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

Let G be a finite group. Denoting by cd(G)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\textrm{cd}(G)$$\end{document} 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 cd(G)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\textrm{cd}(G)$$\end{document}, and two distinct vertices p, q are adjacent if and only if pq divides some number in cd(G)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\textrm{cd}(G)$$\end{document}. 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≅PSL2(2a)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$S\cong \textrm{PSL}_{2}(2^a)$$\end{document} for some integer a≥2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a\ge 2$$\end{document}; the remaining case is addressed in the present paper.


Introduction
The character degree graph ∆(G) of a finite group G is a very useful tool for studying the arithmetical structure of the set cd(G) = {χ(1) : χ ∈ 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 ∆(G).
In the papers [5] and [6] we considered the problem of classifying the finite non-solvable groups G such that ∆(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 PSL 2 (t a ) (where t a is a prime power greater than 3), the Suzuki group Sz(2 a ) (where 2 a − 1 is a prime number), PSL 3 (4), the Mathieu group M 11 , and the first Janko group J 1 .The aforementioned papers carry out an analysis (and provide a complete classification) of all the possibilities, except for the case S ∼ = PSL 2 (2 a ) when ∆(G) is connected; the present work addresses the remaining case, thus completing the classification of these groups.We refer the reader to [5] and [6] for a thorough description of the problem and, in particular, for the full statements of the relevant theorems (see the introductions of [5] and [6], and Section 2 of [6]).
The situation that remains to be studied is treated in the following Theorem 1 and Theorem 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 = SL 2 (t a ) (where t a is a prime power), an H-module V over the field F t of order t is called the natural module for H if V is isomorphic to the standard module for SL 2 (t a ), or any of its Galois conjugates, seen as an 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 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 ∼ = SL 2 (2 a ), with a ≥ 3.Then, ∆(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) = π(G/R) ∪ {p} and one of the following holds.
(a) K ∼ = S is a minimal normal subgroup of G; also, either p = 2 and V(G/K) ∪ π(G/KR) = {2}, or p = 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 ∼ = S and L is the natural module for K/L; also, p = 2, V(G/K) = {p}, G/KR has odd order and, for a Sylow 2-subgroup T of G, we have In all cases, p is a complete vertex and the unique cut-vertex of ∆(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 ∼ = SL 2 (4).Then, ∆(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} ∪ {p} and one of the following holds.
(a) K is isomorphic either to SL 2 (4) or to SL 2 (5), and V(G/K) = {p}; if p = 5, then K ∼ = SL 2 (4) and G = K × 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 = 2 and V(G/K) = {p}, (ii) or L is isomorphic to the restriction to K/L, embedded as Ω − 4 (2) into SL 4 (2), of the standard module of SL 4 (2).Moreover p = 5, G = K × R 0 where R 0 = C G (K), and V(R 0 ) = V(G/K) ⊆ {5}.(c) K contains a minimal normal subgroup L of G such that K/L is isomorphic to SL 2 (5), and (i) either L is the natural module for K/L, p = 5 and V(G/K) = {p}, (ii) or L is isomorphic to the restriction to K/L, embedded in SL 4 (3), of the standard module of SL 4 (3), p = 2 and V(G/K) ⊆ {2}.
In all cases, p is a complete vertex and the unique cut-vertex of ∆(G).
To conclude this introduction, we display in Table 1 the graphs related to the groups as in Theorem 1 and Theorem 2, so, all the possible connected graphs having a cut-vertex p, of the form ∆(G) where G is a finite group with a composition factor isomorphic to SL 2 (2 a ), a ≥ 2. The first row of the table shows the graphs arising from Theorem 1, whereas the second row shows the graphs arising from Theorem 2 in 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 occur for groups as in Theorem 2(a) (it is enough to consider the direct product SL 2 (4) × R where R is a non-abelian q-group, for q ∈ {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.

Preliminaries
Given a group G, we denote by ∆(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 [9]).
For a positive integer n, the set of prime divisors of n will be denoted by π(n), and we simply write π(G) for π(|G|).If q is a prime power, then the symbol F q will denote the field of order q.
We start by recalling some structural properties of the groups SL 2 (2 a ).(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 |SL 2 (2 a )|; (iv) SL 2 (2 b ), where b is a proper divisor of a.
When dealing with subgroups of 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 + ).Next, some properties of the degree graph of simple and almost-simple groups.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.
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 PSL 2 (t a ), for a prime t with t a > 4 and t a = 9.Then, denoting by K the solvable residual of G, one of the following conclusions holds.
and every non-principal irreducible character of L/L ′ is not invariant in K.
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 ∼ = SL 2 (2 a ) with a ≥ 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 ∩ R and T 1 = T ∩ K.
Then L ≤ Z(T 0 ).Furthermore, every non-principal T -invariant character λ ∈ Irr(L) extends to T 1 and, assuming that T 0 /L is abelian, λ extends to T if and only if Proof.Observe that L = K ∩ R is an elementary abelian 2-group of order 2 2a , T 0 is normalized by K and T = T 0 T 1 .As Z(T 0 ) ∩ L is non-trivial and normal in K, by the irreducibility of L as a K-module it follows that L ≤ Z(T 0 ).It is well known that N K (T 1 )/L = N K/L (T 1 /L) is a subgroup of type (ii) of K/L whose order is 2 a • (2 a − 1) (in fact, N K/L (T 1 /L) can be identified with the subgroup of lower-triangular matrices of SL 2 (2 a )); thus we write N K (T 1 ) = T 1 D, 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 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 L/Z.They are clearly T -invariant and they extend to T 1 , because T 1 /Z is abelian.
