Groups with three projective characters

Abstract Working over the field of complex numbers, the inequivalent irreducible projective representations of a finite group G with 2-cocycle α are considered. The greatest common divisor of the degrees of the α-characters of these representations is used to obtain information about the degrees of the irreducible αS-characters of a Sylow subgroup S of G. Suppose now that G has exactly three irreducible α-characters. Then the potential form that the three degrees can take is found if G has a nontrivial cyclic Sylow subgroup. However, the main result is to show that G is solvable when the degrees are of the form a, ra and ka with certain restrictions on r and k, most notably that r = 1 or k. This involves detailed analysis of the possible actions of G on the irreducible αN-characters of a normal subgroup N of G.


Introduction
Throughout this paper G will denote a finite group.
Obviously if x ∈ G is α-regular, then it is also α k -regular for any integer k.Now let β ∈ [α], then x is α-regular if and only if it is β-regular and any conjugate of x is also α-regular (see [23,Lemma 2.6.1]),so that we may refer to the α-regular conjugacy classes of G. Finally if o(x) and o ([α]) are relatively prime, then x is α-regular from [17,Lemma 1.2 (b)].
An α-representation P is also called a projective representation of G with 2-cocycle α, its trace function ξ is its α-character and ξ (1), which is the dimension of P, is called the degree of ξ .
To avoid repetition all α-representations of G in this paper are defined over C. We now let Proj(G, α) denote the set of all irreducible α-characters of G (see [23, p. 184] for details), but the relationship between Proj(G, α) and α-representations is much the same as that between Irr(G) and (ordinary) representations of G. Now x ∈ G is α-regular if and only if ξ(x) = 0 for some ξ ∈ Proj(G, α) (see [24,Proposition 1.6.4])and | Proj(G, α)| is the number of α-regular conjugacy classes of G (see [24,Theorem 1.3.6]).
Now from [7,Lemma 4.5] there exists α ∈ [β] such that α has order equal to that of [β] and α is classpreserving, that is the elements of Proj(G, α) are class functions.Throughout this paper we will assume without loss of generality that the initial choice of 2-cocycle α has these two properties.Under these assumptions the "standard" inner product , may be defined on α-characters of G and for which the "normal" orthogonality relations hold (see [24,Section 1.11.D]).
Thus f α (g, ) is a one-dimensional α/α g -representation of G.
Next let N be a normal subgroup of G. Then G acts on Proj(N, α N ) by for ζ ∈ Proj(N, α N ), g ∈ G and all x ∈ N; Clifford's theorem for projective characters applies to this action (see [24,Theorem 2.2.1]).We need some terminology and results that arise from Clifford's theorem as it applies to α-characters.(1), so that degree ratios are preserved using the bijection from Proj(I/N, βζ ) onto Proj(G/ζ , α).Finally by considering f α (g, ) restricted to N, we observe that Lemma 1.4.Let α be a 2-cocycle of G and let N be a normal subgroup of G. Then for each g ∈ C G (N), f α (g, ) restricted to N is a linear representation of N. In particular if N is perfect, C α (N) = C G (N).
Proof.We observe that if g ∈ C G (N), then α N /α g N is trivial.So if N is in addition perfect we conclude that f α (g, ) restricted to N is the trivial representation of N and so C α (N) = C G (N). Lemma 1.5.Let α be a 2-cocycle of G. Suppose that N is a normal subgroup of G that is generated by α-regular elements of G. Then C α (N) = C G (N).
Proof.If N is generated by α-regular elements of G, then f α (g, ) restricted to N is the trivial representation of N for all g ∈ C G (N) and so C α (N) = C G (N).
Lemma 1.6.Let α be a 2-cocycle of G and let N be a normal subgroup of G. Then the number of orbits of G on Proj(N, α N ) is equal to the number of α-regular conjugacy classes of G contained in N.
The set of degrees of the elements of Proj(G, α) is denoted by cd α (G), although if [α] is trivial we use the standard notation cd(G).We now let c α (G) and s α (G) denote the greatest common divisor and minimum respectively of the elements of cd α (G).Now s α (G) = 1 if and only if [α] is trivial, because of the trivial character.In general c α (G) is described in [9, Proposition 1], where n p denote the pth part of n for p a prime number:  (1), | and e i is the degree of the element of Proj( Ī, βζ ) that corresponds to ξ i under the bijections described earlier.Thus and using the bijections in Clifford's theorem for projective characters, d 2 = t|N|e 2 .

Degree questions
We conjecture that if | Proj(G, α)| = 3, then G is solvable.This problem appears to be quite difficult in generality, but we will present some evidence to support the conjecture.First we consider a very easy case in which C n denotes the cyclic group of order n and S n the symmetric group on n letters.
Then the number of elements of Proj(G, α) of degree s α (G) is divisible by p and in particular G is of pα-central type when k = p.

Lemma 2.5. Let α be a 2-cocycle of a nilpotent group G and let π(|G|) = {p
where n is divisible by p i for some i.In particular G is of p i α-central type when k = p i .
Proof.For the part when k = p i we have from the proof of [13,Corollary 1.3] that | Proj(G, α)| is the product of | Proj(S j , α S j | for 1 ≤ j ≤ r, where S j is a Sylow p j -subgroup of G. Consequently G is of p i αcentral type if and only if S i is of p i α S i -central type and S j is of α S j -central type for j = i and the result now follows from Lemma 2.4.Proof.Let Proj(G, α) = {ξ i : i = 1, 2, 3}.We may assume without loss of generality that p ξ 1 (1). 2 , so that either two or three of the degrees must be coprime to p.In the case of two, we may assume that p | ξ 3 (1).For convenience we will always assume that a ≤ b ≤ c and in this paper we will be concentrating on the case when a divides b and c, so that a = c α (G) and a 2 | |G|.The proof of this result follows from the observation that p d.In the particular case when G is of (a, a, ka)-type in Lemma 2.8 we have that p | k 2 + 2 and so in particular if p | k, then p = 2. On the other hand if G is of (a, ka, ka)-type in Lemma 2.8 then p | 2k 2 + 1 and p k.

