Brauer characters and fields of values

Theorem 1 ([3, Theorem A]).

Let F be a subfield of the complex numbers and suppose that N is an F-free normal subgroup of the finite group G. Then the canonical homomorphism G→G/N defines a bijection from ClF⁡(G) to ClF⁡(G/N).

Theorem 2 ([3, Theorem B]).

Let F be a subfield of the complex numbers and let G be a finite group. Then G contains a unique largest F-free normal subgroup N, and N is contained in the kernel of every irreducible F-character of G.

Theorem A.

Let p be a prime number and F any subfield of the field of complex numbers. Let N◁G and assume that N∘ is F-free in G. Then the canonical homomorphism G→G/N induces a bijection from ClF∘⁡(G) to ClF∘⁡(G/N).

Theorem B.

Let p be a prime number and F any subfield of the field of complex numbers. Then G has a unique largest normal subgroup N with the property that N∘ is F-free in G; N is contained in the kernel of every ϕ∈IBrF⁡(G).

Assumption 1.

For the remainder of this paper we fix a field 𝔽⊆ℂ and a prime number p.

Lemma 2.

Let S be a non-abelian simple group. Then S has a rational element of order 3 or 5 unless S=PSL2⁡(32⁢n+1), n≥1, which has no non-trivial rational elements of odd order.

Proof.

See [6, Theorems 11.1 and 11.2]. ∎

Lemma 3.

Let x∈G and write X=〈x〉. If x is an F-element in G, then X is F-cyclic in G. Conversely, if X is F-cyclic in G, then every element of X is an F-element in G.

Proof.

This is [3, Lemma 2.1]. ∎

Lemma 4.

Let X=〈x〉 and Y=〈y〉 be cyclic groups of order m and n, respectively, and assume that m divides n. If y↦yt is an automorphism in AutF⁡(Y), then x↦xt is an automorphism in AutF⁡(X). Moreover, every element of the group AutF⁡(X) arises in this way.

Proof.

Let 𝕂=𝔽∩ℚn and 𝔼=𝔽∩ℚm=𝕂∩ℚm. By basic Galois theory, restriction to 𝔼 defines a surjection from Gal⁡(ℚn/𝕂) onto Gal⁡(ℚm/𝔼). Now the lemma follows from the definitions of Aut𝔽⁡(X) and Aut𝔽⁡(Y). ∎

Lemma 5.

Let M and N be normal subgroups of the finite group G and assume that N∩M=1. Let G¯=G/M. Then n∈N is an F-element in G if and only if n¯ is an F-element in G¯.

Proof.

Clearly, if n is an 𝔽-element in G, then n¯ is one in G¯.

So assume that n¯ is an 𝔽-element in G¯. Let ψt:n↦nt be any automorphism in Aut𝔽⁡(〈n〉). Because N∩M=1, we have o⁢(n)=o⁢(n¯) and so, by Lemma 4, the map ϕt:n¯↦n¯t is in Aut𝔽⁡(〈n¯〉). Thus there is some g∈G with n¯g¯=n¯t. Then ng=nt⁢m for some m∈M. But then m=ng⁢n-t∈M∩N=1 and so ng=nt. Thus g induces on 〈n〉 the automorphism ψt and we conclude that n is an 𝔽-element of G. ∎

Theorem 1.

Let N◁G and assume that N∘ is F-free in G. Then the canonical homomorphism G→G/N induces a surjection ClF∘⁡(G)→ClF∘⁡(G/N).

Proof.

Identifying the Brauer characters of G/N with a subset of the Brauer characters of G, it is clear that images of 𝔽-elements of G are 𝔽-elements of G/N. And clearly images of p-regular elements are p-regular. Thus, to prove the theorem we must show that if N⁢x∈G/N is a p-regular 𝔽-element, then the coset Nx contains a p-regular 𝔽-element of G. Assume this is false and let G have minimal order among counterexamples. Also assume that N has minimal order among normal subgroups K such that K∘ is 𝔽-free in G and K gives rise to a counterexample.

Suppose that 1<M<N with M◁G and note that M∘ is 𝔽-free in G. We claim that (N/M)∘ is 𝔽-free in G/M. To see this, let M⁢y∈N/M be p-regular and assume My is an 𝔽-element in G/M. By the minimality of N among counterexamples, My contains a p-regular 𝔽-element x of G. Since M⁢y⊆N and N∘ is 𝔽-free in G, it follows that x=1 and My is the identity in G/M, i.e. (N/M)∘ is 𝔽-free in G/M, as claimed.