Let λ ∈ Irr(L) be a non-principal T -invariant character and assume that λ has an extension τ ∈ Irr(T ).Since τ (1) = λ(1) = 1, we have T ′ ≤ ker(τ ).Assuming that T 0 /L is abelian, then T /L = T 0 /L × T 1 /L is abelian and T ′ ≤ L. So, T ′ = T ′ ∩ L ≤ ker(τ L ) = ker(λ) and, as λ = 1 L , hence T ′ < L. Observing that T ′ is normalized by D and that D acts irreducibly on L/Z, we conclude that T ′ ≤ Z and, since then as observed above Z ≤ ker(λ) and, assuming Z = T ′ , clearly λ, seen as a character of L/Z, extends to the abelian group T /Z.Remark 2.7.Let K be a group having a normal subgroup L with K/L ∼ = SL 2 (2 a ) (for a ≥ 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 ∆(K) is disconnected with two connected components, whose vertex sets are {2} and π(K) − {2} respectively, and which are both complete subgraphs of ∆(K).
In fact, by Theorem 2.3, ∆(K/L) has three connected components with vertex sets π(2 a −1), π(2 a +1) and {2} respectively, which are all complete subgraphs of ∆(K/L).On the other hand, if λ is any nonprincipal character in Irr(L), then I K (λ) is a Sylow 2-subgroup of K, and Lemma 2.6 guarantees that λ extends to I K (λ); our claim then easily follows by Clifford's theory.
Theorem 2.8.Let G be a non-solvable group such that ∆(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) = π(G/R) ∪ {p}.
To conclude this preliminary section, we recall the statements of three crucial results proved in [5] and [6], concerning certain module actions of 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 N q if q divides |H : C H (V )| and, for every non-trivial v ∈ V , there exists a Sylow q-subgroup Q of H such that Q C H (v) (see [2]).
If (H, V ) satisfies 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 F r (see Lemma 4 of [17]).Lemma 2.9 ([6, Lemma 3.10]).Let t, q, r be prime numbers, let H = SL 2 (t a ) (with t a ≥ 4) and let V be an F r [H]-module.Then (H, V ) satisfies N q if and only if either t a = 5 and V is the natural module for H/C H (V ) ∼ = SL 2 (4) or V is faithful and one of the following holds.
Theorem 2.10 ([5, Theorem 3.3]).Let V be a non-trivial irreducible module for G = SL 2 (t a ) over the field F q , where t a ≥ 4 and q is a prime number, q = t.For odd primes r ∈ π(t a − 1) and s ∈ π(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 Theorem 3.4]).Let T be a Sylow t-subgroup of G ∼ = SL 2 (t a ) (where t a ≥ 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 C V (G) = 1, consider the sets Then the following conditions are equivalent.
(a) V I and V II are both non-empty and , of the standard module of SL 4 (2).

The structure of the solvable residual
Let G be a group having a composition factor isomorphic to SL 2 (2 a ) (with a ≥ 2), such that ∆(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 ∼ = SL 2 (2 a ), or K ∼ = SL 2 (5), or K contains a minimal normal subgroup L of G such that either K/L ∼ = SL 2 (2 a ) or K/L ∼ = 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 Theorem 3.2 and Theorem 3.4 respectively.Theorem 3.1.Assume that the group G has a composition factor isomorphic to SL 2 (2 a ) with a ≥ 2, and let p be a prime number.Assume also that ∆(G) is connected with cut-vertex p.Then, denoting by K the solvable residual of G, one of the following conclusions holds.
or to 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 SL 2 (4).Moreover, L is isomorphic to the restriction to K/L, embedded as Ω − 4 (2) into SL 4 (2), of the standard module of SL 4 (2).(d) a = 2, and K contains a minimal normal subgroup L of G such that K/L is isomorphic to SL 2 (5).Moreover, L is isomorphic to the restriction to K/L, embedded in SL 4 (3), of the standard module of 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 ≤ b < n.In this case, n is the order of m modulo q, so n divides q − 1.In view of [11,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 θ ∈ Irr(N ), we denote by Irr(G|θ) the set of all irreducible characters of G that lie over θ.Theorem 3.2.Assume that the group G has a composition factor isomorphic to SL 2 (2 a ) with a > 2, and let p be a prime number.Assume also that ∆(G) is connected with cut-vertex p.Then, denoting by K the solvable residual of G, one of the following conclusions holds.
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 SL 2 (2 a ), and V(G) = π(G/R) ∪ {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 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 ξ in Irr(L/L ′ ): as ), its possible structures are described in Remark 2.1.In particular, if 2 is not a divisor of |K : I K (ξ)|, then I K (ξ)/L contains a Sylow 2-subgroup of K/L as a normal subgroup.Assuming for the moment that this happens for every non-principal ξ ∈ Irr(L/L ′ ), Lemma 2.9 (together with the paragraph preceding it) yields that the dual group 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 = O 2 (K) is a minimal normal subgroup of G), and this is what we do next.
For a proof by contradiction assume L ′ = 1, and consider a chief factor L ′ /Z of K.As observed in Remark 2.7, the graph ∆(K/L ′ ) has two connected components having vertex sets {2} and π(K/L ′ ) − {2}, respectively; since the vertex set of ∆(G) is π(G/R) ∪ {p} and, also in view of Theorem 2.4, π(G/R) − {2} is now a clique of ∆(G), we see that the cut-vertex p of ∆(G) cannot be 2, and that p is the unique vertex adjacent to 2 in ∆(G).Now, let λ be a non-principal irreducible character of L ′ /Z, and let χ ∈ Irr(K/Z | λ).If ψ is an irreducible constituent of χ L/Z lying over λ, then clearly ψ(1) = 1, and since L ′ /Z is an abelian normal subgroup of L/Z whose index is a 2-power, we conclude that ψ(1) (whence χ(1)) is a multiple of 2. As a consequence, we get π(|K : We conclude this part of the proof by considering three situations that are exhaustive, and that all lead to a contradiction.