Sylow subgroups
In particular equality holds if and only if for each p ∈ π(|G|/d 2 ) there is exactly one non-trivial α-regular conjugacy class containing elements of p-power order and there are no other non-trivial α-regular elements.
So in the case | Proj(G, α)| = 3 we have that ω(|G|/d 2 ) = 1 or 2. In this case we also note that |G|/d 2 is the sum of three squares with at least one of these an odd number and hence |G|/d 2 ≡ 0 (mod 4).So if G contains an α-regular element of even order then |G|/d 2 = p a or 2p a for p an odd prime number and a > 0.
Corollary 3.4.Let α be a 2-cocycle of G and set d = c α (G).Suppose that |G|/d 2 is square-free and p ∈ π(|G|/d 2 ).Then a Sylow p-subgroup P of G is of α P -central type and all α P -regular elements of P are α-regular.
Proof.Using Lemma 1.7 we may let γ ∈ Proj(P, α P ) with γ (1) = d p .Now |P|/d 2 p = p and so from Lemma 2.3, Proj(P, α P ) must consist of exactly p elements of degree γ (1).Now from Corollary 3.2 there exists a nontrivial x ∈ P that is α-regular.Now {1, x i : i p−1 ≡ 1 (mod o(x))} is a set of α P -regular conjugacy class representatives for P (see [23,Lemma 2.6.1]) and by the same result each such element is α-regular in G .
We note that if |G|/d 2 is a prime number then the Sylow structure of G described in Lemma 3.1 and Corollary 3.4 is sufficient to guarantee that G is of pα-central type for p ≤ 5 from [16,Theorem 2.5].However, even in the case that |G|/d 2 = 6 and the Sylow structure of G is as described, then either G is of 6α-central type or G is of (d, d, 2d)-type with respect to α.
The theory of p-blocks may be easily applied to α-characters in the case that p o([α]), the particular application that we will use below is when G has a cyclic Sylow p-subgroup, details for which may be found in [2,Section 68].In the next result and subsequently we will consider the situation when | Proj(G, α)| = 3 and G has a cyclic Sylow p-subgroup, we will eventually show that such a group is either of 3α-central type or of (a, a, 2ka)-type with respect to α. Proof.Every element of p-power order is α-regular in G, so that G contains one or two p-regular αregular conjugacy classes of G.In the first case |P| = p or p 2 and in the second |P| = p.Let ξ ∈ Proj(G, α) with p ξ(1) and let B be the p-block of G containing ξ , so that it has defect group P.
Now if G has one p-regular α-regular conjugacy class then B contains |P| elements of Proj(G, α) and since it is the only p-block containing elements of Proj(G, α) we conclude that |P| = 3.The decomposition numbers or Brauer graph yield that the elements of Proj(G, α) all have the same degree.
On the other hand if G has two p-regular α-regular conjugacy classes then B contains either one or two irreducible projective Brauer characters.In the first case B contains |P| ≤ 2 elements of Proj(G, α) and we conclude that |P| = 2; the two elements of B ∩ Proj(G, α) both have the same odd degree and the remaining element of Proj(G, α) has even degree (it is of highest type).In the second case B contains 2 + (|P| − 1)/2 ≥ 3 elements of Proj(G, α) and we conclude that |P| = 3.The Brauer graph yields that the degrees of the elements of Proj(G, α) are a, b and a + b for some a, b.In particular at least one of these three numbers is even.Proof.We observe that ξ vanishes off N. Now
Corollary 4.4.Let α be a 2-cocycle of G and let N be a normal subgroup of G. Suppose Proj(G, α) = {ξ 1 , ξ 2 , ξ 3 } and let x, y be non-trivial α-regular conjugacy class representatives of G. Suppose that x ∈ N, but y ∈ N. Then o(y) is even and for suitable labeling ξ 3 (x) = 0, ξ 3 (y) = 0, and We note in Corollary 4.4 that ξ(x) = 0 for all ξ ∈ Proj(G, α), but ξ(y) = 0 for some ξ ∈ Proj(G, α).This observation yields the following result.Proof.Let s 1 denote the number of distinct irreducible constituents of ξ N each occurring with multiplicity e 1 and let s 2 , e 2 denote the corresponding numbers for ξ N .Then (s and 2 , and hence k 2 s 1 = 2s 2 .Thus, 2(s 1 + s 2 ) = s 1 (2 + k 2 ) and hence s 1 and 2 + k 2 divide 2(s 1 + s 2 ).This implies that q = p and c > 0. Thus s 1 = 2 b+1 p d or 2 b p d for some d ≥ 0 depending on whether k is odd or even respectively.The case r = k is similar, but easier using the fact that here 2k 2 s 1 = s 2 .
Corollary 4.7.Let α be a 2-cocycle of G and set n = | Proj(G, α)|.Suppose that N is a normal subgroup of G such that G has n orbits on Proj(N, α N ) and the elements of Proj(N, α N ) all have the same degree ζ (1).
Then the number of distinct irreducible constituents of ξ N is r 2 t, where r = ξ (1)/ξ(1) and t is the number for ξ N .
We note that if we add the assumption that |N|/ζ (1) 2 = p b for p a prime number and b ≥ 0 to the hypothesis in Corollary 4.7, then t is a power of p.

