Feit numbers and p′-degree characters

Let p be a prime and let G be a finite solvable group. Let χ∈Irr⁡(G) be of degree not divisible by p, and let P∈Sylp⁢(G). If χ⁢(1) is odd, then there exists an element g∈NG⁢(P)/P′ such that o⁢(g)=fχ. In particular, the Feit number fχ divides |NG(P):P′|.

Let N◁G be finite groups. Let χ∈Irr⁡(G) and let P∈Sylp⁢(G) for some prime p. If χ⁢(1) is not divisible by p, then some constituent of χN is P-invariant and any two are NG⁢(P)-conjugate.

Proof.

Let θ∈Irr⁡(N) be under χ, and let Gθ be the stabilizer of θ in G. Since |G:Gθ| is not divisible by p, we have that Pg⊆Gθ for some g∈G. Hence φ=θg-1 is a P-invariant constituent of χN. Let τ∈Irr⁡(N) be P-invariant under χ. Then τ=φx for some x∈G by Clifford’s theorem. Hence P,Px⊆Gτ and there exists some t∈Gτ such that P=Px⁢t. Since φx⁢t=τt=τ, the result follows. ∎

Let G be a finite p-solvable group and let P∈Sylp⁢(G). Then restriction yields an injection from the set of p-special characters of G into the set of characters of P. In particular, if χ is p-special, then Q⁢(χ)=Q⁢(χP)⊆Q|G|p and fχ is a power of p.

Proof.

This statement is a particular case of [3, Proposition 6.1]. See also [3, Corollary 6.3]. ∎

Let G be a finite group, let q be a prime and let ζ be a primitive q-th root of unity. Suppose that G is q-solvable and χ∈Irr⁡(G) is q-special. If χ≠1, then ζ∈Q⁢(χ).

Proof.

Let Q be a Sylow q-subgroup of G. By Lemma 1.2, we have that ψ↦ψQ is an injection from the set of q-special characters of G into the set Irr⁡(Q). In particular, ℚ⁢(χ)=ℚ⁢(χQ). Of course, χQ≠1. Thus, we may assume that G is a q-group. We also may assume that χ is faithful by modding out by ker⁢(χ). Choose x∈𝐙⁢(G) of order q. We have that χ〈x〉=χ⁢(1)⁢λ, where λ∈Irr⁡(〈x〉) is faithful. Hence λ⁢(xi)=ζ for some integer i. In particular, ζ∈ℚ⁢(χ). ∎

Suppose that λ is a linear character of a finite group G, and let P∈Sylp⁢(G). Let NG⁢(P)⊆H≤G and let ν=λH. Then o⁢(λ)=o⁢(ν).

Proof.

If λ=1G, there is nothing to prove. We may assume λ is non-principal and hence G′<G. We have that P⊆P⁢G′◁G. By the Frattini argument, we have that G=G′⁢𝐍G⁢(P)=G′⁢H. Since G′⊆ker⁢(λ) and ker⁢(ν)=ker⁢(λ)∩H, the result follows. ∎

Let (G,K,L,θ,φ) be a character five, and suppose that K/L is a q-group for some odd prime q. Then there exist a character ψ of G such that K⊆ker⁢(ψ) and a subgroup U≤G such that

1. U⁢K=G and U∩K=L,

2. ψ⁢(g)≠0 for every g∈G, ψ(1)2=|K:L| and the determinantal order of ψ is a power of q,

3. if K⊆W≤G, then the equation ξW=ψW⁢ξ0 for characters ξ∈Irr⁡(W|θ) and ξ0∈Irr⁡(W∩U|φ) defines a one-to-one correspondence between these two sets,

4. if K⊆W≤G, then ξ∈Irr⁡(W|θ) and ξ0∈Irr⁡(W∩U|φ) correspond in the sense of (c) if and only if ξ0G=ψ¯W⁢ξ, where ψ¯ denotes the complex conjugate of ψ,

5. if K/L is elementary abelian, then ℚ⁢(ψ)⊆ℚq.

Proof.

