G)$ induced by $\pi (G/R)$, we get $q=p$, and it is readily seen that no other prime divisor of |K/L| can be adjacent to 5 in $\Delta (G)$. This implies on one hand that $\xi$ does not have an extension to $I_K(\xi )$ (otherwise, by Gallagher’s theorem, we would get the adjacency between 5 and 2 in $\Delta (G)$), which in turn yields (via [8, 8.16, 11.22, 11.31]) that the order of $L/L'$ is divisible by 2; on the other hand, one among the sets $\pi (2^a-1)$ and $\pi (2^a+1)$ is in fact $\{5,p\}$.

Now, since $2^{2a}-1$ is divisible by 5, we see that a must be even, so $2^2-1=3$ divides $2^a-1$. Assuming for the moment $\pi (2^a-1)=\{5,p\}$, we then get $p=3$, and we also note that $2^a-1$ has a primitive prime divisor (otherwise a would be 6, but $2^6-1=63$ is not divisible by 5). Certainly 3 is not such a divisor, as 3 divides $2^2-1$ and $a>2$; hence 5 is a primitive prime divisor for $2^a-1$, so we get $a=4$ and $K/L\cong \textrm{SL}_{2}(16)$. But in this case, since $|L/L'|$ is even, we can consider a chief factor L/X of K whose order is a 2-power: the dual group of V of L/X is then an irreducible module for $\textrm{SL}_{2}(16)$ over ${{\mathbb {F}}}_2$. It is well known (see [1], for instance) that such modules all have a dimension belonging to $\{8,16,32\}$; if V is the natural module (of dimension 8) for $K/L\cong \textrm{SL}_{2}(16)$, then the centralizer in K/L of every non-trivial element of V is a Sylow 2-subgroup of K/L, yielding the contradiction that 5 is adjacent to 17 in $\Delta (G)$. Also, a direct computation with GAP [14] shows that in the modules of dimensions 16 and 32 there are elements lying in regular orbits for the action of K/L, thus the primes in $\Delta (K/L)$ would be pairwise adjacent in $\Delta (G)$. Only one module is left, which has dimension 8 and is not the natural module: to handle this, we can see via GAP [14] that in all possible isomorphism types of extensions of V by $\textrm{SL}_{2}(16)$ the set of irreducible character degrees is $\{1, 15, 16, 17, 51, 68, 204, 255, 272, 340\}$, so $\pi (K/L)$ would again be a set of pairwise adjacent vertices of $\Delta (G)$.

It remains to consider the case when 5 divides $2^a+1$, hence $\pi (2^a+1)=\{5,p\}$, and again we choose a chief factor L/X of K that is a 2-group. Now, the dual group of L/X can be viewed as a (non-trivial) irreducible K/L-module over ${{\mathbb {F}}}_2$, and if T/L is a Sylow 2-subgroup of K/L, then clearly there exists a non-principal $\mu$ in $\textrm{Irr}(L/X)$ which is fixed by T/L (so, such that $I_K(\mu )/L$ contains T/L); as $I_K(\mu )/L$ is a proper subgroup of K/L, the only possibility for $I_K(\mu )/L$ is to be of type (ii). Moreover, since 5 is only adjacent to p in $\Delta (G)$, no prime divisor of $2^a-1$ lies in $\pi (|K:I_K(\mu )|)$, thus in fact $I_K(\mu )/L=\textbf{N}_{K/L}(T/L)$ has irreducible characters of degree $2^a-1$. Now, $\mu$ does not extend to $I_K(\mu )$, as otherwise (by Gallagher’s theorem and Clifford correspondence) we would get adjacencies in $\Delta (G)$ between 5 and all the primes in $\pi (2^a-1)$; but then, for $\psi \in \textrm{Irr}(I_K(\mu )\,|\,\mu )$ and $\theta$ an irreducible constituent of $\psi _{T/X}$ lying over $\mu$, we have that $\theta (1)$ is a 2-power larger than 1 (otherwise $\psi$ would be an extension of $\mu$ to T, and $\mu$ would then extend to the whole $I_K(\mu )$). As a consequence, 2 divides $\psi (1)$ and we get the adjacency in $\Delta (G)$ between 5 and 2. This is the final contradiction that rules out the case $I_K(\xi )/L\cong A_5$.

(c) For every non-principal $\xi \in \textrm{Irr}(L/L')$, the subgroup $I_K(\xi )/L$ of K/L cannot be isomorphic to $A_4$.

Assume $I_K(\xi )/L\cong A_4$ for some $\xi \in \textrm{Irr}(L/L')$. Then we see at once that the primes in $\pi (K/L)-\{3\}$ are pairwise adjacent in $\Delta (G)$, thus the 3-part of |K/L| is 3 and we get the same conclusions as in the first paragraph of (b) with 3 in place of 5: the cut-vertex p is an odd prime and it is the unique neighbor of 3 in $\Delta (G)$ among the primes in $\pi (K/L)$, and one among the sets $\pi (2^a-1)$ and $\pi (2^a+1)$ is $\{3,p\}$.

Assuming first $\pi (2^a-1)=\{3,p\}$, we see that $a\ne 6$ because the 3-part of $2^6-1$ is $3^2$. Hence $2^a-1$ has a primitive prime divisor, which is necessarily p. Note that a cannot be a prime number, as otherwise it would be odd and K/L would not have subgroups isomorphic to $A_4$; moreover, if k is a divisor of a such that $1<k<a$, then $2^k-1$ divides $2^a-1$ and is coprime to p, so $2^k-1$ must be 3 and k is 2. We conclude that a is 4, so $K/L\cong \textrm{SL}_{2}(16)$, and we reach a contradiction as in the second paragraph of (b).

As regards the case $\pi (2^a+1)=\{3,p\}$, the same argument as in the last paragraph of (b) (replacing 5 with 3) completes the proof.

(d) The subgroups $I_K(\xi )/L$ of K/L, for $\xi$ non-principal in $\textrm{Irr}(L/L')$, cannot be all of type (ii) and of ever order, unless each of them contains a (unique) Sylow 2-subgroup of K/L.

Let us assume that all the subgroups $I_K(\xi )/L$ of K/L (for $\xi$ non-principal in $\textrm{Irr}(L/L')$) are of type (ii) and of even order, but there exists $\xi _0\in \textrm{Irr}(L/L')$ such that 2 divides $|K:I_K(\xi _0)|$. In this setting we observe that $2^a-1$ does not divide $|I_K(\xi _0)/L|$, because $I_K(\xi _0)/L$ is a Frobenius group whose kernel is its unique Sylow 2-subgroup $T_0/L$, and we are assuming $|T_0/L| =2^f<2^a$. Therefore there exists $r\in \pi (2^a-1)\cap \pi (|K:I_K(\xi _0)|)$, and Clifford’s correspondence yields that $\{2,r\}\cup \pi (2^a+1)$ is a set of pairwise adjacent vertices of $\Delta (G)$. It follows that r is adjacent in $\Delta (G)$ to every prime in $\pi (G/R)-\{r\}$, thus r is the cut-vertex p, and no other prime in $\pi (2^a-1)$ can have any neighbor in $\{2\}\cup \pi (2^a+1)$; in particular, no prime in $\pi (2^a-1)-\{p\}$ shows up as a divisor of $|K:I_K(\xi )|$ for any $\xi \in \textrm{Irr}(L/L')$. Note also that a primitive prime divisor of $2^a-1$ cannot lie in $\pi (I_K(\xi _0)/L)$, as otherwise it would divide $2^f-1$ (and $f<a$); so, if $a\ne 6$, p is forced to be the unique primitive prime divisor of $2^a-1$. Observe finally that the $p'$-part of $2^a-1$ is not 1, otherwise p would not be a cut-vertex of $\Delta (G)$. Thus there exists a prime $q\in \pi (2^a-1)-\{p\}$ such that, for every $\xi \in \textrm{Irr}(L/L')$, the subgroup $I_K(\xi )/L$ contains a Sylow q-subgroup of K/L.

Furthermore, the character $\xi _0$ does not extend to $I_K(\xi _0)$, as otherwise we would get characters in $\textrm{Irr}(K\,|\,\xi _0)$ whose degree is divisible by q and every prime in $\{2\}\cup \pi (2^a+1)$, not our case; so $|L/L'|$ is even, and there exists a chief factor L/X of K whose order is a 2-power. Note that, by the conclusion in the paragraph above, the subgroups of the kind $I_K(\xi )/L$ for $\xi$ non-principal in $\textrm{Irr}(L/X)$ are not Sylow 2-subgroups of K/L, thus L/X is not the natural module for K/L; this in turn implies (via Lemma 2.9) that, for some non-principal $\xi \in \textrm{Irr}(L/X)$, $I_K(\xi )/L$ does not contain a full Sylow 2-subgroup of K/L. In other words, we can assume that $\xi _0$ is in fact an irreducible character of $L/L'$ whose kernel has index 2 in L.