Solvable groups
Proof.K is an elementary abelian p-group for some prime number p and G does not act transitively on and let R/C be a chief factor of G, also let K * denote the dual group of K. Then R/C acts faithfully on K * .Now from the fixed-point theorem (see [Lemma 11.14,22]) we obtain that R/C has order relatively prime to p.It follows from [12, Proposition 1.5 and Corollary 2.4] that there exists an R-invariant λ ∈ Proj(K, α K ), and so So now suppose that K contains two α-regular conjugacy classes of G. Thus G has two orbits on Proj(K, α K ) by Lemma 1.6 and we let ζ 1 , ζ 2 be representatives for them.Then from Proposition 4.3 we can assume without loss of generality that I G (ζ i )/K is of i βζ i -central type for the obstruction 2-cocycles βζ i .Using the bijections detailed in Section 1 this means that there is one element of Proj(G, α) vanishing off K and lying over the orbit containing ζ 1 and the remaining two elements of Proj(G, α) are of the same degree and lie over the other orbit.The squares of the degrees of the elements of Proj(G, α) are thus |G : K|t 1 and |G : K|t 2 /2 (twice), where t i is the length of the orbit containing ζ i .Now either t 1 or t 2 is 1 and t 1 + t 2 = |K|.These two possibilities yield that s 2 = d 2 = |G : K| or |G : K|/2 respectively.
Finally suppose that K contains all three α-regular conjugacy classes of G. Thus by Lemma 1.6, G has three orbits on Proj(K, α K ) of lengths t 1 , t 2 , t 3 say.Then from Lemma 4.2 for each ζ ∈ Proj(K, α K ) we have that I G (ζ )/K is of βζ -central type for the obstruction 2-cocycle βζ .It follows from the bijections in Section 1 that the three elements of Proj(G, α) all vanish off K.So the squares of the degrees of the elements of Proj(G, α) are |G : K|t i .Now we know that at least one of t 1 , t 2 and t 3 is one, and so s 2 = d 2 = |G : K|.
We note for future reference that the proof only uses the solvability of G to give that both K and R are solvable.Proof.Let K be a minimal normal subgroup of G, so that K is an elementary abelian p-group for some prime number p.
Suppose K contains no nontrivial α-regular elements.Then using a similar argument to that in the proof of Corollary 5.2 we have that the two statements hold in G.
So now suppose that K contains nontrivial α-regular elements.Then p ∈ π(|G|/d 2 ) and K does not contain all the α-regular elements of G from Lemma 3.1, so we obtain at least two elements of Proj(G, α) of the same degree.Now as described in the proof of Proposition 5.1 there exists λ ∈ Proj(K, α K ) such that I G (λ)/K is of 2 βλ -central type for the obstruction 2-cocycle βλ .Now I G (λ)/K has one nontrivial βλ -regular conjugacy class which consists of elements of 2-power order.It follows that the α-regular class of G not contained in K consists of elements of even order and hence π(|G|/d 2 ) = {2, p}.
We conclude this section by considering the situation when |G| is odd.Theorem 5.4.Let α be a 2-cocycle of G, where G has odd order.Suppose that | Proj(G, α)| = 3.Then G is of type (a, ka, ka)-type with respect to α for some a and k, where 1 + 2k 2 = p r for some prime number p ≡ 3 (mod 8) and r is odd.
Proof.Choose if possible a normal subgroup N of G such that N contains no non-trivial α-regular elements of G. Then by induction, and using the degree ratio preserving bijection in Clifford's theorem detailed in Section 1, I G (ζ )/N is of type (b, kb, kb) with respect to the obstruction 2-cocycle βζ , 1+2k 2 = p r for some prime number p ≡ 3 (mod 8) and r is odd.Set t = |G : I G (ζ )|, then G is of type (tb, ktb, ktb) with respect to α as required.So now let K be a minimal normal subgroup of G. Then we may assume K contains a non-trivial α-regular element of G. From Proposition 5.1 there exists a ), but v and v −1 are in different orbits in the action of G on Irr(K).So λv and λv −1 are in distinct G-orbits of length t = (|K| − 1)/2.Thus the degrees of the elements of Proj(G, α) are a = |G : K| 1/2 and two of degree ka, where k 2 = t.Now |K| = 1 + 2k 2 and k is odd, so that |K| ≡ 3 (mod 16), it follows easily that p ≡ 3 (mod 8) and |K| = p r , where r is odd.
One consequence of Theorem 5.4 is that we can now complete work on the situation when G has a cyclic Sylow 2-subgroup.
Suppose that G has a cyclic Sylow 2-subgroup P. Then |P| = 2 and G is of (a, a, 2ka) type with respect to α, where 1 + 2k 2 = p r for some prime number p ≡ 3 (mod 8) and r odd.
Proof.From Proposition 3.5, |P| = 2 and the elements of Proj(G, α) are two ξ 1 , ξ 2 of the same odd degree and one ξ 3 of even degree.Let H be a Hall 2 -subgroup of G. Then (ξ 3 ) H is the sum of two elements of Proj(H, α H ) of the same odd degree and ξ 3 vanishes off H. Now every element of 2-power order is αregular so we must have that (ξ 1 ) H = (ξ 2 ) H ∈ Proj(H, α H ). Thus | Proj(H, α H )| = 3 and the result is immediate from Theorem 5.4.