is a chief factor of K and it is not centralized by L/L ′ , we deduce that 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 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 λ be any non-principal character in Irr(L ′ /Z): since π(|K : I K (λ)|) ⊆ {2, p}, Gallagher's theorem implies that I K (λ)/L ′ contains a unique Sylow q-subgroup of K/L ′ for every prime q ∈ π(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 ).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)×(L/L ′ ) and it would then be abelian, a clear contradiction.Also, for a non-principal λ in Irr(L ′ /Z), we already observed that I K (λ) is a proper subgroup of K such that π(|K : We claim that I K (λ)/L cannot be a subgroup of type (iv) of K/L unless it is also of type (iii).In fact, assume ) 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 π(|K : I K (λ)|) 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 cite[Proposition 3.1]MW) 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 (ψ), where ψ lies in Irr(L/Z | λ) (recall that ψ(1) is a multiple of 2, and that , and therefore I K (ψ)/L is either the whole I K (λ)/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 (λ)/L is of type (iii) isomorphic to A 4 : then there must be a prime in π(|K : , 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 , 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 π(|K : I K (λ)|) − {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 • 5 = 2 a − 1, i.e.K/L ∼ = SL 2 (2 4 ) and p = 17 is the only vertex adjacent to 2 in ∆(G).But in the latter case, taking ψ ∈ Irr(L/Z | λ), we see that I K (ψ) cannot be a proper subgroup of I K (λ) (otherwise |I K (λ) : I K (ψ)| 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 (ψ) ⊆ I K (λ), we get I K (ψ)/L ∼ = 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 λ ∈ Irr(L ′ /Z), the subgroup I K (λ)/L of K/L is either of type (i) or of type (ii).
Next, assume that I K (λ)/L is a subgroup of type (i + ).Then we get p = 2 a − 1 and, since 2 cannot be adjacent in ∆(G) to any prime in π(2 a + 1), for every non-principal ν ∈ Irr(L ′ /Z) the subgroup I K (ν)/L must be either of type (i + ) containing a unique Hall π(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 ν, by Lemma 2.9; on the other hand, if the latter situation occurs for some non-principal ν ∈ Irr(L ′ /Z), then we reach a contradiction via Theorem 2.11 (recall that L ′ /Z is a 2-group).
If I K (λ)/L is a subgroup of type (i − ) then, as above, for every non-principal ν ∈ Irr(L ′ /Z), the subgroup I K (ν)/L must be either of type (i − ) containing a unique Hall π(2 a − 1)-subgroup of K/L or of type (ii).Observe that if, in the latter case, |K : I K (ν)| is divisible by 2, then I K (ν)/L must contain a Hall π(2 a − 1)-subgroup of K/L; hence, by the structure of the subgroups of type (ii), I K (ν)/L should contain a full Sylow 2-subgroup of K/L as well, against the fact that |K : I K (λ)| is even.Therefore I K (ν)/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 λ ∈ Irr(L ′ /Z), the subgroup I K (λ)/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 (λ)/L is a Sylow 2-subgroup of K/L for every non-principal λ ∈ Irr(L ′ /Z).Considering ψ ∈ Irr(L/Z) lying over such a λ, and recalling once again that ψ(1) is even and I K (ψ) ⊆ I K (λ), Clifford's theory yields that the primes in π(K/L) are pairwise adjacent in ∆(G) and, also in view of Theorem 2.4, every odd prime divisor of |K/L| is a complete vertex of ∆(G).This is clearly not compatible with the existence of a cut-vertex of ∆(G).
(iii) L ′ /Z ⊆ Z(K/Z).As in case (ii), we have that L ′ /Z is a 2-group.If λ is a non-principal irreducible character of L ′ /Z, then λ is fully ramified with respect to the K/Z-chief factor L/L ′ (see Exercise 6.12 of [9]); therefore, the unique The fact that the Schur multiplier of K/L is trivial implies that ψ extends to K, yielding a clear contradiction via Gallagher's theorem.
To conclude the proof, we will show that I K (ξ)/L contains a unique Sylow 2-subgroup of K/L for every non-principal ξ in Irr(L/L ′ ).To this end, we will proceed through a number of steps.
(a) For every non-principal ξ ∈ Irr(L/L ′ ), the subgroup I K (ξ)/L of K/L cannot be of type (iv), unless it is also of type (iii).For a proof by contradiction, let ξ ∈ Irr(L/L ′ ) be such that I K (ξ)/L ∼ = SL 2 (2 b ) for some b > 2 properly dividing a.Thus, 2 is a divisor of |K : I K (ξ)|.Since the Schur multiplier of I K (ξ)/L is trivial, ξ extends to I K (ξ) and this yields (via Clifford's correspondence and Gallagher's theorem) that 2 is adjacent in ∆(G) to every prime in π(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 π(G/R) − π(K/L) is adjacent to every prime in π(K/L) − {2}.Finally, 2 2a − 1 has a primitive prime divisor q because a = 3; this prime q, which clearly belongs to π(2 a + 1), is a divisor of |K : I K (ξ)|, so every prime in π(2 b − 1) is adjacent to q in ∆(G).As easily seen, this setting is not compatible with the existence of a cut-vertex of ∆(G).