Assume for the moment that a is an even number different from 6 (say, $a=2b)$: as $2^b-1$ is coprime to p, we get that $2^b-1$ divides the order of (a Frobenius complement of) $I_K(\xi _0)/L$, and so $|T_0/L|-1=2^f-1$ is a multiple of $2^b-1$. This forces f to be a multiple of b and, since $f<a=2b$, the only possibility is $f=b$; note that $T_0/L$ is then a minimal normal subgroup of $I_K(\xi _0)/L$. The fact that $\xi _0$ does not extend to its inertia subgroup in K also implies that $T_0/\ker {\xi _0}$ is a non-abelian 2-group; thus $L/\ker {\xi _0}$, which has order 2, is in fact the derived subgroup of $T_0/\ker {\xi _0}$. Moreover, the normal subgroup $Z/\ker {\xi _0}=\textbf{Z}(T_0/\ker {\xi _0})$ of $I_K(\xi _0)/\ker {\xi _0}$ cannot be larger than $L/\ker {\xi _0}$, because $T_0/L$ is a minimal normal subgroup of $I_K(\xi _0)/L$ and clearly Z/L is not the whole $T_0/L$. We deduce that $T_0/\ker {\xi _0}$ is an extraspecial 2-group, so (b is even and) an application of [7, II, Satz 9.23] yields the contradiction that $2^b-1$ divides $2^{b/2}+1$.

If $a=6$, then p can be either 3 or 7. In the former case, 7 divides $|I_K(\xi _0)/L|$ and so $T_0/L$ has order $2^3$; the same argument as above shows that $T_0/\ker {\xi _0}$ is an extraspecial 2-group, a clear contradiction. On the other hand, if $p=7$, then $|I_K(\xi _0)/L|$ should be a multiple of 9, but 9 is not a divisor of $2^f-1$ for any $f<6$, contradicting the fact that $I_K(\xi _0)/L$ is a Frobenius group with kernel $T_0/L$.

It remains to treat the case when a is odd. In this case, we start by fixing a Sylow q-subgroup Q of K/L: if a non-principal $\xi \in \textrm{Irr}(L/X)$ is stabilized both by Q and by another $Q_1\in \textrm{Syl}_q(K/L)$, then Q and $Q_1$ are contained in the same subgroup of type (ii) of K/L, whence in the normalizer of a suitable Sylow 2-subgroup of K/L. By Lemma 2.2, Q normalizes precisely two Sylow 2-subgroups of K/L, and since these normalizers contain a total number of $2^a$ Sylow q-subgroups each, there are at most $2(2^a-1)$ choices for $Q_1$. On the other hand, the total number of Sylow q-subgroups of K/L is $2^{a-1}(2^a+1)$, so there certainly exists an element $h\in K/L$ such that no non-trivial element in the dual group ${\widehat{L/X}}$ of L/X is centralized by both Q and $Q^h$. As a consequence, setting $|L/X|=2^d$, we get $|\textbf{C}_{{\widehat{L/X}}}(Q)|\le 2^{d/2}$, and then

$$\begin{aligned}2^d-1\le (2^{d/2}-1)\cdot 2^{a-1}\cdot (2^a+1).\end{aligned}$$

It is easily checked that the above inequality yields $d<4a$ and, since a is odd, Lemma 3.12 in [12] (whose hypotheses require $d\le 3a$, but whose proof works assuming $d<4a$ as well) leaves only one possibility for the isomorphism type of the K/L-module ${\widehat{L/X}}$ over ${{\mathbb {F}}}_2$. First of all, a is a multiple of 3 (say $a=3c$) and $d=8c$; then, denoting by R(1) the natural module for K/L over ${{\mathbb {F}}}_{2^a}$ and by $\omega$ an automorphism of order 3 of ${{\mathbb {F}}}_{2^a}$, we have that ${\widehat{L/X}}$ is a “triality module", which can be described as follows. Start from the K/L-module $V=R(1)\otimes R(1)^\omega \otimes R(1)^{\omega ^2}$ over ${{\mathbb {F}}}_{2^a}$ (or one of its Galois twists), and observe that the field of values of (the character of) V is ${{\mathbb {F}}}_{2^c}$; now, restricting the scalars to ${{\mathbb {F}}}_{2^c}$, V is a homogeneous K/L-module and we take an irreducible constituent of it. This irreducible constituent remains irreducible if the scalars are restricted further to ${{\mathbb {F}}}_2$, and this is the ${{\mathbb {F}}}_{2}[K/L]$-module we are considering. In order to finish the proof for this case, it will be enough to show that there exist non-trivial elements of V whose centralizer in K/L is not a subgroup of type (ii).

Recall that the elements of $\textrm{SL}_{2}(2^a)$ whose order is a divisor of $2^a+1$ are conjugate to elements of the form $x=\left( \begin{array}{cc} 0 &{}\quad 1 \\ 1 &{}\quad \lambda \end{array} \right)$, where $\lambda =\mu +\mu ^{2^a}$ for $\mu \in {{\mathbb {F}}}_{2^{2a}}-\{1\}$ such that $\mu ^{2^a+1}=1$. The action of such an x on V is of course given by the Kronecker product

$$\begin{aligned}\left( \begin{array}{cc} 0 &{}\quad 1 \\ 1 &{}\quad \lambda \end{array} \right) \otimes \left( \begin{array}{cc} 0 &{} 1 \\ 1 &{} \lambda ^{\omega } \end{array} \right) \otimes \left( \begin{array}{cc} 0 &{} 1 \\ 1 &{} \lambda ^{\omega ^2} \end{array} \right) .\end{aligned}$$

Now, setting ${{\mathbb {K}}}={{\mathbb {F}}}_{2^{2a}}$ and $V^{{{\mathbb {K}}}}=V\otimes {{\mathbb {K}}}$, we have $\dim _{{{\mathbb {K}}}}\textbf{C}_{V^{{{\mathbb {K}}}}}(x)=\dim _{{{\mathbb {F}}}_{2^a}}\textbf{C}_{V}(x)$; moreover, $V^{{{\mathbb {K}}}}=R(1)^{{{\mathbb {K}}}}\otimes (R(1)^{{{\mathbb {K}}}})^\omega \otimes (R(1)^{{{\mathbb {K}}}})^{\omega ^2}$, so the action of x on $V^{{{\mathbb {K}}}}$ is expressed by the same Kronecker product as above. But x is conjugate to $\left( \begin{array}{cc} \mu &{} \quad 0 \\ 0 &{} \quad \mu ^{-1} \end{array} \right)$ in $\textrm{SL}_{2}(2^{2a})$, so our aim is in fact to find $\mu$ such that the matrix

$$\begin{aligned}\left( \begin{array}{cc} \mu &{}\quad 0 \\ 0 &{} \quad \mu ^{-1} \end{array} \right) \otimes \left( \begin{array}{cc} \mu ^{2^c} &{}\quad 0 \\ 0 &{} \quad \mu ^{-2^c} \end{array} \right) \otimes \left( \begin{array}{cc} \mu ^{2^{2c}} &{} \quad 0 \\ 0 &{}\quad \mu ^{-2^{2c}} \end{array} \right) \end{aligned}$$

has a nonzero eigenspace for the eigenvalue 1. A direct calculation shows that it is enough to choose $\mu$ of order $2^{2c}-2^c+1$.

(e) The subgroups $I_K(\xi )/L$ of K/L, for $\xi$ non-principal in $\textrm{Irr}(L/L')$, cannot be all of type (ii) of even order or of type (i$_-$) (both types occurring).

If, assuming the contrary, there exists a non-principal $\xi _0\in \textrm{Irr}(L/L')$ such that $I_K(\xi _0)/L$ is of type (ii) and of even order, but not containing a full Sylow 2-subgroup of K/L then, as in (d), we get the following conditions: there exists a prime $q\in \pi (2^a-1)$ such that, for every non-principal $\xi \in \textrm{Irr}(L/L')$, the subgroup $I_K(\xi )/L$ contains a Sylow q-subgroup of K/L, and $\xi _0$ does not extend to $I_K(\xi _0)$. Similarly, no power of $\xi _0$ can extend to its inertia subgroup I in K, if I/L does not contain a full Sylow 2-subgroup of K/L. Since all Sylow q-subgroups of K/L are cyclic for $q \ne 2$, by [8, Theorem 6.26] we can actually assume that the order $o(\xi _0)$ in the dual group of $L/L'$ is a power of 2. Hence $\xi _0$ is in $\textrm{Irr}(L/X)$, for a chief factor L/X of K that is a 2-group, and the rest of the argument in (d) goes through.