Simple groups
Notation concerning simple groups is consistent with that used in [1] and in which relevant projective character tables may be found.), and there exists another prime number q such that q | |G|, but q |M(G)|.Now all elements of p-power and q-power order are α-regular from Section 1.
Suppose that | Proj(G, α)| = 3.Then we conclude from Proposition 3.5 that P has order three.Thus the three α-regular conjugacy classes of G consist of the identity element, the elements of order 3 and the elements of order q.Now let r be any other prime divisor of |G| aside from 3 and q, then r must divide o([α]).So provided ω(|G|) > 3, |M(G)| is divisible by at least two prime numbers ( = 3).The non-abelian simple groups G which have a Sylow 3-subgroup of order three are classified in Theorem A.1 and the Schur multipliers of groups divisible by at least two prime numbers other than three may be found in [22,Table 8.5].The intersection of these two lists is empty.Finally we have to consider the case when ω(|G|) = 3 and G has a Sylow 3-subgroup of order three.From [8, Theorem 1] there are just two possibilities for G, either A 1 (4) ∼ = A 1 (5) ∼ = A 5 or A 1 (7), both with Schur multiplier of order 2.However, using Lemma 2.1 it is sufficient to observe that | Proj(A 5 , α)| = 4 and | Proj(A 1 (7) One consequence of the Classification Theorem is the following result, see [5,Theorem 1.53]: Let G be a simple group and let p be a prime number.Suppose p 2 |G|, then p 2 | Aut(G)|.We have had reason to consider simple groups G whose order is divisible by 3 but not 9, so in this case 9 | Out(G)|.Lemma 6.2.Let G be a simple group with a Sylow 3-subgroup of order three.Then Out(G) is cyclic unless G ∼ = A 1 (q), where q = p f , p is odd and f is even, in which case Out(G) ∼ = C 2 × C f and q ≡ 4 or 7 (mod 9).
Proof.The outer automorphism group of every simple group is known from [1] and is trivial for G ∼ = J 1 and of order 2 for G ∼ = A 5 .
For the classical groups we are considering we have the outer automorphism group is a semi-direct product of two groups: firstly the cyclic group of diagonal automorphisms (of order d) and secondly the direct product of the cyclic group of field automorphisms (of order f ) and the cyclic group of graph automorphisms (of order g).For A 1 (q), d = (2, q − 1), q = p f and g = 1.For A 2 (q), d = (3, q − 1), q = p f and g = 2.For 2 A 2 (q), d = (3, q + 1), q 2 = p f and g = 1.
Now for G ∼ = A 2 (q), q ≡ 2 or 5 (mod 9), we have that d = 1, and f is odd and so the outer automorphism group in this case is cyclic of order 2f .For G ∼ = 2 A 2 (q), q ≡ 4 or 7 (mod 9), we have that d = 1 and so the outer automorphism group is cyclic of order f in this case.Finally for G ∼ = A 1 (q), q ≡ ±2 or ±5 (mod 9), d is 2 if p is odd and is 1 for p = 2, also f even implies that q ≡ 4 or 7 (mod 9).So the outer automorphism group is either cyclic or is C 2 × C f if p is odd and f is even.Lemma 6.3.Let G be a group and suppose K is a non-solvable minimal normal subgroup of G. Then if K is the unique minimal normal subgroup of G, G/K is isomorphic to a subgroup of Out(K) and G is isomorphic to a subgroup of Aut(K).
Proof.We observe that C G (K) is a normal subgroup of G and Z(K) is trivial.So C G (K) = {1}, K ∼ = Inn(K), and hence G/K is isomorphic to a subgroup of Out(K) and G is isomorphic to a subgroup of Aut(K).Secondly assume x ∈ K and y ∈ K or vice versa.Let Proj(G, α) = {ξ 1 , ξ 2 , ξ 3 }.Then two of these α-characters restricted to K have the same irreducible constituents and using the bijections in Clifford's theorem only the third vanishes off K. Suppose now that G is not of 3α-central type.Then from Proposition 3.5, we may assume ξ 1 (1) = a, ξ 2 (1) = b and ξ 3 (1) = a + b for some a, b, and from the proof of the same result G contains exactly one nontrivial α-regular conjugacy class consisting of 3-regular elements.Suppose z ∈ G is both α-regular and 3-regular.Then ξ 3 (z) = ξ 1 (z) + ξ 2 (z) using the Brauer graph.Now the second orthogonality relation gives that and so We note that ξ 1 (z) = 0 if and only if ξ 2 (z) = 0 and consequently ξ 3 (z) = 0 if and only if a = b.So if z ∈ K, the only possibility is that a = b.So suppose z ∈ K and a = b.Then as noted above there exist two elements of Proj(G, α) that have the same restriction to K up to a scalar multiple.All three possibilities lead, using the equalities above, to the impossible equation a 2 + ab + b 2 = 0.The only possibility left is that x, y ∈ K and we can now also assume K is the only minimal normal subgroup of G. Then |K| 3 = 3 and hence K is simple.If |K| = 3, then Proj(G, α) consists of three elements of degree |G : K| 1/2 from Clifford's theorem.Otherwise G/K is isomorphic to a subgroup of Out(K) by Lemma 6.3 and If Out(K) is cyclic then I G (ζ ) = K and each of the three orbits has length |G : K|.Now M(K) is trivial for K ∼ = J 1 , or for the values of q we are considering for K ∼ = 2 A 2 (q), or A 1 (2 m ) for m = 2, or A 2 (q) for q = 2 from [1]. For For the nontrivial cohomology class [α] there are two irreducible α-characters of degree 2, one of degree 4 and one of degree 6, contradicting equality of orbit lengths.Similarly M(A 2 (2)) ∼ = C 2 .For the nontrivial cohomology class [α] there are two irreducible αcharacters of degree 4, one of degree 6 and one of degree 8, again contradicting equality of orbit lengths.
If Out(K) is not cyclic, then either and each of the three orbits either has length |G : K| or |G : K|/4.As shown in the previous paragraph equality of orbit lengths is impossible, so either there are two orbits of length |G : K| and one of length |G : K|/4 or vice versa.
Then using the same information as above we obtain for q ≡ 1 (mod 4) and q > 5 that q = 33, which is impossible.Similarly for q ≡ −1 (mod 4) we obtain that q − 3 = (q + 1)/4, which is also impossible.
(b) We first note that [α K ] must be nontrivial from Lemma 2.2, so in particular we need not consider those groups K with M(K) trivial.We also note that if K has three distinct irreducible projective character degrees, then G has three orbits on Proj(K, α K ) and each orbit contains all the irreducible α K -characters of a specific degree.In this situation suppose that Proj(K, α K ) consists of one element ζ 1 of one degree, two elements ζ 2 , ζ 3 of a second degree and the rest of a third degree.Then I G (ζ i )/K is of βζ i -central type for i = 1, 2 with the only irreducible βζ i -character of degree e i where e The elements of Proj(G, α) all vanish off K and so e 2 1 = |G : K| = 2e 2 2 , which is impossible.This result applies for K ∼ = A 1 (5) or A 1 (7) ∼ = A 2 (2), as detailed in (a).
Suppose K ∼ = A 1 (q) and o([α K ]) = 2. Suppose q ≡ 1 (mod 4).Then using the detailed information in [24, Theorem 5.8.10], the sum γ of the two irreducible elements of Proj(K, α K ) of degree (q − 1)/2 is non-zero on 1 + (q − 1)/4 + 2 conjugacy classes of K.However, there is an element of Proj(G, α) that is a scalar multiple of this character.So these conjugacy classes of K become the three α-regular conjugacy classes of G. Now either from the projective character values of γ or the orders of the elements in the conjugacy classes we conclude that this can only occur if q = 5, which we already know is impossible.Now suppose that q ≡ −1 (mod 4).Then a similar result holds using [24, Theorem 5.8.12], although this time it is the sum γ of two elements of degree (q + 1)/2 that is non-zero on 1 + (q − 3)/4 + 2 conjugacy classes of K, and we conclude that q = 7, which we again know is impossible.Finally we consider K ∼ = A 1 (9) and o([α K ]) = 3 or 6.In the case that o([α K ]) = 3, we have that | cd α K (K)| = 4, which is impossible.If o([α K ]) = 6, Proj(K, α K ) consists of two elements of degree 6 and two of degree 12. Now if G has three orbits on Proj(K, α K ), then G must have one orbit of length 2 and two orbits of length one and we obtain a contradiction as in the first paragraph.If on the other hand G has two orbits on Proj(K, α K ), then both orbits have length 2 and also Proposition 4.3 applies.So if ζ 1 and ζ 2 are orbit representatives then we can assume without loss of generality that there exist obstruction 2-cocycles βζ i for i = 1, 2 such that I G (ζ i )/K is of i βζ i -central type with irreducible βζ i -character degrees e i .Thus Next we consider a simple counting result: Lemma 6.5.Let S i = {a i,1 , . . ., a i,r i } for i = 1, . . ., n be non-empty sets of natural numbers with the elements in each set arranged in increasing order.Define If there exist at least two values of i with r i ≥ 2, then equality holds if and only if for all i with r i ≥ 2 and some fixed d.
Proof.We prove the inequality by induction on n, with the result being trivially true for n = 1.Assume the result is true for n − 1, then we may list the elements of where . Then there is nothing to prove if n = 1, so we may assume that n ≥ 2. Now |S 1 • • • S n−1 | = k.Thus (by permuting the indices) |S i S j | = r i + r j − 1 for any pair of distinct indices i, j from {1, . . ., n}.Also if Consequently we can assume that r i ≥ 2 for all i.
Conversely suppose We will apply Lemma 6.5 to cd α (G), when G is non-abelian and simple.[23,Corollary 1.2.5] and so we may assume without loss of generality that Proj(G, α)