Now set G¯=G/M. Then G/N≃G¯/N¯ and N⁢x¯ is a p-regular 𝔽-element of G¯/N¯. By minimality of G among counterexample groups and since N¯∘ is 𝔽-free in G¯, it follows that there is a p-regular 𝔽-element of G¯ contained in N⁢x¯, call it y¯. Equivalently, My is a p-regular 𝔽-element in G/M. Again by the minimality of N we deduce that My contains a p-regular 𝔽-element z of G. Finally, M⁢y⊆N⁢x, so z∈N⁢x is a p-regular 𝔽-element of G. This contradicts our assumption about N and so we must have that N is minimal normal in G.

Next, we argue that N≤Φ⁢(G), the Frattini subgroup. To see this, let H be a maximal subgroup of G and assume, aiming for a contradiction, that N≰H. Then N⁢H=G and N⁢x=N⁢h for some h∈H. Let D=N∩H and note that the natural isomorphism G/N→H/D sends Nx to Dh. Since Nx is a p-regular 𝔽-element in G/N, it follows that Dh is a p-regular 𝔽-element in H/D. Also, 𝔽-elements of H are automatically 𝔽-elements of G and so it follows that D∘ is 𝔽-free in H. By the minimality of G we conclude that D⁢h⊆N⁢x contains a p-regular 𝔽-element of G, but we are assuming that this is not the case. This contradiction implies that N≤H and so N≤Φ⁢(G). In particular, N is nilpotent and thus, since N is minimal normal in G, N is an elementary abelian ℓ-group for some prime ℓ.

Let U=N⁢〈x〉. We argue that U◁G. Now U/N is 𝔽-cyclic in G/N and thus it is 𝔽-cyclic in L/N, where L/N=NG/N⁢(U/N). Aiming for a contradiction, suppose that L<G. Then by minimality of G, the coset Nx contains a p-regular 𝔽-element of L, which is automatically a p-regular 𝔽-element of G. But we are assuming Nx contains no such element and this contradiction implies that L=G, i.e. U◁G. Also note that U/N is cyclic and N≤Φ⁢(G), whence U is nilpotent (by [2, Lemma 9.19]).

Assume that ℓ=p, i.e. N is a p-group. Since U is nilpotent and x is p-regular, it follows that V:=〈x〉 is the p-complement in U. In particular, V◁G. Since V≃U/N as G-sets (under the conjugation action of G) and U/N is 𝔽-cyclic in G/N, we conclude that V is 𝔽-cyclic in G. By Lemma 3, every element of V is an 𝔽-element. In particular, x is an 𝔽-element in G and x is p-regular.

If instead ℓ≠p, then every element of N is p-regular and N∘=N is 𝔽-free in G. Thus by the Isaacs–Navarro Theorem 1Nx contains an 𝔽-element y of G. Now N⁢x=N⁢y and, since 〈y〉 is 𝔽-cyclic in G, by Lemma 3 it follows that 〈y〉∩N=1. We conclude that o⁢(y)=o⁢(N⁢y), so y is p-regular. This is our final contradiction, and the proof is complete. ∎

Corollary 2.

Let M and N be normal subgroups of the finite group G with M≤N and suppose that N∘ is F-free in G. Then (N/M)∘ is F-free in G/M.

Proof.

Suppose that Mx is a p-regular 𝔽-element of G/M, where M⁢x∈N/M. Then by Theorem 1Mx contains a p-regular 𝔽-element y of G. Since M⁢x⊆N and N∘ is 𝔽-free in G, it follows that y=1 and thus Mx is the identity of G/M, as required. ∎

Corollary 3.

Assume that M and N are normal subgroups of G and that M∘ and N∘ are F-free in G. Then M⁢N◁G and (M⁢N)∘ is F-free in G. In particular, G has a unique largest normal subgroup K such that K∘ is F-free in G.

Proof.

The second statement follows from the first by taking K to be the subgroup of G generated by all normal subgroups N of G such that N∘ is 𝔽-free in G.