On the other hand, if all the inertia subgroups of type (ii) and of even order contain a full Sylow 2-subgroup of K/L, and there is an inertia subgroup of type (i$_-$) whose index in K/L is divisible by a prime in $\pi (2^a-1)$ (that is necessarily p), then again we are in the same situation as in (d): for every $\xi \in \textrm{Irr}(L/L')$, the subgroup $I_K(\xi )/L$ contains a Sylow q-subgroup of K/L for a suitable prime $q\in \pi (2^a-1)-\{p\}$, and $L/L'$ has even order. Taking a chief factor L/X of K that is a 2-group, we are in a position to apply Theorem 2.11 together with Lemma 2.9, and we get a contradiction. Finally, if all the inertia subgroups of type (ii) and even order contain a Sylow 2-subgroup of K/L, and all those of type (i$_-$) have order divisible by $2^a-1$, then Theorem 2.10 (applied to the action of K/L on any chief factor $V=L/X$ of K of odd order) yields that $L/L'$ is a 2-group, and now Theorem 2.11 yields a contradiction.

(f) The subgroups $I_K(\xi )/L$ of K/L, for $\xi$ non-principal in $\textrm{Irr}(L/L')$, cannot be all of type (ii) of even order or of type (i$_+$) (both types occurring).

If, assuming the contrary, there exists a non-principal $\xi _0\in \textrm{Irr}(L/L')$ such that $I_K(\xi _0)/L$ is of type (ii) but not containing a full Sylow 2-subgroup of K/L, then 2 is adjacent in $\Delta (G)$ to every prime in $\pi (K/L)-\{2\}$; however, there also exists a prime $r\in \pi (2^a-1)$ such that r divides $|K:I_K(\xi _0)|$, and this r is adjacent in $\Delta (G)$ to every other prime in $\pi (G/R)$; this is incompatible with the existence of a cut-vertex of $\Delta (G)$. The case when all the inertia subgroups of type (ii) contain a full Sylow 2-subgroup of K/L, and there is an inertia subgroup of type (i$_+$) whose index in K/L is divisible by a prime in $\pi (2^a+1)$ (which must be p), yields the following situation: every inertia subgroup of type (i$_+$) contains a Sylow q-subgroup of K/L for a suitable prime $q\in \pi (2^a+1)-\{p\}$, and every inertia subgroup of type (ii) is a full normalizer of a Sylow 2-subgroup of K/L. Moreover, $|L/L'|$ is even, and again we reach a contradiction via Theorem 2.11. Finally, if all the inertia subgroups of type (ii) contain a Sylow 2-subgroup of K/L, and all those of type (i$_+$) have order divisible by $2^a+1$, then Theorem 2.10 (applied to the action of K/L on any chief factor $V=L/X$ of K of odd order) yields that $L/L'$ is a 2-group, and again Theorem 2.11 yields a contradiction.

(g) The subgroups $I_K(\xi )/L$ of K/L, for $\xi$ non-principal in $\textrm{Irr}(L/L')$, cannot be all of type (ii) with even order, of type (i$_+$), or of type (i$_-$) (all types occurring).

Let us assume the contrary. Then 2 is adjacent in $\Delta (G)$ to all the primes in $\pi (K/L)-\{2\}$, and any inertia subgroup $I_K(\xi _0)/L$ of type (ii) is the full normalizer of a Sylow 2-subgroup $T_0/L$ of K/L, i.e., a Frobenius group of order $2^a\cdot (2^a-1)$.

Note that $T_0/L$ is then a minimal normal subgroup of $I_K(\xi _0)/L$. Moreover, $\xi _0$ does not extend to $I_K(\xi _0)$, as otherwise we would get adjacencies between primes in $\pi (2^a-1)$ and primes in $\pi (2^a+1)$; hence $L/L'$ has even order and (as already observed) for a chief factor L/X of K having 2-power order, there exists a character in $\textrm{Irr}(L/X)$ whose stabilizer in K/L contains a Sylow 2-subgroup of K/L. In other words, we can assume that $\xi _0$ lies in $\textrm{Irr}(L/X)$, so $|L/\ker {\xi _0}|=2$. Now, $T_0/\ker {\xi _0}$ is a non-abelian 2-group, thus $L/\ker {\xi _0}$ is the derived subgroup of $T_0/\ker {\xi _0}$. Moreover, the normal subgroup $Z/\ker {\xi _0}=\textbf{Z}(T_0/\ker {\xi _0})$ of $I_K(\xi _0)/\ker {\xi _0}$ cannot be larger than $L/\ker {\xi _0}$, because Z/L is not the whole $T_0/L$. We deduce that $T_0/\ker {\xi _0}$ is an extraspecial 2-group, so (a is even and) an application of [7, II, Satz 9.23] yields the contradiction that $2^a-1$ divides $2^{a/2}+1$.

(h) The subgroups $I_K(\xi )/L$ of K/L, for $\xi$ non-principal in $\textrm{Irr}(L/L')$, cannot be all of type (i$_+$) or of type (i$_-$) (both types occurring).

Assuming the contrary, as in the previous case we see that 2 is adjacent in $\Delta (G)$ to all the primes in $\pi (K/L)-\{2\}$; moreover, all the inertia subgroups are forced to contain either a subgroup of order $2^a-1$ or a subgroup of order $2^a+1$. Now, let L/X be a chief factor of K; by Lemma 2.9, both types (i$_+$) and (i$_-$) occur for the inertia subgroups even if we only consider the characters in $\textrm{Irr}(L/X)$, but then Theorem 2.10 yields that L/X is a 2-group, which is impossible because no non-trivial element of $\widehat{L/X}$ is centralized by a Sylow 2-subgroup of K/L.

(i) The subgroups $I_K(\xi )/L$ of K/L, for $\xi$ non-principal in $\textrm{Irr}(L/L')$, cannot be all of type (i$_+$).

Let us assume the contrary, and let L/X be a chief factor of K. If there exists $\xi _0\in \textrm{Irr}(L/X)$ such that $I_K(\xi _0)/L$ does not contain a subgroup of order $2^a+1$, which means that there exists $r\in \pi (2^a+1$) dividing $|K:I_K(\xi _0)|$, then r is a complete vertex of $\Delta (G)$ and it is in fact p. Now $\pi (2^a+1)-\{p\}$ is forced to contain at least one prime q, and this q cannot show up in the index of any inertia subgroup $I_K(\xi )$ in K. In other words, for every non-principal $\xi$ in $\textrm{Irr}(L/X)$, the inertia subgroup $I_K(\xi )/L$ contains a Sylow q-subgroup of K/L (as a normal subgroup).

Of course the conclusion of the previous paragraph holds if $I_K(\xi )/L$ does contain a subgroup of order $2^a+1$ for every $\xi \in \textrm{Irr}(L/X)$. Thus, in any case, Lemma 2.9 applies and we get a contradiction.

(j) The subgroups $I_K(\xi )/L$ of K/L, for $\xi$ non-principal in $\textrm{Irr}(L/L')$, cannot be all of type (i$_-$).

This is totally analogous to (i).

As we saw, the only possibility that is left is the desired one: $I_K(\xi )/L$ contains a unique Sylow 2-subgroup of K/L for every non-principal $\xi$ in $\textrm{Irr}(L/L')$. The proof is complete. $\square$

Next, we conclude the proof of Theorem 3.1 addressing the remaining case, i.e., when $a=2$. We start by introducing some notation and a few facts concerning a relevant set of modules.

• We denote by $V_0$ the natural module for $S = \textrm{SL}_{2}(4)$. We have $|V_0| = 2^4$, $|\textbf{C}_{S}(v)| = 2^2$ for all non-trivial $v \in V_0$, and the cohomology group $\textrm{H}^2(S,V_0)$ is trivial (whereas $\textrm{H}^1(S,V_0) \ne 0$).

• We denote by $V_1$ the restriction to $S = \textrm{SL}_{2}(4)$, embedded as $\Omega _4^-(2)$ into $\textrm{SL}_4(2)$, of the standard module of $\textrm{SL}_4(2)$. We have $|V_1| = 2^4$; moreover, S has two orbits $O_1$ and $O_2$ on $V_1-\{0\}$, and $\textbf{C}_{S}(v) \cong S_3$ for $v \in O_1$, while $\textbf{C}_{S}(v) \cong A_4$ for $v \in O_2$. As for the relevant cohomology groups, we have $\textrm{H}^1(S,V_1) = 0 = \textrm{H}^2(S,V_1)$.

• We denote by W the restriction to $S_1 = \textrm{SL}_{2}(5)$, seen as a subgroup of $\textrm{SL}_4(3)$, of the standard module of $\textrm{SL}_4(3)$. We have $|W| = 3^4$ and $|\textbf{C}_{S_1}(v)| = 3$ for all non-trivial $v \in W$; moreover, $\textrm{H}^2(S_1,W) = 0$.

• We denote by U the natural module for $S_1 = \textrm{SL}_{2}(5)$. We have $|U| = 5^2$ and $|\textbf{C}_{S_1}(v)| = 5$ for all non-trivial $v \in U$; moreover, $\textrm{H}^2(S_1,U) = 0$.