For parts (a)–(c) see [9, Theorem 3.1]. Part (d) follows from [4, Corollary 9.2] (since the complement U provided by [9] is “good” not only for G/L but also for every W/L where K⊆W≤G). For part (e), by [4, Theorem 9.1] and [4, discussion at the end of p. 619], the values of the character ψ are ℚ-linear combinations of products of values of the bilinear multiplicative symplectic form ≪⋅,⋅≫φ:K×K→ℂ× associated to φ (defined in [4, beginning of Section 2]). The values of ≪⋅,⋅≫φ are values of linear characters of cyclic subgroups of K/L. Since K/L is q-elementary abelian, we do obtain that ℚ⁢(ψ)⊆F. ∎

Let p be a prime and let G be a finite solvable group. Let χ∈Irr⁡(G) be of degree not divisible by p, and let P∈Sylp⁢(G). If χ⁢(1) is odd, then there exists an element g∈NG⁢(P)/P′ such that o⁢(g)=fχ. In particular, the Feit number fχ divides |NG(P):P′|.

Proof.

By the Amit–Chillag theorem [1], we may assume that p divides |G|. We proceed by induction on |G|.

Let N◁G. If θ∈Irr⁡(N) is P-invariant and lies under χ, then we may assume that θ is G-invariant. Let ψ∈Irr⁡(Gθ|θ) be the Clifford correspondent of χ. By the character formula for induction, ℚ⁢(χ)⊆ℚ⁢(ψ) and χ(1)=|G:Gθ|ψ(1). Thus the character ψ satisfies the hypotheses of the theorem in Gθ and fχ divides fψ. If Gθ<G, then by induction there exists some g∈𝐍Gθ⁢(P)/P′≤𝐍G⁢(P)/P′ (notice that the P-invariance of θ implies P⊆Gθ) such that o⁢(g)=fψ. Hence, some power of g has order fχ and we may assume Gθ=G.

We claim that we may assume that χ is primitive. Otherwise, suppose that χ is induced from ψ∈Irr⁡(H) for some H<G. In particular, p does not divide |G:H| and so H contains some Sylow p-subgroup of G, which we may assume is P. Again by the character formula for induction, the degree ψ⁢(1) is an odd p′-number and fχ divides fψ. By induction there is g∈𝐍H⁢(P)/P′≤𝐍G⁢(P)/P′ such that o⁢(g)=fψ. Thus some power of g has order fχ, as claimed.

By [5, Theorem 2.6] the primitive character χ factorizes as a product

where the χq are q-special characters of the group G for distinct primes q. Let σ∈Gal⁢(ℚ|G|/ℚ⁢(χ)). Then

By using the uniqueness of the product of special characters (see [3, Proposition 7.2]), we conclude that χqσ=χq for every q. Hence fχq divides fχ for every q, and since the integers fχq are coprime also ∏qfχq divides fχ. Notice that ℚ(χ)⊆ℚ(χq:q)⊆ℚ∏qfχq by elementary Galois theory. This implies the equality

Now, consider K=𝐎p′,p⁢(G)<G. Notice that P⁢K=𝐎p⁢(G)◁G. By the Frattini argument G=P⁢K⁢𝐍G⁢(P)=K⁢𝐍G⁢(P). If K=1, then P◁G and we are done in this case. We may assume that K>1. Let K/L be a chief factor of G. Then K/L is an abelian p′-group. If H=𝐍G⁢(P)⁢L, then G=K⁢H and K∩H=L, by a standard group theoretical argument. Furthermore, all the complements of K in G are G-conjugate to H. Finally, notice that 𝐂K/L⁢(P)=1 using that H∩K=L.

We claim that for every q, there exists some q-special χq*∈Irr⁡(H) such that fχq* is equal to fχq and χq*⁢(1) is an odd p′-number.

If q∈{2,p}, then the character λ=χq is linear (because χ has odd p′-degree). Let λ*=λH. Then λ* is q-special (since λ is linear and q-special, this is straight-forward from the definition) and fλ*=fλ by Lemma 1.4.

Let q≠p be an odd prime and write η=χq. We work to find some character η*∈Irr⁡(H) of odd p′-degree with fη*=fη. By Lemma 1.1, let θ∈Irr⁡(K) be some P-invariant constituent of ηK and let φ∈Irr⁡(L) be some P-invariant constituent of ηL. By the second paragraph of the proof, we know that both θ and φ are G-invariant and hence φ lies under θ. By [8, Theorem 6.18] one of the following holds:

1. θL=∑i=1tφi, where the φi∈Irr⁡(L) are distinct and t=|K:L|,

2. θL∈Irr⁡(L), or

3. θL=e⁢φ, where φ∈Irr⁡(L) and e2=|K:L|.