(b) For every non-principal ξ ∈ Irr(L/L ′ ), the subgroup I K (ξ)/L of K/L cannot be isomorphic to A 5 .Assume the contrary, and take ξ ∈ Irr(L/L ′ ) such that I K (ξ)/L ∼ = A 5 .Working with character triples, we observe that Irr(K | ξ) contains characters whose degrees are divisible by every prime in π(|K : I K (ξ)|) ∪ {3}, which contains π(K/L) − {5} (note that 2 divides |K : I K (ξ)| because 2 a > 4); thus the 5-part of |K/L| is 5, otherwise the primes in π(K/L) would be pairwise adjacent in ∆(G), easily contradicting the existence of a cut-vertex of ∆(G).Observe also that, since neither 2 a − 1 nor 2 a + 1 can be 5, there exists an odd prime q in π(K/L) − {5} that is adjacent to 5 in ∆(K/L); as q is now a complete vertex in the subgraph of ∆(G) induced by π(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 ∆(G).This implies on one hand that ξ does not have an extension to I K (ξ) (otherwise, by Gallagher's theorem, we would get the adjacency between 5 and 2 in ∆(G)), which in turn yields (via [9, 8.16, 11.22, 11.31]) that the order of L/L ′ is divisible by 2; on the other hand, one among the sets π(2 a − 1) and π(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 π(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 ∼ = 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 SL 2 (16) over 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 ∼ = 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 ∆(G).Also, a direct computation with GAP [7] 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 ∆(K/L) would be pairwise adjacent in ∆(G).Only one module is left, which has dimension 8 and is not the natural module: to handle this, we can see via GAP [7] that in all possible isomorphism types of extensions of V by SL 2 (16) the set of irreducible character degrees is {1, 15, 16, 17, 51, 68, 204, 255, 272, 340}, so π(K/L) would again be a set of pairwise adjacent vertices of ∆(G).
It remains to consider the case when 5 divides 2 a + 1, hence π(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 (nontrivial) irreducible K/L-module over F 2 , and if T /L is a Sylow 2-subgroup of K/L, then clearly there exists a non-principal µ in Irr(L/X) which is fixed by T /L (so, such that I K (µ)/L contains T /L); as I K (µ)/L is a proper subgroup of K/L, the only possibility for I K (µ)/L is to be of type (ii).Moreover, since 5 is only adjacent to p in ∆(G), no prime divisor of 2 a − 1 lies in π(|K : I K (µ)|), thus in fact I K (µ)/L = N K/L (T /L) has irreducible characters of degree 2 a − 1.Now, µ does not extend to I K (µ), as otherwise (by Gallagher's theorem and Clifford correspondence) we would get adjacencies in ∆(G) between 5 and all the primes in π(2 a −1); but then, for ψ ∈ Irr(I K (µ) | µ) and θ an irreducible constituent of ψ T /X lying over µ, we have that θ(1) is a 2-power larger than 1 (otherwise ψ would be an extension of µ to T , and µ would then extend to the whole I K (µ)).As a consequence, 2 divides ψ(1) and we get the adjacency in ∆(G) between 5 and 2. This is the final contradiction that rules out the case (c) For every non-principal ξ ∈ Irr(L/L ′ ), the subgroup I K (ξ)/L of K/L cannot be isomorphic to A 4 .Assume I K (ξ)/L ∼ = A 4 for some ξ ∈ Irr(L/L ′ ).Then we see at once that the primes in π(K/L) − {3} are pairwise adjacent in ∆(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 neighbour of 3 in ∆(G) among the primes in π(K/L), and one among the sets π(2 a − 1) and π(2 a + 1) is {3, p}.
Assuming first π(2 a −1) = {3, p}, we see that a = 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 ∼ = SL 2 (16), and we reach a contradiction as in the second paragraph of (b).
As regards the case π(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 (ξ)/L of K/L, for ξ non-principal in 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 (ξ)/L of K/L (for ξ non-principal in Irr(L/L ′ )) are of type (ii) and of even order, but there exists ξ 0 ∈ Irr(L/L ′ ) such that 2 divides |K : I K (ξ 0 )|.In this setting we observe that 2 a − 1 does not divide |I K (ξ 0 )/L|, because I K (ξ 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 ∈ π(2 a − 1) ∩ π(|K : I K (ξ 0 )|), and Clifford's correspondence yields that {2, r} ∪ π(2 a + 1) is a set of pairwise adjacent vertices of ∆(G).It follows that r is adjacent in ∆(G) to every prime in π(G/R)− {r}, thus r is the cut-vertex p, and no other prime in π(2 a − 1) can have any neighbour in {2} ∪ π(2 a + 1); in particular, no prime in π(2 a − 1) − {p} shows up as a divisor of |K : I K (ξ)| for any ξ ∈ Irr(L/L ′ ).Note also that a primitive prime divisor of 2 a − 1 cannot lie in π(I K (ξ 0 )/L), as otherwise it would divide 2 f − 1 (and f < a); so, if a = 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 ∆(G).Thus there exists a prime q ∈ π(2 a − 1) − {p} such that, for every ξ ∈ Irr(L/L ′ ), the subgroup I K (ξ)/L contains a Sylow q-subgroup of K/L.
Furthermore, the character ξ 0 does not extend to I K (ξ 0 ), as otherwise we would get characters in Irr(K | ξ 0 ) whose degree is divisible by q and every prime in {2} ∪ π(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 (ξ)/L for ξ non-principal in 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 ξ ∈ Irr(L/X), I K (ξ)/L does not contain a full Sylow 2-subgroup of K/L.In other words, we can assume that ξ 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 (ξ 0 )/L, and so 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 (ξ 0 )/L.The fact that ξ 0 does not extend to its inertia subgroup in K also implies that T 0 /ker(ξ 0 ) is a non-abelian 2-group; thus L/ker(ξ 0 ), which has order 2, is in fact the derived subgroup of T 0 /ker(ξ 0 ).Moreover, the normal subgroup Z/ker(ξ 0 ) = Z(T 0 /ker(ξ 0 )) of I K (ξ 0 )/ker(ξ 0 ) cannot be larger than L/ker(ξ 0 ), because T 0 /L is a minimal normal subgroup of I K (ξ 0 )/L and clearly Z/L is not the whole T 0 /L.We deduce that T 0 /ker(ξ 0 ) is an extraspecial 2-group, so (b is even and) an application of [8, 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 (ξ 0 )/L| and so T 0 /L has order 2 3 ; the same argument as above shows that T 0 /ker(ξ 0 ) is an extraspecial 2-group, a clear contradiction.On the other hand, if p = 7, then |I K (ξ 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 (ξ 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 ξ ∈ Irr(L/X) is stabilized both by Q and by another Q 1 ∈ 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 ∈ K/L such that no non-trivial element in the dual group L/X of L/X is centralized by both Q and Q h .As a consequence, setting |L/X| = 2 d , we get |C L/X (Q)| ≤ 2 d/2 , and then It is easily checked that the above inequality yields d < 4a and, since a is odd, Lemma 3.12 in [13] (whose hypotheses require d ≤ 3a, but whose proof works assuming d < 4a as well) leaves only one possibility for the isomorphism type of the K/L-module L/X over 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 F 2 a and by ω an automorphism of order 3 of F 2 a , we have that L/X is a "triality module", which can be described as follows.Start from the K/L-module V = R(1) ⊗ R(1) ω ⊗ R(1) ω 2 over F 2 a (or one of its Galois twists), and observe that the field of values of (the character of) V is F 2 c ; now, restricting the scalars to 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 F 2 , and this is the 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 SL 2 (2 a ) whose order is a divisor of 2 a + 1 are conjugate to elements of the form x = 0 1 1 λ , where λ = µ + µ 2 a for µ ∈ F 2 2a − {1} such that µ 2 a +1 = 1.The action of such an x on V is of course given by the Kronecker product , so the action of x on V K is expressed by the same Kronecker product as above.But x is conjugate to ), so our aim is in fact to find µ such that the matrix has a non-zero eigenspace for the eigenvalue 1.A direct calculation shows that it is enough to choose µ of order 2 2c − 2 c + 1.