Note that all the above modules are self-dual: this follows from [12, Lemma 3.10] for $V_0$, $V_1$ and U, and for W by observing that $\textrm{GL}_4(3)$ has a unique conjugacy class of subgroups isomorphic to $\textrm{SL}_{2}(5)$.

Finally, let B be an abelian group and A a group acting on B via automorphisms: we will denote by $\Delta _{orb}(B)$ the graph whose vertex set is the set of the prime divisors of the set of orbit sizes $\{ |A:\textbf{C}_{A}(b)|: b \in B\}$ of the action of A on B, and such that two (distinct) vertices p and q are adjacent if and only if there exists $b \in B$ such that the product pq divides $|A:\textbf{C}_{A}(b)|$.

Lemma 3.3

Let q be a prime number and V an elementary abelian q-group.

• (a) If V is a non-trivial irreducible $\textrm{SL}_{2}(4)$-module and the graph $\Delta _{orb}(V)$ is not a clique with vertex set $\{2,3,5\}$, then $q=2$ and V is isomorphic either to $V_0$ or to $V_1$;

• (b) If V is a faithful irreducible $\textrm{SL}_{2}(5)$-module and the graph $\Delta _{orb}(V)$ is not a clique with vertex set $\{2,3,5\}$, then either $q=3$ and V is isomorphic to W or $q=5$ and V is isomorphic to U.

Proof

By Theorem 2.3 of [9], both $\textrm{SL}_{2}(4)$ and $\textrm{SL}_{2}(5)$ always have regular orbits on a faithful module of characteristic $p \ge 7$. The remaining cases, of characteristic $p \in \{2,3,5\}$, can be settled by direct computation using GAP [14]. $\square$

Theorem 3.4

Assume that the group G has a composition factor isomorphic to $\textrm{SL}_{2}(4)\cong \textrm{PSL}_{2}(5)$, and let p be a prime number. Assume also that $\Delta (G)$ is connected and that it has a cut-vertex p. Then, denoting by K the solvable residual of G, one of the following conclusions holds.

• (a) K is isomorphic to $\textrm{SL}_{2}(4)$ or to $\textrm{SL}_{2}(5)$.

• (b) K contains a minimal normal subgroup L of G such that K/L is isomorphic either to $\textrm{SL}_{2}(4)$ or to $\textrm{SL}_{2}(5)$ and L is the natural module for K/L.

• (c) K contains a minimal normal subgroup L of G such that K/L is isomorphic to $\textrm{SL}_{2}(4)$. Moreover, L is isomorphic to the restriction to K/L, embedded as $\Omega _4^-(2)$ into $\textrm{SL}_4(2)$, of the standard module of $\textrm{SL}_4(2)$.

• (d) K contains a minimal normal subgroup L of G such that K/L is isomorphic to $\textrm{SL}_{2}(5)$. Moreover, L is isomorphic to the restriction to K/L, embedded in $\textrm{SL}_4(3)$, of the standard module of $\textrm{SL}_4(3)$.

Proof

By Lemma 2.5 (applied with $t^a=5$) either (a) holds, or K has a non-trivial normal subgroup L such that K/L is isomorphic to $\textrm{SL}_{2}(4)$ or to $\textrm{SL}_{2}(5)$ and every non-principal irreducible character of $L/L'$ is not invariant in K. In the latter case, consider a chief factor L/X of K and set V to be its dual group; then, taking into account that ${V}(G)={V}(K)\cup \{p\}$, the hypothesis of p being a cut-vertex for $\Delta (G)$ implies that the subgraph of $\Delta (G)$ induced by the set of vertices $\{2,3,5\}$ is not a clique. Moreover, V is a non-trivial irreducible module for K/L, and Clifford’s theory yields that $\Delta _{orb}(V)$ is not a clique as well. Therefore Lemma 3.3 applies, and the K/L-module V is isomorphic to $V_0$ or to $V_1$ if $K/L\cong \textrm{SL}_{2}(4)$ whereas it is isomorphic to W or to U if $K/L\cong \textrm{SL}_{2}(5)$. Note that $L/X = \textbf{F}(K/X)$ is a chief factor of G as well, and our proof is complete if $X=1$.

Working by contradiction, we assume $X\ne 1$ and we consider a chief factor X/Y of K: in this situation, we first show that X/Y is the unique minimal normal subgroup of K/Y. In fact, let M/Y be another minimal normal subgroup of K/Y. Setting $N/L=\textbf{Z}(K/L)$ (and observing that N is contained in the solvable radical R of G), we have that K/N is the unique non-solvable chief factor of K; so, if M/Y is non-solvable, then we get $M/Y\cong K/N$ and hence $K/Y=M/Y\times N/Y$, contradicting the fact that K is perfect. Therefore, M/Y is abelian, so the normal subgroup MX/X of K/X lies in $\textbf{F}(K/X)=L/X$, and we conclude that M/Y is contained in L/Y. As a consequence, the K/L-module M/Y is isomorphic to L/X, i.e., to one of the K/L-modules $V_0$, $V_1$, W and U. Now $L/Y\cong M/Y\times X/Y$ can be regarded as a K/L-module which is the direct sum of two modules in $\{V_0,V_1\}$ or two modules in $\{W,U\}$ (depending on whether $K/L\cong \textrm{SL}_{2}(4)$ or $K/L\cong \textrm{SL}_{2}(5)$, respectively); but it is easy to see that K/L has regular orbits on (the duals of) such modules, and this leads via Clifford’s theory to the contradiction that $\{2,3,5\}$ is a clique of $\Delta (G)$.

Next, suppose that L/Y is nilpotent. Since K/Y has a unique minimal normal subgroup, clearly L/Y must be a group of prime-power order and, since |L/X| is a q-power for $q\in \{2,3,5\}$, the same holds for |L/Y|. Furthermore, we have $X/Y \le \textbf{Z}(L/Y)$ and, in particular, $I_{K}(\lambda )\subseteq I_{K}(\mu )$ for every $\mu \in \textrm{Irr}(X/Y)$ and $\lambda \in \textrm{Irr}(L/Y\,|\,\mu )$.

If $q \ne 2$, then $|N/L|=2$ and we claim that X/Y is a non-trivial K/L-module. In fact, assuming the contrary, we get $X/Y \subseteq \textbf{Z}(K/Y)$ and $|X/Y| = q$. Observe that $\textbf{C}_{L/Y}(N/L)$ is a normal subgroup of K/Y which contains X/Y but is not the whole L/Y, so, as L/X is a chief factor of K, we have $\textbf{C}_{L/Y}(N/L)=X/Y$. Now if L is abelian, then by coprime action we get $L = X/Y \times [L/Y,N/L]$, contradicting the uniqueness of X/Y as a minimal normal subgroup of K/Y. On the other hand, if L is non-abelian, then $X/Y = (L/Y)' = \textbf{Z}(L/Y)$ and L/Y is an extraspecial q-group.

So, every nonlinear irreducible character of L/Y is K-invariant and, since K/L has cyclic Sylow q-subgroups, it extends to K. It easily follows that $\{2,3,5\}$ is a clique of $\Delta (G)$, a contradiction. Thus the claim is proved, and Lemma 3.3 applies: our assumption that q is not 2 yields then $X/Y \cong L/X \cong U$, or $X/Y \cong L/X \cong W$, as K/L-modules. By the fact that $\{ 2,3,5\}$ cannot be a clique and by the observation in the last sentence of the previous paragraph, it follows that $I_{K/L}(\lambda )$ is a Sylow q-subgroup of K/L for every $\lambda \in \textrm{Irr}(L/Y)$, a contradiction by the paragraph preceding Lemma 2.9.

So we can assume $q=2$ and $L = N$. One can check with GAP [14] that the perfect groups of order $2^5\cdot |\textrm{SL}_{2}(4)|$ always have irreducible characters whose degrees are multiple, respectively, of 6, 10 and 15: it follows that X/Y is not the trivial K/L-module. Hence by Clifford’s theory, together with the fact that $I_{K}(\lambda )\subseteq I_{K}(\mu )$ for every $\mu \in \textrm{Irr}(X/Y)$ and $\lambda \in \textrm{Irr}(L/Y\,|\,\mu )$, the assumptions of Lemma 3.3 are satisfied for the action of K/L on X/Y. As a result, X/Y is isomorphic either to $V_0$ or to $V_1$ as a K/L-module and, in particular, we get $|X/Y| = 2^4 = |L/X|$. But again, a direct check via GAP [14]shows that the perfect groups of order $2^8\cdot |\textrm{SL}_{2}(4)|$ all have irreducible characters whose degrees are multiples of 6, 10, 15, yielding the same contradiction as above.