Proof. The result is true for
The result is now immediate from these two observations and Lemma 6.5.

Minimal normal subgroups
This section will aim to prove the existence of a solvable minimal normal subgroup of a group G of (a, b, c)-type with respect to some 2-cocycle α under certain restrictions on the values of a, b and c.
In a similar vein to Lemma 6.3 we have Lemma 7.1.Let α be a 2-cocycle of G and set n = | Proj(G, α)|.Suppose that K is a non-solvable minimal normal subgroup of G and that G has n orbits on Proj(K, α K ).Now suppose K must be trivial using the facts that K is non-abelian and minimal.Now suppose that [ βζ ] is trivial.We conclude from Lemma 6.7 that K is simple and ζ 1 (1) = ζ 3 (1).Now, from [25, Theorem 3.1] (see also [18, p.556]), there exists a p ∈ π(|K|) such that K has a cyclic Sylow p-subgroup and p | Out(K)|.Thus, using Lemma 7.1, G has a cyclic Sylow p-subgroup and we conclude from Proposition 3.5 and Corollary 6.4 that p = 3, |K| 3 = 3 and G is of (a, a, 2a)-type with respect to α.
However, M(K) is either trivial or of order 2 using Theorem A.1 and the known Schur multipliers of simple groups.On the other hand if q ∈ π(ζ 1 (1)), then q | |M(K)|.It follows that all the elements of Proj(K, α K ) have degree a power of 2, which would imply that K is solvable from [10, Theorem B].

G). Then G has a solvable minimal normal subgroup unless there exists an odd prime number p such that p o([α]) and
Proof.Let x and y be representatives of the nontrivial α-regular conjugacy classes of G. Suppose first that there exists a normal subgroup N of G such that x, y ∈ N and let K be a minimal normal subgroup of G with K ≤ N. Then G acts transitively on Proj(K, α K ) and so the elements of Proj(K, α K ) all have the same degree.Thus K is abelian from Lemma 6.7.Now suppose that there does not exist a normal subgroup N of G with x, y ∈ N. Let K be a minimal normal subgroup of G, then K is the unique minimal normal subgroup of G from Corollary 4.5.Now suppose K ∼ = H × • • • × H, where H is a non-abelian simple group.Thus Lemma 6.3 applies and G/K is isomorphic to a subgroup of Out(K), moreover as in the proof of Corollary 7.2 there exists a prime number p such that H has a nontrivial cyclic Sylow p-subgroup Q and p does not divide | Out(H)|.
Label the isomorphic copies of H from 1 to n so that . Now G has two or three orbits on Proj(K, α K ) and so | cd α K (K)| ≤ 3. Thus n ≤ 2 from Lemma 6.7.
If n = 1, we have that p |G : K|.Thus Q is isomorphic to a Sylow p-subgroup of G and hence |Q| = 3 from Proposition 3.5, but Proposition 6.4 (b) then gives a contradiction.
Finally suppose n = 2. Now Aut(K) Consequently all elements of p-power order in G are α-regular.Let x i ∈ Q(i) be nontrivial for i = 1, 2, where Q(i) is a Sylow p-subgroup of H(i).Then x 1 , x 2 and x 1 x 2 are α-regular in G, so |Q| = p and nontrivial α-regular class representatives for G must be x 1 and x 1 x 2 .This in turn implies that |G|/d 2 = p 2 , otherwise there would be further nontrivial α-regular elements of order coprime to p from Lemma 3.1 (b).
We note that if G is of (a, a, ka)-type with respect to α, then k 2 + 2 = p 2 for p a prime number and so G contains a solvable minimal normal subgroup from Theorem 7.3.