(e) The subgroups I K (ξ)/L of K/L, for ξ non-principal in 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 ξ 0 ∈ Irr(L/L ′ ) such that I K (ξ 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 ∈ π(2 a − 1) such that, for every non-principal ξ ∈ Irr(L/L ′ ), the subgroup I K (ξ)/L contains a Sylow q-subgroup of K/L, and ξ 0 does not extend to I K (ξ 0 ).Similarly, no power of ξ 0 can extend to its inertia subgroup I in K, if I/L it does not contain a full Sylow 2subgroup of K/L.Since all Sylow q-subgroups of K/L are cyclic for q = 2, by [9, Theorem 6.26] we can actually assume that the order o(ξ 0 ) in the dual group of L/L ′ is a power of 2. Hence ξ 0 is in 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 π(2 a − 1) (that is necessarily p), then again we are in the same situation as in (d): for every ξ ∈ Irr(L/L ′ ), the subgroup I K (ξ)/L contains a Sylow q-subgroup of K/L for a suitable prime q ∈ π(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 (ξ)/L of K/L, for ξ non-principal in 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 ξ 0 ∈ Irr(L/L ′ ) such that I K (ξ 0 )/L is of type (ii) but not containing a full Sylow 2-subgroup of K/L, then 2 is adjacent in ∆(G) to every prime in π(K/L) − {2}; however, there also exists a prime r ∈ π(2 a − 1) such that r divides |K : I K (ξ 0 )|, and this r is adjacent in ∆(G) to every other prime in π(G/R); this is incompatible with the existence of a cut-vertex of ∆(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 π(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 ∈ π(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 (ξ)/L of K/L, for ξ non-principal in 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 ∆(G) to all the primes in π(K/L) − {2}, and any inertia subgroup I K (ξ 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 • (2 a − 1).
(h) The subgroups I K (ξ)/L of K/L, for ξ non-principal in 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 ∆(G) to all the primes in π(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 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 L/X is centralized by a Sylow 2-subgroup of K/L.
(i) The subgroups I K (ξ)/L of K/L, for ξ non-principal in 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 ξ 0 ∈ Irr(L/X) such that I K (ξ 0 )/L does not contain a subgroup of order 2 a + 1, which means that there exists r ∈ π(2 a + 1) dividing |K : I K (ξ 0 )|, then r is a complete vertex of ∆(G) and it is in fact p.Now π(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 (ξ) in K.In other words, for every non-principal ξ in Irr(L/X), the inertia subgroup I K (ξ)/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 (ξ)/L does contain a subgroup of order 2 a + 1 for every ξ ∈ Irr(L/X).Thus, in any case, Lemma 2.9 applies and we get a contradiction.
(j) The subgroups I K (ξ)/L of K/L, for ξ non-principal in 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 (ξ)/L contains a unique Sylow 2-subgroup of K/L for every non-principal ξ in Irr(L/L ′ ).The proof is complete.
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 W the restriction to S 1 = SL 2 (5), seen as a subgroup of SL Note that all the above modules are self-dual: this follows from [13, Lemma 3.10] for V 0 , V 1 and U , and for W by observing that GL 4 (3) has a unique conjugacy class of subgroups isomorphic to SL 2 (5).
Finally, let B be an abelian group and A a group acting on B via automorphisms: we will denote by ∆ orb (B) the graph whose vertex set is the set of the prime divisors of the set of orbit sizes {|A : C A (b)| : b ∈ 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 a b ∈ B such that the product pq divides |A : 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 SL 2 (4)-module and the graph ∆ orb (V ) is not a clique with vertex set {2, 3, 5}, then q = 2 and V is isomorphic either to 5)-module and the graph ∆ 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 .
Theorem 3.4.Assume that the group G has a composition factor isomorphic to SL 2 (4) ∼ = PSL 2 (5), and let p be a prime number.Assume also that ∆(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.
L is isomorphic to the restriction to K/L, embedded in SL 4 (3), of the standard module of 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 SL 2 (4) or to 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) ∪ {p}, the hypothesis of p being a cut-vertex for ∆(G) implies that the subgraph of ∆(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 ∆ orb (V ) is not a clique as well.Therefore Lemma 3.3 applies, and the K/L-module V is isomorphic to is a chief factor of G as well, and our proof is complete if X = 1.