Finally, we assume that L/Y is non-nilpotent. Thus we have $X/Y= \textbf{F}(L/Y) = \textbf{F}(K/Y)$, and |X/Y| is coprime to |L/X|. Observe that ${\varvec{\Phi }}(K/Y) \le \textbf{F}(K/Y) = X/Y$ and that ${\varvec{\Phi }}(K/Y) \ne X/Y$, because otherwise K/Y modulo its Frattini subgroup would be isomorphic to K/X and would have a trivial Fitting subgroup, not our case. Since X/Y is a minimal normal subgroup of K/Y, we deduce that ${\varvec{\Phi }}(K/Y)$ is trivial and hence X/Y has a complement $K_0/Y$ in K/Y; in particular, every $\mu \in \textrm{Irr}(X/Y)$ extends to its inertia subgroup $I_K(\mu )$. Let Z/Y be an irreducible L/Y-submodule of X/Y (i.e., a minimal normal subgroup of L/Y contained in X/Y). Set $C/Y=\textbf{C}_{L/Y}(Z/Y)$: as L/X is an elementary abelian q-group (where q is a suitable prime in $\{2,3,5\}$), the factor group L/C is a cyclic group of order q acting fixed-point freely on Z/Y. Writing the completely reducible L/Y-module X/Y as $(Z/Y) \times (Z_1/Y)$ for a suitable L/Y-module $Z_1/Y$, we consider the character $\mu = \mu _0 \times 1_{Z_1/Y} \in \textrm{Irr}(X/Y)$, where $\mu _0$ is a non-principal irreducible character of Z/Y. We observe that $I_{L/Y}(\mu ) = C/Y$ and that every $\chi \in \textrm{Irr}(K/Y|\mu )$ has a degree divisible by q. We also remark that, setting $L_0/Y = (L/Y) \cap (K_0/Y)$, if $L_0/Y \cong L/X$ is isomorphic (as a K/L-module) either to $V_0$, $V_1$ or W, then $|I_{L_0/Y}(\mu )| =|C/X| =|L_0/Y|/q> |L_0/Y|^{1/2}$. We claim that, as a consequence, for every prime divisor $r \ne q$ of |K/L|, either r divides $|K:I_K(\mu )|$ or r divides the degree of some irreducible character of $I_{K/X}(\mu )$ that lies over $\mu$. In fact, fixing $R_0/Y\in \textrm{Syl}_r(K_0/Y)$, it is not difficult to see that there exists another Sylow r-subgroup $R_1/Y$ of $K_0/Y$ with $\langle R_0L_0/L_0, R_1L_0/L_0\rangle =K_0/L_0$ and, since no non-trivial element of $L_0/Y$ is centralized by the whole $K_0/L_0$, the dimension over ${{\mathbb {F}}}_{q}$ of the vector space $\textbf{C}_{L_0/Y}(R_0L_0/L_0)$ cannot be larger than a half of $\dim _{{{\mathbb {F}}}_q}(L_0/Y)$. Now, if $I_{K_0/Y}(\mu )$ (which is isomorphic to $I_{K/X}(\mu )$) contains a Sylow r-subgroup $R_0/Y$ of $K_0/Y$ as a normal subgroup, then $R_0/Y$ centralizes $I_{L_0/Y}(\mu )$ because $I_{L_0/Y}(\mu )$ and $R_0/Y$ are normal subgroups of coprime order of $I_{K_0/Y}(\mu )$, and this is not possible as $|I_{L_0/Y}(\mu )| >|L_0/Y|^{1/2}$. By Gallagher’s theorem, it hence follows that $\{2,3,5\}$ is a clique of $\Delta (G)$, a contradiction.

It only remains the case $L_0/Y \cong U$ (as $K_0/L_0$-module); but in this case $q=5$ divides $\chi (1)$ for every $\chi \in \textrm{Irr}(K\,|\, \mu )$, and the Sylow 2-subgroups and 3-subgroups of $K_0/L_0$ act fixed point freely on $L_0/Y$. Recalling that $I_{L_0/Y}(\mu )$ is normal in $I_{K_0/Y}(\mu )$ and that $|I_{L_0/Y}(\mu )| = 5$, we hence see that 6 divides $[K_0:I_{K_0}(\mu )]$, and again $\{2,3,5\}$ is a clique of $\Delta (G)$, a contradiction. $\square$

4 Proof of Theorem 1

We are ready to prove Theorem 1, that was stated in the Introduction and that is stated again here, for the convenience of the reader, as Theorem 4.2.

Lemma 4.1

Let K be a normal subgroup of the group G with $K \cong \textrm{SL}_{2}(2^a)$, $a \ge 2$. Let R be the solvable radical of G and assume that ${V}(G) = \pi (G/R) \cup \{ p \}$ for a suitable prime p. Then

• (a) The primes in ${V}(R)$ (if any) are complete vertices of $\Delta (G)$.

• (b) If $a \ge 3$ and $2 \in \pi (G/KR)$, then 2 is a complete vertex of $\Delta (G)$.

• (c) If $2 \not \in \pi (G/KR) \cup {V}(R)$, then 2 is adjacent in $\Delta (G)$ to a vertex q if and only if $q \in {V}(G/K)$.

Proof

We start by proving claim (a). Let $q\in {V}(R)$; as $KR = K \times R$, q is adjacent in $\Delta (G)$ to all vertices $\ne q$ in ${V}(K) = \pi (K) = \pi (KR/R)$. For $t \in \pi (G/R)- \pi (KR/R)$, by part (a) of Proposition 2.10 of [4] there exists a character $\theta \in \textrm{Irr}(K)$ such that t divides $|G:I_G(\theta )|$. Take $\varphi \in \textrm{Irr}(R)$ such that q divides $\varphi (1)$ and let $\psi = \theta \times \varphi \in \textrm{Irr}(KR)$. Since $I_G(\psi ) \le I_G(\theta )$, tq divides $\chi (1)$ for every $\chi \in \textrm{Irr}(G)$ that lies over $\theta$. Finally, if $p \in {V}(G)$ but $p \not \in \pi (G/R)$, then $p \in {V}(R)$; so if $q \ne p$, then $q \in \pi (G/R)$ by the assumption on ${V}(G)$, and hence by what we have just proved q is adjacent to p as well. So, q is a complete vertex of $\Delta (G)$.

We now move to claim (b). Assuming $2 \in \pi (G/KR)$ and $a \ge 3$, by Theorem 2.4 we get that 2 is adjacent in $\Delta (G)$ to all primes $\ne 2$ of $\pi (G/R)$; so to p as well if $p \ne 2$ and $p \in \pi (G/R)$. On the other hand, if $p \in {V}(G) - \pi (G/R)$, so $p\ne 2$, then $p \in {V}(R)$ and p is adjacent to 2 in $\Delta (G)$ by part (a). Hence, 2 is a complete vertex of $\Delta (G)$.

Finally, we prove claim (c). Assume that $2 \not \in \pi (G/KR) \cup {V}(R)$. Then every character $\chi \in \textrm{Irr}(G)$ such that $\chi (1)$ is even lies over a character $\psi \in \textrm{Irr}(KR)$ with $\psi (1)$ even. Writing $\psi = \alpha \times \beta$ with $\alpha \in \textrm{Irr}(K)$ and $\beta \in \textrm{Irr}(R)$, since $2 \not \in {V}(R)$ we deduce that $\alpha$ has even degree, and hence $\alpha$ is the Steinberg character of K. Thus $\alpha$ extends to G (see for instance [13]) and hence $\chi (1) = \alpha (1) \gamma (1) = 2^a\gamma (1)$ for a suitable $\gamma \in \textrm{Irr}(G/K)$, concluding the proof. $\square$

Theorem 4.2

Let R and K be, respectively, the solvable radical and the solvable residual of the group G and assume that G has a composition factor $S \cong \textrm{SL}_{2}(2^a)$, with $a \ge 3$. Then, $\Delta (G)$ is a connected graph and it has a cut-vertex p if and only if G/R is an almost simple group with socle isomorphic to S, ${V}(G) = \pi (G/R) \cup \{ p\}$ and one of the following holds.

• (a) K is a minimal normal subgroup of G, $K \cong S$ and either $p=2$ and ${V}(G/K) \cup \pi (G/KR) = \{ 2\}$, or $p\ne 2$, ${V}(G/K) = \{p\}$ and G/KR has odd order.

• (b) K contains a minimal normal subgroup L of G such that $K/L \cong S$, L is the natural module for K/L, $p \ne 2$, ${V}(G/K) = \{p\}$, G/KR has odd order and, for a Sylow 2-subgroup T of G, $T' = (T \cap K)'$.

In all cases, p is is a complete vertex and the only cut-vertex of $\Delta (G)$.

Proof

We start by proving the “only if” part of the statement, assuming that $\Delta (G)$ is connected and that has a cut-vertex p. Then, by Theorem 2.8G/R is an almost-simple group and ${V}(G) = \pi (G/R) \cup \{p\}$. As a consequence, we have that the socle M/R of G/R is isomorphic to S. Let $L = K \cap R$; since $KR = M$, we see that $K/L \cong S$.