For the first statement, obviously M⁢N◁G. Let ψ be the natural G-isomorphism from M⁢N/M to N/(N∩M). We make the following general observation: If C/M is 𝔽-cyclic in G/M, with M≤C≤M⁢N, and if C/M has order coprime to p, then ψ⁢(C/M) is 𝔽-cyclic in G/(M∩N) and so all the elements of ψ⁢(C/M) are p-regular 𝔽-elements of G/(M∩N).

Now let x∈M⁢N be a p-regular 𝔽-element of G. We must show that x=1. Note that Mx is a p-regular 𝔽-element in G/M and we let C/M be the cyclic subgroup of G/M generated by Mx. By the previous paragraph, ψ⁢(C/M) is 𝔽-cyclic in G/(M∩N) and ψ⁢(C/M) has order coprime to p. But ψ⁢(C/M) is contained in N/(M∩N) and (N/(M∩N))∘ is 𝔽-free in G/(M∩N), by Corollary 2. It follows that ψ⁢(C/M) is the identity in G/(M∩N), so C/M is the identity in G/M, i.e. C=M. Thus x∈M and since M∘ is 𝔽-free in G we conclude x=1, as needed. ∎

Lemma 1.

In the above notation, there is a bijection ϕ↦Cϕ between irreducible 2-Brauer characters of the group S=PSL2⁡(32⁢n+1) and conjugacy classes of odd-order elements in S that is compatible with the actions of Aut⁡(S) and Gal⁡(Q⁢(ϵ)/Q), in the following sense. If Cϕ=xS for ϕ∈IBr2⁡(S) and x∈S, then the following statements hold:

1. for any τ∈Aut⁡(S), Cϕτ=(xτ)S,

2. for any σ∈Gal⁡(ℚ⁢(ϵ)/ℚ), Cϕσ=(xσ)S,

3. for any τ∈Aut⁡(S), σ∈Gal⁡(ℚ⁢(ϵ)/ℚ), and ϕ∈IBr2⁡(S), ϕσ=ϕτ if and only if Cϕσ=Cϕτ.

Proof.

The existence of the bijection and statements (i)–(ii) are [7, Lemma 3.4], and statement (iii) is an immediate consequence of (i) and (ii). ∎

Corollary 2.

Let N be a minimal normal subgroup of G and suppose that N=S1×…×Sr≃Sr, where S=PSL2⁡(32⁢n+1). For ϕ=ϕ1×…×ϕr∈IBr2⁡(N), let x=(x1,…,xr)∈N, where Cϕi=(xi)Si in the notation of Lemma 1. Set Cϕ:=xN. Then the map ϕ↦Cϕ is a bijection between IBr2⁡(N) and Cl∘⁡(N) that is compatible with the actions of G and Gal⁡(Q⁢(ϵ)/Q), in the sense of (i)–(iii) in Lemma 1.

Proof.

See [7, Corollary 3.5]. As in Lemma 1, statement (iii) follows at once from (i) and (ii). ∎

Corollary 3.

Let N be a minimal normal subgroup of G and suppose that N=S1×…×Sr≃Sr, where S=PSL2⁡(32⁢n+1). Let ϕ∈IBr2⁡(N) and let Cϕ be as in Corollary 2. Then Q⁢(ϕ)=Q⁢(Cϕ).

Proof.

Let ϕ=ϕ1×…×ϕr, Cϕ=(x1,…,xr)N. Now ℚ⁢(ϕ)=ℚ⁢(ϕ1,…,ϕr) (by evaluating ϕ⁢(y) for elements y∈N with all but one component equal to 1∈S) and likewise ℚ⁢(Cϕ)=ℚ⁢(x1S,…,xrS) (by evaluating η⁢(x) for characters η∈IBr2⁡(N) with all but one component equal to 1S∈IBr2⁡(S)). Thus it suffices to prove the claim for r=1. This follows by inspecting the character table of SL2⁡(32⁢n+1), which can be found e.g. in [1, Chapter 38], and the explicit description of the map ϕ↦Cϕ, as given in [7]. ∎

Proposition 4.

Assume that N=S1×…×Sr≃Sr is a minimal normal subgroup of G, where S=PSL2⁡(32⁢n+1) for some n≥1. Assume that θ∈IBr2,F⁡(G) and N≰ker⁡θ. Then N∘ contains a non-trivial F-element of G.

Proof.