Working by contradiction, we assume X = 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 = 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 ∼ = K/N and hence K/Y = M/Y × N/Y , contradicting the fact that K is perfect.Therefore M/Y is abelian, so the normal subgroup M X/X of K/X lies in 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 ∼ = M/Y × 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 ∼ = SL 2 (4) or K/L ∼ = 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 ∆(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 ∈ {2, 3, 5}, the same holds for |L/Y |.Furthermore, we have X/Y ≤ Z(L/Y ) and, in particular, I K (λ) ⊆ I K (µ) for every µ ∈ Irr(X/Y ) and λ ∈ Irr(L/Y | µ).
If q = 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 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 ) ′ = Z(L/Y ) and L/Y is an extraspecial q-group.So, every non-linear 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 ∆(G), a contradiction.Thus the claim is proved, and Lemma 3.3 applies: our assumption that q is not 2 yields then X/Y ∼ = L/X ∼ = U , or X/Y ∼ = L/X ∼ = 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 (λ) is a Sylow q-subgroup of K/L for every λ ∈ 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 [7] that the perfect groups of order 2 5 • |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 (λ) ⊆ I K (µ) for every µ ∈ Irr(X/Y ) and λ ∈ Irr(L/Y | µ), 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 [7]shows that the perfect groups of order 2 8 • |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 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 Φ(K/Y ) is trivial and hence X/Y has a complement K 0 /Y in K/Y ; in particular, every µ ∈ Irr(X/Y ) extends to its inertia subgroup I K (µ).Let Z/Y be an irreducible L/Y -submodule of X/Y a minimal normal subgroup of L/Y contained in X/Y ).Set C/Y = 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 ) × (Z 1 /Y ) for a suitable L/Y -module Z 1 /Y , we consider the character µ = µ 0 × 1 Z1/Y ∈ Irr(X/Y ), where µ 0 is a nonprincipal irreducible character of Z/Y .We observe that I L/Y (µ) = C/Y and that every χ ∈ Irr(K/Y |µ) has a degree divisible by q.We also remark that, setting . We claim that, as a consequence, for every prime divisor r = q of |K/L|, either r divides |K : I K (µ)| or r divides the degree of some irreducible character of I K/X (µ) that lies over µ.In fact, fixing R 0 /Y ∈ 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 R 0 L 0 /L 0 , R 1 L 0 /L 0 = 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 F q of the vector space C L0/Y (R 0 L 0 /L 0 ) cannot be larger than a half of dim Fq (L 0 /Y ).Now, if I K0/Y (µ) (which is isomorphic to I K/X (µ)) contains a Sylow r-subgroup R 0 /Y of K 0 /Y as a normal subgroup, then R 0 /Y centralizes I L0/Y (µ) because I L0/Y (µ) and R 0 /Y are normal subgroups of coprime order of I K0/Y (µ), and this is not possible as . By Gallagher's theorem, it hence follows that {2, 3, 5} is a clique of ∆(G), a contradiction.

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.
We now move to claim (b).Assuming 2 ∈ π(G/KR) and a ≥ 3, by Theorem 2.4 we get that 2 is adjacent in ∆(G) to all primes = 2 of π(G/R); so to p as well if p = 2 and p ∈ π(G/R).On the other hand, if p ∈ V(G) − π(G/R), so p = 2, then p ∈ V(R) and p is adjacent to 2 in ∆(G) by part (a).Hence, 2 is a complete vertex of ∆(G).
Finally, we prove claim (c).Assume that 2 ∈ π(G/KR)∪V(R).Then every character χ ∈ Irr(G) such that χ(1) is even lies over a character ψ ∈ Irr(KR) with ψ(1) even.Writing ψ = α × β with α ∈ Irr(K) and β ∈ Irr(R), since 2 ∈ V(R) we deduce that α has even degree, and hence α is the Steinberg character of K. Thus α extends to G (see for instance [14]) and hence χ(1) = α(1)γ(1) = 2 a γ(1) for a suitable γ ∈ Irr(G/K), concluding the proof.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 ∼ = SL 2 (2 a ), with a ≥ 3.Then, ∆(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) = π(G/R) ∪ {p} and one of the following holds.In all cases, p is is a complete vertex and the only cut-vertex of ∆(G).
Proof.We start by proving the "only if" part of the statement, assuming that ∆(G) is connected and that has a cut-vertex p.Then, by Theorem 2.8 G/R is an almost-simple group and V(G) = π(G/R) ∪ {p}.
As a consequence, we have that the socle We observe that by Theorem 2.4 every prime in π(G/KR) is adjacent in ∆(G) to every other vertex in ∆(G), except possibly 2 and p.Moreover, part (a) of Lemma 4.1 yields that V(R) ⊆ {p}.
We consider first the situation arising when L = 1.Assuming p = 2, then V(G) = π(G/R) and by the above observation we deduce that G/KR is a 2-group and that V(R) ⊆ {2}.If G = KR = K × R, then, as ∆(G) is connected and 2 is a cut-vertex of ∆(G), it immediately follows that V(R) = {2}.So, in any case, V(G/K) ∪ π(G/KR) = {2}.Assuming instead p = 2, then (since no vertex in V(G) − {p} can be complete in ∆(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) ⊆ {p} and p = 2, part (c) of Lemma 4.1 yields that 2 is adjacent in ∆(G) to all primes in V(G/K), and to them only.As ∆(G) is connected, it follows that V(G/K) is non-empty.If q ∈ V(G/K) and q = p, then q divides |G/KR| (because V(KR/K) = V(R) ⊆ {p}) and hence, by Theorem 2.4, q (being adjacent also to 2) would be a complete vertex of ∆(G), a contradiction.Hence, V(G/K) = {p}.