The solvability of groups of (a, ra, ka)-type
In this section it will be shown that a group G of (a, ra, ka)-type with respect to a 2-cocycle is solvable under certain restrictions on the values of r and k.We begin by recapping some information about simple groups that have a solvable subgroup of p-power index for p a prime number, these results are a summary of [25,Theorem 3.2] (although this result omitted PSL (3,3)), [6, Theorem 1] and [3, Theorems 2.1 and 2.2].It is more convenient to use the notation PSL(n, q) rather than A n−1 (q) in this section.Theorem 8.1.Let S be a non-abelian simple group with a solvable subgroup T of p-power index for p a prime number.Then, up to isomorphism, firstly S and secondly T are as follows: Note that, up to isomorphism, PSL(2, 7) is the only non-abelian simple group in Theorem 8.1 that has a solvable subgroup of prime power index for more than one prime number.The following is also a direct result of [6, Theorem 1].Corollary 8.2.Let S be a non-abelian simple group and let p be a fixed prime number.Suppose that there exist T 1 , T 2 < S of p-power index in S. Then |S : T 1 | = |S : T 2 |, so in particular all such subgroups are maximal in S.
A consequence of [19, Satz II.1.5],which will be utilized at least twice in the rest of this section, is that if a group acts transitively on a set , then every orbit of a subnormal subgroup acting on has length dividing | |.Lemma 8.3.Let G be a group of (a, ra, ka)-type with respect to the 2-cocycle α with r = 1 and k even or r = k.Suppose K is a solvable minimal subgroup of G and let H be a subnormal subgroup of G. Then there exists an H-orbit on Proj(K, α K ) of p-power length, where p is the prime number dividing |K|.
Proof.It is sufficient to show that G has an orbit on Proj(K, α K ) of p-power length.
Suppose that K contains no nontrivial α-regular elements of G. Then G acts transitively on Proj(K, α K ) and hence for any Suppose G has two orbits on Proj(K, α K ).Then the orbit lengths are p d and k 2 p d /2 for r = 1 and k even; p d and 2k 2 p d for r = k from Corollary 4.6.
Finally suppose G has three orbits on Proj(K, α K ).Then the orbit lengths are p d , r 2 p d , and k 2 p d from Corollary 4.7 and the note following it.
We will prove that a group G of (a, ra, ka)-type with respect to a 2-cocycle α is solvable under the following conditions: (a) (1) r = 1, k even and 2 + k 2 is not of the form 2(2 m + 1) n , where 2 m + 1 is a prime number; or (2) r = k and 1 + 2k 2 is not of the form (2 m + 1) n , where 2 m + 1 is a prime number; also 1 + 2k 2 is not the square of a prime number p for p a. (b) k ≡ ±1 (mod 3), k ≡ 0 (mod 7) and either (i) r = 1, k ≡ 2 (mod 4); or (ii) r = k, and either ω(1 + 2k 2 ) = 2 or k is odd.
We note that if G is of the type described then ω(|G|/a 2 ) ≤ 2 with equality certainly being obtained in cases (a) (1) and (b) with (i).All these conditions ensure that there exists a solvable minimal normal subgroup of G from Theorem 7.3.Also note that 2 + k 2 and 1 + 2k 2 are not congruent to 0 modulo 5, 7 or 13 for any value of k.Now assume henceforward that G is a non-solvable group of minimal order of (a, ra, ka)-type with respect to the 2-cocycle α with r and k as specified.The forthcoming results follow the argument used in [25, Section 2] and we give due acknowledgment to the authors of that paper.

Lemma 8.4. There exists a solvable minimal normal subgroup
Proof.Suppose there exists a minimal normal subgroup K of G that contains no nontrivial α-regular elements of G. Then K is solvable and I G (ζ )/K is of (b, rb, kb)-type for some b with respect to βζ , using the notation of Lemma 1.
contrary to the assumption that G is not solvable.
It remains to show that C G (K) is maximal with respect to being solvable and normal.Suppose that R = R/C G (K) is a solvable chief factor of G. Then the argument given in the first paragraph of Proposition 5.1 yields that there is a G-invariant element of Proj(K, α K ) contrary to the assumption that G is non-solvable.
where H is a non-abelian simple group.We now examine some properties of the group H, which we know has a solvable subgroup of p-power index from Lemmas 8.3 and 8.4 with such groups being listed in Theorem 8.1.