Notice that the situation described in case (a) cannot occur here, because φ is G-invariant.

In the case described in (b), we have φ=θL∈Irr⁡(L). Then restriction defines a bijection between the set of irreducible characters of G lying over θ and the set of irreducible characters of H lying over φ (by [7, Corollary (4.2)]). Write ξ=ηH. By [7, Theorem A], we know that ξ is q-special. We claim that ℚ⁢(η)=ℚ⁢(ξ). Clearly, ℚ⁢(ξ)⊆ℚ⁢(η). If σ∈Gal⁢(ℚ⁢(η)/ℚ⁢(ξ)), then notice that φ is σ-invariant because ξL is a multiple of φ. Now, φ is P-invariant, and because 𝐂K/L⁢(P)=1, there is a unique P-invariant character over φ (by [8, Problem 13.10]). By uniqueness, we deduce that θσ=θ. Now, ησ lies over θ and restricts to ξ, so we deduce that ησ=η, by the uniqueness in the restriction. Thus ℚ⁢(η)=ℚ⁢(ξ). We write η*=ξ.

Finally, we consider the situation described in (c). Since θL is not irreducible, then |K:L| is not a q′-group, by [8, Corollary 11.29]. Hence K/L is q-elementary abelian and e is a power of q. By Theorem 1.5 (and using that all the complements of K/L in G/L are conjugate), there exists a (not necessarily irreducible) character ψ of G such that:

1. ψ contains K in its kernel, ψ⁢(g)≠0 for every g∈G, ψ⁢(1)=e and the determinantal order of ψ is a power of q,

2. if K⊆W≤G and ξ∈Irr⁡(W|θ), then ξW∩H=ψW∩H⁢ξ0 for a unique irreducible character ξ0 of W∩H,

3. the values of ψ lie on ℚq.

In particular, ηH=ψ⁢η0, so that η0∈Irr⁡(H|φ) (where we are viewing ψ as a character of H). We claim that η0 is q-special. First notice that η0⁢(1)=η⁢(1)/e is a power of q. Now, we want to show that whenever S is a subnormal subgroup of H, the irreducible constituents of (η0)S have determinantal order a power of q. Since (η0)L is a multiple of φ, which is q-special, we only need to control the irreducible constituents of (η0)S when L⊆S◁◁H, by using [3, Proposition 2.3]. We have that K⊆S⁢K◁◁G. Write

where the γi∈Irr⁡(S⁢K) are q-special because η is q-special and ai∈ℕ0. By using property (ii) of ψ, we have that ηS=ψS⁢(η0)S also decomposes as

Since ψ never vanishes on G, we conclude that (η0)S=a1⁢(γ1)0+⋯+ar⁢(γr)0. It suffices to see that o⁢((γi)0) is a power of q for every γi constituent of ηS⁢K. Just notice that

Since o⁢(ψ), o⁢(γi), γi⁢(1) and e are powers of q, we easily conclude that also the determinantal order of (γi)0 is a power of q. This proves that η0 is q-special. We claim that ℚ⁢(η)=ℚ⁢(η0) so that the two Feit numbers are the same. Let ζ be a primitive q-th root of unity and write F=ℚ⁢(ζ). Then the values of ψ lie in F. We next see that η and η0 are non-principal. This is obvious because θ and φ are fully ramified. Suppose that σ∈Gal⁢(ℚ|G|/F) stabilizes η. Then

Using that ψ is never zero, we conclude that η0σ=η0. Now, by part (d) of Theorem 1.5, we have that ξ and ξ0 correspond (as in part (c) of Theorem 1.5) if and only if (ξ0)G=ψ¯⁢ξ. Hence, if σ∈Gal⁢(ℚ|G|/F) and η0σ=η0, then

(since ℚ⁢(ψ¯) is also contained in F). This implies that ησ=η. By Galois theory, we have that F⁢(η)=F⁢(η0). By Lemma 1.3, this implies ℚ⁢(η)=ℚ⁢(η0). We set η*=η0. The claim follows.

Now, we define χ*=∏qχq* which has odd p′-degree. The character χ* is irreducible by [3, Proposition 7.2]. Also fχ*=∏qχq* as in the fourth paragraph of this proof. Hence

By the inductive hypothesis, there exists g∈𝐍H⁢(P)/P′≤𝐍G⁢(P)/P′ such that o⁢(g)=fχ* and we are done. ∎