We assume now L = 1.Then, by Theorem 3.2, L is a minimal normal subgroup of G and L is the natural module for K/L ∼ = S.By Remark 2.7, the subgraph of ∆(G) induced by the vertex set V(G) − {2, p} is a complete graph.Hence, the assumptions on ∆(G) imply that p = 2 and that 2 is adjacent only to p in ∆(G).Moreover, recalling that ∆(G/L) is a subgraph of ∆(G), by part (a) and part (b) of Lemma 4.1 we deduce that 2 ∈ V(R/L) ∪ π(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 ≤ KR.Setting T 0 = T ∩ R, we observe that T 0 /L is an abelian normal Sylow 2-subgroup of R/L because 2 ∈ V(R/L).Let T 1 = T ∩ K and assume, working by contradiction, that T ′ = T ′ 1 .Let λ ∈ Irr(L) be a non-principal character; by Lemma 2.6 L ≤ Z(T 0 ), so λ is T 0 -invariant and, since L is a self-dual K/L-module, I K (λ)/L is a Sylow 2-subgroup of K/L.Hence, since T = T 0 T 1 , we can assume (up to conjugation) that λ is T -invariant.So, by Lemma 2.6 λ has no extension to T .As I K (λ)/L = T 1 /L and KR/L = K/L × R/L, T /L is a normal subgroup of I KR (λ) and hence 2 divides the degree of every irreducible character ψ of I KR (λ) that lies over λ.By Clifford correspondence, it follows that 2 is adjacent in ∆(G) to all primes in π(2 2a − 1) = π(|K : I K (λ)|), a contradiction.Hence, T ′ = (T ∩ 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) = π(G/R) ∪ {p} for some prime p.
Suppose first that (a) holds, so K is a minimal normal subgroup of G and K ∼ = S. Hence, KR = K×R.Assume that p = 2, so V(G) = π(G/R), and that V(G/K) ∪ π(G/KR) = {2}.If KR < G, then 2 is a complete vertex of ∆(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 χ ∈ Irr(G) and an irreducible constituent ψ of χ KR , the odd parts of χ(1) and of ψ(1) coincide by [9,Corollary 11.29], so by part (a) of Theorem 2.3 the graph ∆(G) − 2, obtained by deleting the vertex 2 and all incident edges, has two complete connected components, with vertex sets π(2 a − 1) and π(2 a + 1).So, 2 is a cut-vertex of ∆(G) and, being a complete vertex of ∆(G), it is the unique cut-vertex of ∆(G).If p = 2, V(G/K) = {p} and G/KR has odd order, then (as R ∼ = KR/K G/K) 2 ∈ V(R) and by part (c) of Lemma 4.1 the vertex 2 is adjacent only to p in ∆(G).Hence, p is a cut-vertex of ∆(G).We also observe that p is a complete vertex of ∆(G): this is a consequence of Theorem 2.4 if p ∈ π(G/KR), while if p ∈ π(G/KR) the assumption V(G/K) = {p} implies that p ∈ V(R) and hence the claim follows by part (a) of Lemma 4.1.Thus, p is the unique cut-vertex of ∆(G).
We assume now that (b) holds, so K contains a minimal normal subgroup L of G such that K/L ∼ = S and L is the natural module for K/L.Moreover, p = 2, V(G/K) = {p}, G/KR has odd order and, for any Sylow 2-subgroup T of G, T ′ = (T ∩ K) ′ .For a non-principal λ ∈ Irr(L), the argument used in the fourth paragraph of this proof shows that I = I G (λ) contains a Sylow 2-subgroup T of G, and T /L is abelian and normal in I/L.By Lemma 2.6 λ extends to T and hence λ extends to I G (λ) by [9,Theorem 6.26].So, Gallagher's theorem implies that every irreducible character of G that lies over λ has odd degree.We hence deduce that if χ ∈ Irr(G) has even degree, then χ ∈ Irr(G/L).Then, by part (c) of Lemma 4.1, 2 is adjacent only to p in ∆(G).So, by Remark 2.7, the graph obtained by removing the vertex p from ∆(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 ∆(G/L), hence of ∆(G); thus, p is the only cut-vertex of ∆(G).

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 ∼ = SL 2 (4) or K ∼ = SL 2 (5).Let R be the solvable radical of G, N = K ∩ R and assume that V(G) = {2, 3, 5, p} for a suitable prime p. Then Proof.(a): By part (a) of Lemma 4.1 the primes in V(KR/K) = V(R/N ) are complete vertices of ∆(G).Let q ∈ V(G/K) − V(KR/K); then for χ ∈ Irr(G/K) such that q divides χ(1) and an irreducible constituent θ of χ KR/K , q divides χ(1)/θ( 1 In all cases, p is is a complete vertex and the only cut-vertex of ∆(G).(x): N is the natural module for K/N : then 3 and 5 are adjacent in ∆(G) (see Remark 2.7), and hence p = 2, as othewise ∆(G) would be a complete graph.We show that G = KR: in fact, if this is not the case, then G/R ∼ = S 5 and the Sylow 2-subgroups of G/N are non-abelian.For a non-principal λ ∈ Irr(L) and I = I G (λ), 15 divides |G : I|.Hence, recalling Theorem A of [12], independently on the parity of |G : I| there exists χ ∈ Irr(G), lying above λ, that has degree 30, a contradiction.Finally, we observe that if G/K ∼ = R/N is abelian, then 2 is an isolated vertex of ∆(G), because by part (a) of Lemma 5.2 every χ ∈ Irr(G) of even degree is a character of G/N = K/N × R/N .So, V(G/K) = {p} and we have case (b)(i).The action of G on N gives an embedding φ of G = G/C in G = GL 4 (2).One can check (for instance by GAP [7]) that N G (φ(K)) ∼ = S 5 , and that if φ(G) ∼ = S 5 then ∆(G/R 0 ), which is a subgraph of ∆(G), has a complete subgraph with vertex set {2, 3, 5}, a contradiction.So, φ(G) = φ(K), and hence G = K × R 0 , giving case (b)(ii).
As the final case, we assume that G has a minimal normal subgroup L, such that L ≤ N and K/L ∼ = SL 2 (5).We have two possible cases: (y): L is the natural module for K/L.Then ∆(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 = 5.Moreover, part (b) of Lemma 5.2 yields that 5 is adjacent in ∆(G) only to the primes in V(G/K).Thus, as ∆(G) is connected, V(G/K) = ∅, 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 GL 4 (3).So, ∆(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.In case (b)(ii), it is clear that ∆(G) = ∆(K) is the graph 2 − 5 − 3.