Lemma 8.5.
There exists a solvable minimal normal subgroup K of G as in Lemma 8.4 such that all the H-orbits on Proj(K, α K ) have the same length, p d > 1, where π(|K|) = {p}.Suppose also that G satisfies the conditions in (a), then every solvable minimal normal subgroup of G contains no non-trivial α-regular elements of G.
Proof.From Lemma 8.4 we have that H has a solvable subgroup of p-power index.Suppose that G has two orbits on Proj(K, α K ).Then the orbit lengths are of the form p d and k 2 p d /2 for r = 1 and k even or p d and 2k 2 p d for r = k as in Corollary 4.6.So |K| = p d (1 + k 2 /2) or p d (1 + 2k 2 ) respectively.So 2 + k 2 is of the form 2p n in the first case and hence p is odd, whereas in the second case 1 + 2k 2 is of the form p n .Now suppose that G has three orbits on Proj(K, α K ).Then similarly the orbit lengths are of the form p d , r 2 p d , and k 2 p d and so 1 + r 2 + k 2 is a power of p and r = k.
Under the conditions specified in (a), this is impossible from Lemma 8.4 and Theorem 8.1.So in this case G must act transitively on Proj(K, α K ).On the other hand if G does act transitively on Proj(K, α K ), then for each ζ ∈ Proj(K, α K ) we have I H (ζ ) is a solvable subgroup of H of p-power index, moreover this index is uniquely determined from Corollary 8.2.
So now suppose that every solvable minimal normal subgroup of G contains a non-trivial α-regular element of G. Then K is unique and the conditions in (b) give that p = 3.Thus H ∼ = PSL (2,8) and one orbit length of H acting on Proj(K, α K ) is 9 from Theorem 8.1, moreover all subgroups I of H with C G (K) ≤ I = H are solvable.
In the two orbit case let ζ ∈ Proj(K, α K ) be in the G-orbit of length k 2 3 d /2 for r = 1 and k ≡ 2 (mod 4) or 2k 2 3 d for r = k and k odd.Then I H (ζ ) is a solvable subgroup of H of index l dividing k 2 3 d /2 or 2k 2 3 d .Given the assumptions on k, we must have that l ≡ 0 (mod 4).So I H (ζ )/C G (K) is a subgroup of the maximal subgroup of H of index 9 and the only possibilities for l are 9 and 18.Now a maximal subgroup of PSL (2,8) of index 9 is a Frobenius group of order 56 with a Frobenius kernel of order 8, and so l = 18.The same argument applies in the three orbit case starting with any G-orbit of length k 2 p d .
Using the first paragraph in the proof of Lemma 8.5 we note that G cannot have three orbits on Proj(K, α K ) for r = 1 and k even, whereas G can only have two (or three) orbits on Proj(K, α K ) in the case r = k when ω(1 + 2k 2 ) = 1.
We continue with K as specified in Lemma 8.5.
Note that R acts faithfully on K * as in the proof of Lemma 5.1.In particular no nontrivial element h ∈ H fixes every element of K * and so H acts faithfully on K * and on W = K * /C K * ( H).Thus there exists an injective mapping, , from H into Aut(W) and Aut(W) ∼ = GL(d, p) by [26,Theorem 9.14], where |W| = p d .
Suppose that H ∼ = PSL (2, q), where q = 2 m − 1 is a prime number and m > 3. Then | H : Ī| = 2 m from Theorem 8.1 and Ī ∼ = C q C 2 m−1 −1 .Now the highest power of q dividing | H| is q itself, so that Ī is the normalizer of a cyclic Sylow q-subgroup Q of H. Now [4, Lemma 15.1.1]applies and Ī is a Frobenius group with a cyclic complement acting irreducibly on Q.So Q is a maximal abelian subgroup of Ī.
Suppose that Q fixes λ ∈ K * .We know that Ī fixes exactly one element in each H-orbit of K * , so Ī h fixes λ for some h ∈ H. Then Q, Ī h fixes λ, but since Ī and Ī h are maximal in H we have that either Clearly Ī fixes λ for the first possibility.For the second possibility Q is a normal Sylow q-subgroup of Ī h, since Ī h ∼ = Ī.In particular we have Ī h ≤ N H ( Q) = Ī, and so we have equality.Hence any λ fixed by Q is also fixed by Ī and therefore and no nontrivial element of Z is fixed by Q.So, since the size of any orbit of Q on Z divides | Q| = q = 2 m − 1, the only option is that Q has just two orbits on Z with one being the trivial one.In particular Z is a simple Q module and [20,Theorem 31.7]applies, so | Ī : Q| = 2 m−1 − 1 divides m.However 2 m−1 − 1 > m for any m ≥ 4 and so we rule out these groups.
We have thus disposed of the case that G does not contain a minimal normal subgroup K containing no nontrivial α-regular elements of G.So we can now assume that G acts transitively on Proj(K, α K ).From the previous result we need to consider H ∼ = PSL (2,7) with | H : Ī| = 8, so that K is a 2-group.Let F = F(G) denote the Fitting subgroup of G. Proposition 8.7.F = O 2 (G) and I G (τ ) is a proper solvable subgroup of 2-power index in G for all τ ∈ Proj(F, α F ).
Proof.Suppose O q (G) is nontrivial for q an odd prime number and let L ≤ O q (G) be a minimal normal subgroup of G. Then the previous results of this section show that L must contain two or three α-regular classes of G. Now Lemma 8.3 yields that there exists ζ ∈ Proj(L, α L ) such that |G : I G (ζ )| is a power of q.Then by Lemma 4.2,Propositions 4.3 and [11,Theorem B] we also obtain that I G (ζ ) is solvable.It follows from Lemmas 8.3 and 8.4 applied to L and K respectively that H has a solvable subgroup of q-power index and hence q = 7.By assumption G is of (a, ra, ka)-type with respect to α, where either r = 1 or k; however, as previously noted k 2 + 2 and 2k 2 + 1 ≡ 0 (mod 7).
Finally by [18,Theorem 3.7] there exists a bijection : A → C, where if γ = (ψ) then ψ(1)/γ (1) = τ (1)/φ (1).We conclude that | Proj(N G (Q), α N G (Q) )| = 3. Moreover the degree ratios in Proj(G, α) are preserved by these bijections, so N G (Q) is of (b, rb, kb)-type with respect to α N G (Q) for some b and so N G (Q) must be solvable.However, G/F = FN G (Q)/F ∼ = N G (Q)/N G (Q) ∩ F and thus G/F is solvable giving a contradiction.We have thus proved: Theorem 8.10.Let G be a group of (a, ra, ka)-type with respect to the 2-cocycle α.Then G is solvable under conditions (a) or (b).Corollary 8.11.Suppose that k satisfies the conditions in (a) (2) or (b) with (ii), but with ω(1 + 2k 2 ) = 2 in either case.Then there does not exist a group G of (a, ka, ka)-type with respect to any 2-cocycle α of G.
Proof.Suppose such a group G exists.Then G is solvable from Theorem 8.10, but then Corollary 5.3 gives a contradiction.
To summarize we have produced considerable evidence to support the conjecture that if G is a finite group with a 2-cocycle α such that | Proj(G, α)| = 3, then G is solvable.The techniques used rely on

Definition 2 . 7 .
Suppose that there exists a 2-cocycle α of G such that | Proj(G, α)| = 3 and the elements of Proj(G, α) have degrees a, b, c.Then we say that G is of (a, b, c)-type with respect to α.

Proposition 3 . 5 .
Let α be a 2-cocycle of G such that | Proj(G, α)| = 3 and let p ∈ π(|G|).Suppose that G has a cyclic Sylow p-subgroup P. Then either (a) |P| = 3, and G is of 3α-central type or |G| is even and G is of (a, b, a + b)-type with respect to α for some a, b; or (b) |P| = 2, and the elements of Proj(G, α) consist of two of the same odd degree and one of even degree.
One of the purposes of this section is to examine some properties of a normal subgroup N of a group G with | Proj(G, α)| = 3.Our results use Lemma 1.5 and the bijections in Section 1.We first consider in general what happens if I G (ζ )/N is of βζ -central type for ζ ∈ Proj(N, α N ).

Lemma 4 . 1 .
Let α be a 2-cocycle of G and let N be a normal subgroup of G. Suppose G is not of α-central type, but there exists ζ ∈ Proj(N, α N ) such that I G (ζ )/N is of βζ -central type.Let ξ ∈ Proj(G/ζ , α) and ξ ∈ Proj(G, α).Then there exists a non-trivial x ∈ N such that ξ(x)ξ (x) = 0.

Lemma 4 . 2 .
Let α be a 2-cocycle of G and let N be a normal subgroup of G. Suppose G has n obits on Proj(N, α N ).Then | Proj(G, α)| = n if and only if I G (ζ )/N is of βζ -central type for all ζ ∈ Proj(N, α N ).Proof.From the bijections in Section 1, we have that | Proj(I G (ζ )/N, βζ )| = | Proj(G/ζ , α)|.In the context of Lemma 4.2 if I G (ζ )/N is of n βζ -central type then the n elements of Proj(G/ζ , α) all have the same degree.