Let ϕ∈IBr2⁡(N) be a non-trivial constituent of the restriction θN. Let N=exp(S)2′, n=|G|, 𝕂=𝔽∩ℚN, and 𝔼=𝔽∩ℚn. Then 𝕂=𝔼∩ℚN and so restriction to ℚN defines a surjection from Gal⁡(ℚn/𝔼) onto Gal⁡(ℚN/𝕂). Note that θ is an 𝔼-character and that x∈N∘ is an 𝔽-element in G if and only if it is a 𝕂-element.

Now let σ∈Gal⁡(ℚN/𝕂) and choose τ∈Gal⁡(ℚn/𝔼) whose restriction to ℚN is σ. Since θτ=θ, it follows that ϕτ=ϕσ also enters θN (ϕσ∈IBr2⁡(N) by [7, Proposition 3.2 (ii)]) and thus there is some g∈G with ϕg=ϕσ. Let Cϕ be the conjugacy class of N corresponding to ϕ under the bijection of Corollary 2. Then by Corollary 2 we have Cϕg=Cϕσ and, by Corollary 3, we conclude that any x∈Cϕ is a 𝕂-element in G, whence it is an 𝔽-element. Since ϕ≠1N, Cϕ is non-trivial and so the proof is complete. ∎

Theorem 5.

Let ϕ∈IBrF⁡(G) and assume that N is a normal subgroup of G and that N∘ is F-free in G. Then N≤ker⁡ϕ.

Proof.

Assume that the theorem is false. Arguing as in [3, Theorem 4.1], we may assume N is minimal normal in G. If N is not solvable, then, by the remarks following Lemma 2, we must have p=2 and N≃Sr, where S=PSL2⁡(32⁢n+1). Applying Proposition 4, we obtain a contradiction.

So N is solvable and thus N is an elementary abelian ℓ-group for some prime ℓ. We have ℓ≠p, as otherwise N≤Op⁢(G)≤ker⁡ϕ, e.g. by [5, Lemma 2.32]. Now the remainder of the proof of [3, Theorem 4.1] goes through unchanged (the point is, since N is a p′-group, we can identify IBr⁡(N) with Irr⁡(N)). ∎

Lemma 1.

Suppose that M,N◁G, M∩N=1, and N∘ is F-free in G. Let x,y∈G∘ be F-elements such that N⁢x=N⁢y∈G/N. If x∈M, then x=y.

Proof.

We have x⁢n-1=y for some n∈N. Also, [M,N]≤M∩N=1 and so [x,n]=1. Thus

It follows that o(n)∣o(y) and thus n∈N∘. Now A:=NG⁢(〈y〉) induces on 〈y〉 a group of automorphisms containing Aut𝔽⁡(〈y〉). Let n↦nt be any automorphism in Aut𝔽⁡(〈n〉). By Lemma 4 we may express t in such a way that some g∈A induces on 〈y〉 the map y↦yt. Then

so (n-1)g⁢nt=(x-1)g⁢xt∈N∩M=1. We conclude that ng=nt, so n is an 𝔽-element in G. Thus n=1 and x=y. ∎

Lemma 2.

Fix an integer n and assume that x∈H≤Sym⁡(n) has prime order ℓ and is an F-element in H. Let t=|Qℓ:F∩Qℓ|. Then there is some h∈H such that the natural action of 〈h〉 on n points has an orbit of length t.

Proof.

We will identify {0,1,…,(ℓ-1)}=𝔽ℓ, the field with ℓ elements.

Since x is an 𝔽-element, there is some g∈H with g⁢x⁢g-1=xa, where a has order t in 𝔽ℓ×. Also, since x has order ℓ, the non-trivial orbits of 〈x〉 all have length ℓ. Let Ω1,…,Ωm denote these orbits.

Now g permutes the Ωi. We may assume that

(possibly j=1). We label the points of Ωi by αi, where α∈𝔽ℓ, and choose the labeling in such a way that x acts on Ωi via the cycle (0i,1i,…,(ℓ-1)i) and also

Thus x acts on Ω:=⋃i=1jΩi via

Now on the one hand we have, on Ω,

and on the other we have

Thus we see that for any αi∈Ω,

Note that for any β∈Ω and any integer s, if gs⁢(β)=β, then j∣s; also, the length of the 〈g〉-orbit of β is the minimal s such that gs⁢(β)=β. Thus, every 〈g〉-orbit on Ω has length a multiple of j. Finally, observe that to prove the lemma it actually suffices to show that 〈g〉 has an orbit on Ω with length a multiple of t (as then, for some power h of g, 〈h〉 has an orbit of exact length t). This is what we now prove.