(c): We assume that there exists L G, L ≤ K, such that K/L ∼ = SL 2 (5).In case (c)(i), by part (b) of Lemma 5.2 the vertex 5 is adjacent only to p (p = 5) in ∆(G) and, by part (a) of the same lemma, p is a complete vertex of ∆(G).
In case (c)(ii), we prove that ∆(G) = ∆(K), so ∆(G) is the graph 3 − 2 − 5. To this end, it is enough to show that 3 and 5 are non-adjacent in ∆(G).Since |G/KR| ≤ 2, KR contains a Sylow 3-subgroup Q of G; moreover, as V(R/N ) ⊆ V(G/K) ⊆ {2} and |N/L| = 2, it easily follows that, setting Q 0 = Q ∩ R, Q 0 /L is abelian and normal in R/L, and hence in G/L.Let λ ∈ Irr(L) be a non-principal character and let I = I G (λ).An application of the Frattini argument, as in the proof of part (b) of Lemma 5.2, proves that G splits over L, so λ extends to I. By [5, Lemma 2.6], L ≤ Z(Q 0 ) and hence, since I ∩ K is a Sylow 3-subgroup of K, we can assume Q ≤ 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 χ ∈ Irr(G) that lies over λ has degree not divisible by 3. Hence, if χ ∈ Irr(G) and 3 divides χ(1), then L ≤ ker(χ) and χ ∈ Irr(G/L).Now, an application of part (c) of Lemma 5.1 yields that 5 not adjacent to 3 in ∆(G/L), and hence 3 and 5 are not adjacent in ∆(G).
So, in every case, p is a cut-vertex of ∆(G) and, as p is also a complete vertex of ∆(G), there are no other cut-vertices in ∆(G).The proof is complete.

Lemma 2 . 2 .
Let G ∼ = SL 2 (2 a ), where a ≥ 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 ∈ N − {0}.Then U lies in the normalizer in G of precisely two Sylow 2-subgroups of G. Proof.See [5, Lemma 2.2].

Lemma 4 . 1 .
Let K be a normal subgroup of the group G with K ∼ = SL 2 (2 a ), a ≥ 2. Let R be the solvable radical of G and assume that V(G) = π(G/R) ∪ {p} for a suitable prime p. Then (a) The primes in V(R) (if any) are complete vertices of ∆(G).(b) If a ≥ 3 and 2 (a) K is a minimal normal subgroup of G, K ∼ = S and either p = 2 and V(G/K) ∪ π(G/KR) = {2}, or p = 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 ∼ = S, L is the natural module for K/L, p = 2, V(G/K) = {p}, G/KR has odd order and, for a Sylow 2-subgroup T of G, T ′ = (T ∩ K) ′ .

Proof.
We start by proving the "only if" part of the statement, assuming that ∆(G) is connected and that it has a cut-vertex p.Then, by Theorem 2.8 G/R is an almost-simple group and V(G) = π(G/R) ∪ {p}.So, the socle M/R of G/R is isomorphic to SL 2 (4), and V(G) = {2, 3, 5, p}.Hence, the subgraph of ∆(G) induced by the set of vertices {2, 3, 5} cannot be a clique.As N = K ∩ R and KR = M , then K/N ∼ = M/R ∼ = SL 2 (4) and |G/KR| ≤ 2. Since no vertex of ∆(G) different from p can be complete, part (a) of Lemma 5.1 implies that V(G/K) ⊆ {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 SL 2 (4) or to SL 2 (5) (i.e.|N | ≤ 2), then part (c) of Lemma 5.1 implies (as 5 cannot be an isolated vertex of ∆(G)) that V(G/K) is non-empty, so V(G/K) = {p}.By part (b) of Lemma 5.1 p = 5 when K ∼ = SL 2 (5) or KR = 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 ∼ = SL 2 (4), |N | = 2 4 and we have two cases:

(
xx): N is the restriction to K/L, embedded as Ω − 4 (2) into SL 4 (2), of the standard module of SL 4 (2).Then ∆(K) is the graph 2 − 5 − 3 and hence necessarily p = 5.Let R 0 = C G (K) and C = C G (N ).So, N ≤ CG and R 0 ≤ C ≤ R, since K/N is the only non-solvable composition factor of G, and it acts non-trivially on N .As H 2 (K/N, N ) = 0, K splits over N ; let K 0 be a complement ofN in K.Note that R 0 = C C (K) = C C (K 0 ).We prove that C = N × R 0 .It is enough to show that C = N R 0 , since Z(K) = 1.As [K, R] ≤ N , in particular [K 0 , C] ≤ N and hence K c 0 ≤ K 0 N = K for every c ∈ C. Since H 1 (K 0 , N ) = 0, all complements of N in K are conjugate in K.It follows that there exists an element b ∈ N such that K c 0 = K b 0 , so d = bc −1 ∈ N C (K 0 ) and hence [K 0 , d] ≤ K 0 ∩ C = 1, as K 0 ∼ = K/N acts faithfully on N .Thus, d ∈ R 0 .So, C = N R 0 = N × R 0 .
(a): If G ∼ = SL 2 (4) × R with V(R) = V(G/K) = {5}, then clearly ∆(G) is the graph 2 − 5 − 3.If p = 5, then 5 is adjacent only to p in ∆(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 ∆(G).(b): We assume that G = KR and that N = K ∩ R is a normal in G of order 2 4 .In case (b)(i), since G/N = K/N × R/N and V(R/N ) = V(G/K) = {p} for some prime p = 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 ∆(G), so p is a cut-vertex of ∆(G).By part (a) of Lemma 5.1, p is a complete vertex, and hence the only cut-vertex, of ∆(G).