We observe that by Theorem 2.4 every prime in $\pi (G/KR)$ is adjacent in $\Delta (G)$ to every other vertex in $\Delta (G)$, except possibly 2 and p. Moreover, part (a) of Lemma 4.1 yields that ${V}(R) \subseteq \{p\}$.

We consider first the situation arising when $L=1$. Assuming $p=2$, then ${V}(G) = \pi (G/R)$ and by the above observation we deduce that G/KR is a 2-group and that ${V}(R) \subseteq \{2\}$. If $G = KR = K \times R$, then, as $\Delta (G)$ is connected and 2 is a cut-vertex of $\Delta (G)$, it immediately follows that ${V}(R) = \{2\}$. So, in any case, ${V}(G/K) \cup \pi (G/KR) = \{ 2\}$. Assuming instead $p \ne 2$, then (since no vertex in ${V}(G) -\{p \}$ can be complete in $\Delta (G)$) part (b) of Lemma 4.1 implies that |G/KR| is odd and it only remains to show that ${V}(G/K) = \{p\}$. As ${V}(R) \subseteq \{p\}$ and $p \ne 2$, part (c) of Lemma 4.1 yields that 2 is adjacent in $\Delta (G)$ to all primes in ${V}(G/K)$, and to them only. As $\Delta (G)$ is connected, it follows that ${V}(G/K)$ is non-empty.

If $q \in {V}(G/K)$ and $q\ne p$, then q divides |G/KR| (because ${V}(KR/K) = {V}(R) \subseteq \{ p \}$) and hence, by Theorem 2.4, q (being adjacent also to 2) would be a complete vertex of $\Delta (G)$, a contradiction. Hence, ${V}(G/K) = \{p\}$.

We assume now $L \ne 1$. Then, by Theorem 3.2, L is a minimal normal subgroup of G and L is the natural module for $K/L \cong S$. By Remark 2.7, the subgraph of $\Delta (G)$ induced by the vertex set ${V}(G) - \{ 2,p\}$ is a complete graph. Hence, the assumptions on $\Delta (G)$ imply that $p \ne 2$ and that 2 is adjacent only to p in $\Delta (G)$. Moreover, recalling that $\Delta (G/L)$ is a subgraph of $\Delta (G)$, by part (a) and part (b) of Lemma 4.1 we deduce that $2 \not \in {V}(R/L) \cup \pi (G/KR)$ and hence, by part (c) of the same lemma, that ${V}(G/K)= \{p\}$. Let now T be a Sylow 2-subgroup of G; as |G/KR| is odd, then $T \le KR$. Setting $T_0 = T \cap R$, we observe that $T_0/L$ is an abelian normal Sylow 2-subgroup of R/L because $2 \not \in {V}(R/L)$. Let $T_1 = T \cap K$ and assume, working by contradiction, that $T' \ne T_1'$. Let $\lambda \in \textrm{Irr}(L)$ be a non-principal character; by Lemma 2.6$L \le \textbf{Z}(T_0)$, so $\lambda$ is $T_0$-invariant and, since L is a self-dual K/L-module, $I_{K}(\lambda )/L$ is a Sylow 2-subgroup of K/L. Hence, since $T = T_0T_1$, we can assume (up to conjugation) that $\lambda$ is T-invariant. So, by Lemma 2.6$\lambda$ has no extension to T. As $I_{K}(\lambda )/L = T_1/L$ and $KR/L = K/L \times R/L$, T/L is a normal subgroup of $I_{KR}(\lambda )$ and hence 2 divides the degree of every irreducible character $\psi$ of $I_{KR}(\lambda )$ that lies over $\lambda$. By Clifford correspondence, it follows that 2 is adjacent in $\Delta (G)$ to all primes in $\pi (2^{2a} -1) = \pi (|K:I_K(\lambda )|)$, a contradiction. Hence, $T' = (T\cap K)'$.

We proceed now to prove the “if” part of the statement and we assume that G/R is an almost simple group and that ${V}(G) = \pi (G/R) \cup \{p\}$ for some prime p.

Suppose first that (a) holds, so K is a minimal normal subgroup of G and $K\cong S$. Hence, $KR = K \times R$. Assume that $p=2$, so ${V}(G) = \pi (G/R)$, and that ${V}(G/K) \cup \pi (G/KR) = \{ 2\}$. If $KR < G$, then 2 is a complete vertex of $\Delta (G)$ by part (b) of Lemma 4.1, and if $G = KR$, then the same is true because in this case ${V}(R) = {V}(G/K) = \{2\}$. For $\chi \in \textrm{Irr}(G)$ and an irreducible constituent $\psi$ of $\chi _{KR}$, the odd parts of $\chi (1)$ and of $\psi (1)$ coincide by [8, Corollary 11.29], so by part (a) of Theorem 2.3 the graph $\Delta (G)-2$, obtained by deleting the vertex 2 and all incident edges, has two complete connected components, with vertex sets $\pi (2^a-1)$ and $\pi (2^a+1)$. So, 2 is a cut-vertex of $\Delta (G)$ and, being a complete vertex of $\Delta (G)$, it is the unique cut-vertex of $\Delta (G)$. If $p\ne 2$, ${V}(G/K) = \{p\}$ and G/KR has odd order, then (as $R \cong KR/K \trianglelefteq \,G/K$) $2 \not \in {V}(R)$ and by part (c) of Lemma 4.1 the vertex 2 is adjacent only to p in $\Delta (G)$. Hence, p is a cut-vertex of $\Delta (G)$. We also observe that p is a complete vertex of $\Delta (G)$: this is a consequence of Theorem 2.4 if $p \in \pi (G/KR)$, while if $p \not \in \pi (G/KR)$ the assumption ${V}(G/K) = \{p\}$ implies that $p \in {V}(R)$ and hence the claim follows by part (a) of Lemma 4.1. Thus, p is the unique cut-vertex of $\Delta (G)$.

We assume now that (b) holds, so K contains a minimal normal subgroup L of G such that $K/L \cong S$ and L is the natural module for K/L. Moreover, $p \ne 2$, ${V}(G/K) = \{p\}$, G/KR has odd order and, for any Sylow 2-subgroup T of G, $T' = (T \cap K)'$. For a non-principal $\lambda \in \textrm{Irr}(L)$, the argument used in the fourth paragraph of this proof shows that $I = I_G(\lambda )$ contains a Sylow 2-subgroup T of G, and T/L is abelian and normal in I/L. By Lemma 2.6$\lambda$ extends to T and hence $\lambda$ extends to $I_G(\lambda )$ by [8, Theorem 6.26]. So, Gallagher’s theorem implies that every irreducible character of G that lies over $\lambda$ has odd degree. We hence deduce that if $\chi \in \textrm{Irr}(G)$ has even degree, then $\chi \in \textrm{Irr}(G/L)$. Then, by part (c) of Lemma 4.1, 2 is adjacent only to p in $\Delta (G)$. So, by Remark 2.7, the graph obtained by removing the vertex p from $\Delta (G)$ has two connected components: the single vertex 2 and the complete graph with vertex set ${V}(G) -\{2,p\}$. By the discussion of case (a), we know that p is a complete vertex of $\Delta (G/L)$, hence of $\Delta (G)$; thus, p is the only cut-vertex of $\Delta (G)$. $\square$

5 Proof of Theorem 2

The last section of this paper is devoted to the proof of Theorem 2, that we state again (in a slightly different form, for technical reasons) as Theorem 5.3.

Lemma 5.1

Let K be a normal subgroup of the group G with $K \cong \textrm{SL}_{2}(4)$ or $K \cong \textrm{SL}_{2}(5)$. Let R be the solvable radical of G, $N = K \cap R$ and assume that ${V}(G) = \{2,3,5, p \}$ for a suitable prime p. Then

• (a) The primes in ${V}(G/K)$ (if any) are complete vertices of $\Delta (G)$.

• (b) If $N \ne 1$ or $KR \ne G$, then 2 is adjacent to 3 in $\Delta (G)$.

• (c) If $5 \not \in {V}(G/K)$, then 5 is adjacent in $\Delta (G)$ exactly to the primes in ${V}(G/K)$.