Corollary 4 . 5 .Corollary 4 . 6 .
Let α be a 2-cocycle of G such that | Proj(G, α)| = 3 and let x and y be non-trivial α-regular conjugacy class representatives of G. Suppose there exists a normal subgroup N of G such that x ∈ N, but y ∈ N. Then there does not exist a normal subgroup M of G such that y ∈ M, but x ∈ M.The next results will become relevant when we consider an abelian minimal normal subgroup of an (a, b, c)-group G with respect to α for suitable values of b and c.Let G be an (a, ra, ka)-group with respect to α, where r ≤ k.Suppose that N is a normal subgroup of G such that (i) G has two orbits on Proj(N, α N ), (ii) the elements of Proj(N, α N ) all have the same degree ζ(1) and (iii) |N|/ζ (1) 2 is of the form 2 b q c with q an odd prime number and b, c ≥ 0. Let ξ , ξ ∈ Proj(G, α) with (ξ(1), ξ (1)) = (a, ka) and ξ N , ξ N = 0. Then r = 1 or k and the number of distinct irreducible constituents of ξ N and ξ N are respectively: (a) 2 b+1 p d and 2 b k 2 p d for r = 1 and k odd; (b) 2 b p d and 2 b−1 k 2 p d for r = 1 and k even; (c) 2 b p d and 2 b+1 k 2 p d for r = k; where d ≥ 0 and p is the odd prime divisor of |G|/a 2 .

Proposition 6 . 4 .
Let α be a 2-cocycle of G such that | Proj(G, α)| = 3.(a) Suppose G has a cyclic Sylow 3-subgroup.Then G is of 3α-central type or of (a, a, 2a)-type with respect to α for some a.(b) Suppose that K is a non-abelian simple normal subgroup of G. Then K is neither isomorphic to one of the groups listed in Theorem A.1 nor A 1 (q).Proof.(a) From Proposition 3.5 a Sylow 3-subgroup of G has order three.Let x and y be non-trivial αregular conjugacy class representatives.Let K be a minimal normal subgroup of G. Then either |K| 3 = 3 or 3 |K|.Also G has 1, 2 or 3 orbits on Proj(K, α K ) from Lemma 1.6 according as to whether x, y ∈ K, x ∈ K, y ∈ K or vice versa, or x, y ∈ K respectively.Firstly suppose x, y ∈ K. Then 3 |K|.Let ζ ∈ Proj(K, α K ).Then I G (ζ )/K has a cyclic Sylow 3subgroup and there exists an obstruction 2-cocycle βζ of I G (ζ )/K such that | Proj(I G (ζ )/K, βζ )| = 3.By induction I G (ζ )/K is of 3 βζ -central type or of (b, b, 2b)-type with respect to βζ for some b.Using the standard bijections in Clifford's theorem detailed in Section 1 yields respectively that G is of 3α-central type or of (a, a, 2a)-type with respect to α, where a = b|G : I G (ζ )|.
8 and the degree ratio preserving bijections of Section 1. Thus I G (ζ ) is solvable.If the supposition is false then G contains a unique minimal normal subgroup K using Corollary 4.5.In this case G has two or three orbits on Proj(K, α K ) and hence I G (ζ ) is solvable from Proposition 4.3 and Corollary 4.4.Now [α K ] is trivial from [14, Lemma 1.1].From the same result and its proof either then by the same argument as in Case 1 there exists ζ in the H-orbit of δ which is fixed by I H (ζ ).Moreover we can choose ζ ∈ Proj(K, α K ) to maximize the number of H-orbits containing an element of Proj(K, α K ) that is fixed by I H (ζ ).Then by a similar argument to that in Case 1 and using the same notation we obtain | Proj(K, α K )| ≤ 2|F||H : I H (ζ )|.Following the same argument we obtain |K * : C K * (I H (ζ ))| ≤ 2| H : Ī|.Now since | H : Ī| = 13 and |K * : C K * (I H (ζ ))| is a power of 13, this simplifies to |K * : C K * (I H (ζ ))| ≤ | H : Ī|, which yields |K * : C K * ( H)| ≤ | H : Ī| 2 .

Corollary 3.2. Let
Lemma 3.1.Let α be a 2-cocycle of G, let P be a Sylow p-subgroup of G and set d = c α (G).Then (a) P is of α P -central type if and only if p |G|/d 2 ; (b) P contains a non-trivial α-regular element of G if and only if p | |G|/d 2 .Proof.Let γ ∈ Proj(P, α P ) of smallest degree, so that γ (1) = d p from Lemma 1.7.Firstly suppose p |G|/d 2 .Then d 2 p = |P| and so P is of α P -central type.On the other hand suppose p | |G|/d 2 and that P contains no nontrivial α-regular elements.Let ξ ∈ Proj(G, α) such that p ξ(1)/d p .Then Sylow p-subgroup of G is either of α P -central type or it contains a non-trivial α-regular element of G.The following corollaries are immediate from Lemma 3.1.α be a 2-cocycle of G and let P be a Sylow p-subgroup of G. Then P contains a non-trivial α-regular element if and only if it contains a non-trivial α P -regular element.

Corollary 3.3. Let
α be a 2-cocycle of G and set d is a power of p independent of the choice of ζ .Suppose H is isomorphic to one of the groups listed in Theorem 8.1.To avoid repetition, we begin by building a relationship between the groups C K * ( H) and C K * ( Ī), where Ī = I H (ζ )/C G (K). Now Ī is a maximal subgroup of H by Corollary 8.2, so N H . However 9 divides |PSL(3, 3)| but does not divide |SL(2, 13)|.Case 3: Suppose H ∼ = PSL(2, 7) and | H : Ī| = 7. Similarly to Case 2, H has two orbits on the set of subgroups of H of index 7 and we can choose ζ ∈ Proj(K, α K ) such that | Proj(K, α K )| ≤ 2|F||H : I H (ζ )|.Following the same argument with this time | H : Ī| and |K * : C K * (I H (ζ ))| being powers of 7, we obtain |K * : C K * ( H)| ≤ | H : Ī| 2 and we can define W and as before.Now |W| ≤ 7 2 and PSL(2, 7) is isomorphic to a subgroup of GL