If aj=1, then t∣j (recall that a has order t in 𝔽ℓ×). By the previous paragraph any 〈g〉-orbit on Ω has length a multiple of t, and so we are finished. Thus we may assume that aj≠1. For any α∈𝔽ℓ we compute

so for any integer i we have

Set c=aj≠1 and choose α1∈Ω1 with α≠b1-c. Choose i minimal with gi⁢j⁢(α1)=α1. The displayed equation implies that

and thus

Since α≠b1-c, we have ai⁢j=ci=1 and thus t∣ij. Since i was chosen minimally, ij is the 〈g〉-orbit length of α1 and so again by the previous paragraph we are finished. ∎

Theorem 3.

Let p=2. Let N=S1×…×Sr≃Sr, where S=PSL2⁡(32⁢n+1) for some n≥1. Assume that N is a minimal normal subgroup of G, CG⁢(N)=1, and N∘ is F-free in G. Let x∈G∘ be an F-element with odd prime order ℓ and assume that |Qℓ:F∩Qℓ|=t. Then there is some g∈G such that 〈g〉 has an orbit of length t on the factors {S1,…,Sr}.

Proof.

By the minimality of N, any g∈G permutes the factors {S1,…,Sr} so the claim makes sense. Let x↦xi have order t in Aut⁡(〈x〉), so that x↦xi is a generator of Aut𝔽⁡(〈x〉). Since N∘ is 𝔽-free in G, x¯∈G¯:=G/N has order ℓ and of course it is still an 𝔽-element. Let g∈G satisfy xg=xi, so also x¯g¯=x¯i.

Since CG⁢(N)=1, we may view

Write g¯=h⁢π, where h∈Out(S)r and π∈Sym⁡(r). Suppose first that x fixes every factor Sm, i.e. that x¯∈Out(S)r, and write x¯=(x1,…,xr), where each xi∈Out⁡(Si). Assume, without loss, that o⁢(x1)=ℓ. As Out⁡(S) is abelian, we have

and, for any integer j,

Since o⁢(x1)=ℓ=o⁢(x), the map x1↦x1i has order t. Thus the x1ij are all distinct for 1≤j≤t and we conclude that the 〈g〉-orbit of S1 has length at least t. But since x¯gt=x¯it=x¯, clearly this orbit has length at most t.

Thus we may assume that x moves some factor Sm. So the image of x under the composition of natural maps

is an 𝔽-element of order ℓ. Let H denote the image of G under this map, so by Lemma 2 there is some h∈H such that 〈h〉 has an orbit of length t on the factors {S1,…,Sr}; equivalently, for any preimage g∈G of h, 〈g〉 has an orbit of length t on {S1,…,Sr}. This completes the proof. ∎

Theorem 4.

Fix a prime ℓ. Let N=S1×…×Sr≃Sr, where S=PSL2⁡(32⁢n+1) for some n≥1. Further, assume that N is a minimal normal subgroup of G with CG⁢(N)=1. Assume that some g∈G permutes the factors {S1,…,Sr} with an orbit of length t, where t=|Qℓ:F∩Qℓ|. If ℓ divides |N|, then there is some F-element of G with order ℓ contained in N.

Proof of Theorem 4.

We have

Write g=h⁢π, where h∈Aut(S)r and π∈Sym⁡(r). Reordering the factors, if necessary, we may assume that

We will show that there is an element v in S1×…×St with order ℓ and such that 〈g〉 induces on 〈v〉 a group of automorphisms containing Aut𝔽⁡(〈v〉). Thus, there is no loss to assume that r=t. Since Aut⁡(S) is split over S, by [4], we may choose a complement O≤Aut(S)t for N and identify O with Out(S)t. Therefore, by replacing g by ng for some appropriate n∈N, we may assume that h∈O and that h acts on N via (h1,…,ht)∈Out(S)t. Clearly this new choice of g still has an orbit of length t on the factors Si. Recall that Out⁡(S) is abelian.