Proof

(a): By part (a) of Lemma 4.1 the primes in ${V}(KR/K) = {V}(R/N)$ are complete vertices of $\Delta (G)$. Let $q \in {V}(G/K) - {V}(KR/K)$; then for $\chi \in \textrm{Irr}(G/K)$ such that q divides $\chi (1)$ and an irreducible constituent $\theta$ of $\chi _{KR/K}$, q divides $\chi (1)/\theta (1)$ by Clifford’s theorem and $\chi (1)/\theta (1)$ divides |G/KR| by [8, Corollary 11.29]. As $|G/KR| \le 2$, we have $q = |G/KR| = 2$ and $\chi = \theta ^{G/K}$. Seeing by inflation $\theta \in \textrm{Irr}(KR/N)$ with $K/N \le \ker {\theta }$, we write $\theta = 1_{K/N} \times \psi$, with $\psi \in \textrm{Irr}(R/N)$ and $I_{G/N}(\psi ) = I_{G/N}(\theta ) = KR/N$. So, for every $\varphi \in \textrm{Irr}(K/N)$, $\varphi \times \psi \in \textrm{Irr}(KR/N)$ and $I_{G/N}(\varphi \times \psi ) = KR/N$, hence 2 is adjacent to both 3 and 5 in $\Delta (G)$. If $p \not \in \{2,3,5\}$, then (since $|N| \le 2$ and $p \in {V}(G)$) $R/N \cong KR/K$ cannot have a normal abelian Sylow p-subgroup, so $p \in {V}(KR/K)$ is adjacent to 2 in $\Delta (G)$ and $q= 2$ is a complete vertex of $\Delta (G)$.

Part (b) is clear, as both $\textrm{SL}_{2}(5)$ and $\textrm{Aut}(\textrm{SL}_{2}(4)) \cong S_5$ have an irreducible character of degree 6.

(c): Since $|G/KR| \le 2$, KR contains every Sylow 5-subgroup of G and, as $5\not \in {V}(R)\subseteq {V}(G/K)$, if $\chi \in \textrm{Irr}(G)$ has degree divisible by 5, then $\chi$ lies (both if $K \cong \textrm{SL}_{2}(4)$, as well as if $K \cong \textrm{SL}_{2}(5)$) over the unique character $\alpha \in \textrm{Irr}(K)$ such that 5 divides $\alpha (1)$. It is easily seen that $\alpha$ extends to G. By Gallagher’s theorem, we conclude that 5 is adjacent only to the vertices of ${V}(G/K)$ in $\Delta (G)$. $\square$

Lemma 5.2

Let R and K be, respectively, the solvable radical and the solvable residual of the group G, and let $N = R \cap K$.

• (a) If $2\not \in {V}(G/K)$, $G = KR$ and N is the natural module for $K/N \cong \textrm{SL}_{2}(4)$, then $N \le \ker {\chi }$ for every $\chi \in \textrm{Irr}(G)$ such that $\chi (1)$ is even.

• (b) Let $L \trianglelefteq \,G$, $L \le N$, be such that $K/L\cong \textrm{SL}_{2}(5)$ and L is the natural module for K/L. If $5 \not \in {V}(G/K)$, then 5 is adjacent in $\Delta (G)$ exactly to the primes in ${V}(G/K)$.

Proof

(a): Assume that $2\not \in {V}(G/K)$, $G = KR$ and that N is the natural module for $K/N \cong \textrm{SL}_{2}(4)$. Let $\lambda \in \textrm{Irr}(N)$ be a non-principal character and let $I = I_G(\lambda )$, T a Sylow 2-subgroup of I, $T_0 = T \cap R$ and $T_1 = T \cap K$. Since, by Lemma 2.6, I contains a Sylow 2-subgroup of R, we see that $T_0\in \textrm{Syl}_2(R)$; moreover, as $2 \not \in {V}(G/K) = {V}(R/N)$, $T_0/N$ is abelian and $T_0 \trianglelefteq \,R$. For $B/N \in \textrm{Syl}_3(K/N)$, as $N \le \textbf{Z}(T_0)$ and $[B/N, T_0/N] =1$ by coprimality we get $T_0 = N \textbf{C}_{T_0}(B) = N \times \textbf{C}_{T_0}(B)$, because $\textbf{C}_{N}(B)= 1$; in particular, $T_0$ is abelian. Write $C = \textbf{C}_{T_0}(B)$ and $D = \textbf{C}_{T_0}(K)$; so $D \trianglelefteq \,C$. Since $I\cap K$ is a Sylow 2-subgroup of K, we have $T \in \textrm{Syl}_2 (G)$. As $T = T_1T_0$, we have $T' = T_1' [T_1, T_0] T_0' = T_1'[T_1, T_0]$. We claim that $[T_1, T_0] \le T_1'$. Observing that $[T_1, T_0] = [T_1, N][T_1, C]$, it is enough to prove that $[T_1, C/D]\le T_1'$. Identifying C/D with a normal subgroup of $\mathrm{{Out}}(K)$, one can check (for instance by GAP [14], as $K = \text {SmallGroup(960,11357)}$) that

$$\begin{aligned}{}[T_1, C/D] \le [T_1, \textbf{O}_{2}(\mathrm{{Out}}(K))] \le T_1', \end{aligned}$$

so the claim follows. Hence, $T' = T_1'$ and by Lemma 2.6$\lambda$ extends to T. Thus, by [8, Theorem 6.26] $\lambda$ extends to I. As I/N has odd index in G/N and has a normal abelian Sylow 2-subgroup, it follows that every irreducible character of G lying over $\lambda$, where $\lambda$ is any non-principal character of N, has odd degree.

(b):We observe that G splits over L. In fact, if X is a Sylow 2-subgroup of N (so, $|X| = |N/L|= 2$), then by the Frattini argument $G = L \textbf{C}_{G}(X)$ and, as X acts fixed-point-freely on L, $L \cap \textbf{C}_{G}(X) = 1$.

Let $Q_0 \in \textrm{Syl}_5(R)$; since $R/N \cong KR/K \trianglelefteq \,G/K$, ${V}(R/N) \subseteq {V}(G/K)$ and $5 \not \in {V}(R/N)$, so $Q_0N/N$ is abelian and normal in R/N. As $N/L\trianglelefteq \,R/L$ and $|N/L|= 2$, N/L is central in R/L and it follows that $Q_0/L \trianglelefteq \,G/L$, so $Q_0 \trianglelefteq \,G$. For a non-principal $\lambda \in \textrm{Irr}(L)$, $I_K(\lambda ) = Q_1 \in \textrm{Syl}_5(K)$. So, as $|G/KR| \le 2$, $Q = Q_0Q_1 \in \textrm{Syl}_5(\,)G$ and $Q \le I = I_G(\lambda )$. Since G splits over L, $\lambda$ extends to I and, as $Q/L = Q_1/L \times Q_0/L$ is abelian and normal in I/L, by Gallagher’s theorem and Clifford correspondence it follows that 5 does not divide $\chi (1)$ for every $\chi \in \textrm{Irr}(G)$ that lies over $\lambda$. Thus, L is contained in the kernel of every irreducible character of G with degree divisible by 5, and part (c) of Lemma 5.1 applied to G/L yields that 5 is adjacent in $\Delta (G)$ exactly to the primes in ${V}(G/K)$. $\square$

Theorem 5.3

Let R and K be, respectively, the solvable radical and the solvable residual of the group G and assume that G has a composition factor $S \cong \textrm{SL}_{2}(4)$. Let $N = K \cap R$. Then, $\Delta (G)$ is a connected graph and has a cut-vertex p if and only if G/R is an almost simple group with socle isomorphic to S, ${V}(G) = \{2,3,5\} \cup \{ p\}$ and one of the following holds.

• (a) K is isomorphic either to $\textrm{SL}_{2}(4)$ or to $\textrm{SL}_{2}(5)$ and ${V}(G/K) =\{p\}$; if $p=5$, then $K \cong \textrm{SL}_{2}(4)$ and $G= K\times R$.

• (b) $K/N \cong \textrm{SL}_{2}(4)$, $|N| = 2^4$, $G = KR$ and one of the following:

  • (i) N is the natural module for K/N, $p \ne 2$, ${V}(G/K)=\{p\}$.

  • (ii) N isomorphic to the restriction to K/L, embedded as $\Omega _4^-(2)$ into $\textrm{SL}_4(2)$, of the standard module of $\textrm{SL}_4(2)$. Moreover, $p = 5$, $G = K \times R_0$, where $R_0 = \textbf{C}_{G}(K)$, and ${V}(R_0) = {V}(G/K) \subseteq \{ 5\}$;

• (c) There exists $1\not = L \le N$, L normal in G, with $K/L \cong \textrm{SL}_{2}(5)$ and one of the following:

  • (i) $|L| = 5^2$, L is the natural module for $\textrm{SL}_{2}(5)$, $p\not =5$ and ${V}(G/K)=\{p\}$.

  • (ii) $|L| = 3^4$, L is the natural module for $\textrm{SL}_{2}(5)$ seen as a subgroup of $\mathrm{{GL}}_4(3)$, $p=2$ and ${V}(G/K)\subseteq \{2\}$.

In all cases, p is is a complete vertex and the only cut-vertex of $\Delta (G)$.

Proof

We start by proving the “only if” part of the statement, assuming that $\Delta (G)$ is connected and that it has a cut-vertex p. Then, by Theorem 2.8G/R is an almost-simple group and ${V}(G) = \pi (G/R) \cup \{p\}$. So, the socle M/R of G/R is isomorphic to $\textrm{SL}_{2}(4)$, and ${V}(G) = \{ 2,3,5,p\}$. Hence, the subgraph of $\Delta (G)$ induced by the set of vertices $\{2,3,5\}$ cannot be a clique. As $N = K \cap R$ and $KR = M$, then $K/N\cong M/R \cong \textrm{SL}_{2}(4)$ and $|G/KR|\le 2$. Since no vertex of $\Delta (G)$ different from p can be complete, part (a) of Lemma 5.1 implies that ${V}(G/K) \subseteq \{p\}$.

We now apply Theorem 3.4, considering the possible structure types for the solvable residual K of G.

If K is isomorphic either to $\textrm{SL}_{2}(4)$ or to $\textrm{SL}_{2}(5)$ (i.e., $|N| \le 2$), then part (c) of Lemma 5.1 implies (as 5 cannot be an isolated vertex of $\Delta (G)$) that ${V}(G/K)$ is non-empty, so ${V}(G/K) = \{ p\}$. By part (b) of Lemma 5.1$p \ne 5$ when $K \cong \textrm{SL}_{2}(5)$ or $KR \ne G$; so we have case (a).

Assume now that $|N| > 2$, and that N is a minimal normal subgroup of G. Then, by Theorem 3.4$K/N \cong \textrm{SL}_{2}(4)$, $|N| = 2^4$ and we have two cases:

(x): N is the natural module for K/N: then 3 and 5 are adjacent in $\Delta (G)$ (see Remark 2.7), and hence $p \ne 2$, as othewise $\Delta (G)$ would be a complete graph. We show that $G = KR$: in fact, if this is not the case, then $G/R \cong S_5$ and the Sylow 2-subgroups of G/N are non-abelian. For a non-principal $\lambda \in \textrm{Irr}(L)$ and $I = I_G(\lambda )$, 15 divides |G : I|. Hence, recalling Theorem A of [11], independently on the parity of |G : I| there exists $\chi \in \textrm{Irr}(G)$, lying above $\lambda$, that has degree 30, a contradiction. Finally, we observe that if $G/K \cong R/N$ is abelian, then 2 is an isolated vertex of $\Delta (G)$, because by part (a) of Lemma 5.2 every $\chi \in \textrm{Irr}(G)$ of even degree is a character of $G/N = K/N \times R/N$. So, ${V}(G/K) = \{p\}$ and we have case (b)(i).

(xx): N is the restriction to K/L, embedded as $\Omega _4^-(2)$ into $\textrm{SL}_4(2)$, of the standard module of $\textrm{SL}_4(2)$. Then $\Delta (K)$ is the graph $2-5-3$ and hence necessarily $p = 5$.

Let $R_0 = \textbf{C}_{G}(K)$ and $C = \textbf{C}_{G}(N)$. So, $N \le C \trianglelefteq \,G$ and $R_0 \le C \le R$, since K/N is the only non-solvable composition factor of G, and it acts non-trivially on N. As $\mathrm{{H}}^2(K/N, N) = 0$, K splits over N; let $K_0$ be a complement of N in K. Note that $R_0 = \textbf{C}_{C}(K) = \textbf{C}_{C}(K_0)$. We prove that $C = N \times R_0$. It is enough to show that $C = N R_0$, since $\textbf{Z}(K) = 1$. As $[K, R] \le N$, in particular $[K_0, C] \le N$ and hence $K_0^c \le K_0N = K$ for every $c \in C$. Since $\textrm{H}^1(K_0, N) = 0$, all complements of N in K are conjugate in K. It follows that there exists an element $b \in N$ such that $K_0^c = K_0^b$, so $d = bc^{-1} \in \textbf{N}_{C}(K_0)$ and hence $[K_0, d] \le K_0 \cap C = 1$, as $K_0 \cong K/N$ acts faithfully on N. Thus, $d \in R_0$. So, $C = N R_0 = N \times R_0$.

The action of G on N gives an embedding $\phi$ of $\overline{G} = G/C$ in ${\widehat{G}} = \textrm{GL}_4(2)$. One can check (for instance by GAP [14]) that $\textbf{N}_{{\widehat{G}}}(\phi (\overline{K})) \cong S_5$, and that if $\phi (\overline{G}) \cong S_5$ then $\Delta (G/R_0)$, which is a subgraph of $\Delta (G)$, has a complete subgraph with vertex set $\{2,3,5\}$, a contradiction. So, $\phi (\overline{G}) = \phi (\overline{K})$, and hence $G = K \times R_0$, giving case (b)(ii).

As the final case, we assume that G has a minimal normal subgroup L, such that $L \le N$ and $K/L \cong \textrm{SL}_{2}(5)$. We have two possible cases:

(y): L is the natural module for K/L. Then $\Delta (K)$ is the graph with vertex set {2,3,5} where 5 is an isolated vertex and 2, 3 are adjacent, so we deduce that $p \ne 5$. Moreover, part (b) of Lemma 5.2 yields that 5 is adjacent in $\Delta (G)$ only to the primes in ${V}(G/K)$. Thus, as $\Delta (G)$ is connected, ${V}(G/K) \ne \emptyset$, so ${V}(G/K) = \{p \}$ and we have case (c)(i).

(yy): L is the natural module for K/L seen as a subgroup of $\textrm{GL}_4(3)$. So, $\Delta (K)$ is the graph $3-2-5$ and consequently $p = 2$ and we have case (c)(ii).

We now prove the “if” part of the statement, going through the various cases.

(a): If $G \cong \textrm{SL}_{2}(4) \times R$ with ${V}(R) = {V}(G/K) = \{5\}$, then clearly $\Delta (G)$ is the graph $2-5-3$. If $p \ne 5$, then 5 is adjacent only to p in $\Delta (G)$ by part (c) of Lemma 5.1. By part (a) of Lemma 5.1, p is a complete vertex, and hence the only cut-vertex, of $\Delta (G)$.

(b): We assume that $G = KR$ and that $N = K \cap R$ is a normal in G of order $2^4$.

In case (b)(i), since $G/N = K/N \times R/N$ and ${V}(R/N) = {V}(G/K) = \{p\}$ for some prime $p\ne 2$, part (a) of Lemma 5.2 and part (a) of Theorem 2.3 yield that the vertex 2 is adjacent only to p in $\Delta (G)$, so p is a cut-vertex of $\Delta (G)$. By part (a) of Lemma 5.1, p is a complete vertex, and hence the only cut-vertex, of $\Delta (G)$.

In case (b)(ii), it is clear that $\Delta (G) = \Delta (K)$ is the graph $2-5-3$.

(c): We assume that there exists $L \trianglelefteq \,G$, $L \le K$, such that $K/L \cong \textrm{SL}_{2}(5)$.

In case (c)(i), by part (b) of Lemma 5.2 the vertex 5 is adjacent only to p ($p \ne 5$) in $\Delta (G)$ and, by part (a) of the same lemma, p is a complete vertex of $\Delta (G)$.

In case (c)(ii), we prove that $\Delta (G) = \Delta (K)$, so $\Delta (G)$ is the graph $3-2-5$. To this end, it is enough to show that 3 and 5 are non-adjacent in $\Delta (G)$. Since $|G/KR| \le 2$, KR contains a Sylow 3-subgroup Q of G; moreover, as ${V}(R/N) \subseteq {V}(G/K) \subseteq \{2\}$ and $|N/L| = 2$, it easily follows that, setting $Q_0 = Q \cap R$, $Q_0/L$ is abelian and normal in R/L, and hence in G/L. Let $\lambda \in \textrm{Irr}(L)$ be a non-principal character and let $I = I_G(\lambda )$. An application of the Frattini argument, as in the proof of part (b) of Lemma 5.2, proves that G splits over L, so $\lambda$ extends to I. By [5, Lemma 2.6], $L \le \textbf{Z}(Q_0)$ and hence, since $I \cap K$ is a Sylow 3-subgroup of K, we can assume $Q \le I$. So, Q/L is an abelian Sylow 3-subgroup of G/L and it is normal in I/L. Thus, by Gallagher’s theorem we deduce that every $\chi \in \textrm{Irr}(G)$ that lies over $\lambda$ has degree not divisible by 3. Hence, if $\chi \in \textrm{Irr}(G)$ and 3 divides $\chi (1)$, then $L \le \ker {\chi }$ and $\chi \in \textrm{Irr}(G/L)$. Now, an application of part (c) of Lemma 5.1 yields that 5 not adjacent to 3 in $\Delta (G/L)$, and hence 3 and 5 are not adjacent in $\Delta (G)$.

So, in every case, p is a cut-vertex of $\Delta (G)$ and, as p is also a complete vertex of $\Delta (G)$, there are no other cut-vertices in $\Delta (G)$. The proof is complete. $\square$