Odd-degree Rational Irreducible Characters

Abstract

We study finite groups whose rational-valued irreducible characters are all of odd degrees. We conjecture that in such groups, all rational elements must be 2-elements.

1 Introduction

Let G be a finite group and let $\mathbb K$ be a field with $\mathbb Q\subseteq \mathbb K\subseteq \mathbb C$. We denote by $\operatorname {Irr}_{\mathbb K}(G)$ the set of complex irreducible characters χ ∈ Irr(G) with values in $\mathbb K$, and refer to any character in $\operatorname {Irr}_{\mathbb R}(G)$ as a real character, respectively to any character in $\operatorname {Irr}_{\mathbb Q}(G)$ as a rational character.

The Itô-Michler theorem for the prime p = 2 states that the degrees of all complex irreducible characters of a finite group G are even if and only if G has a normal abelian Sylow 2-subgroup. A version of this theorem for real characters was obtained in [5] and another refinement to strongly real characters was proved in [15]. In both versions, it was shown that if all real (or strongly real) irreducible characters of a finite group G have odd degrees, then G has a normal Sylow 2-subgroup. Clearly, these groups have no non-trivial real elements of odd order. Unfortunately, when restricted to rational characters, a similar conclusion fails. The simple groups L₂(3^2f+ 1), where f ≥ 1 is an integer, have exactly two rational irreducible characters which are the trivial character and the Steinberg character of degree 3^f.

However, Navarro and Sanus [16] managed to show that if all rational irreducible characters of G are linear, then G has a normal Sylow 2-subgroup. In this case, clearly, G has no non-trivial rational element of odd order. Note that all rational elements are real elements and since any power of rational elements remains rational, the two conditions G have no non-trivial rational element of odd order and all rational elements of G are 2-elements coincide. Navarro and Tiep show in [17] that if G has exactly two rational irreducible characters, then G has exactly two rational classes and since |G| is even, all non-trivial rational elements of G are involutions. These results seem to suggest the following.

Conjecture 1

Let G be a finite group. If every rational irreducible character of G has odd degree, then all rational elements of G are 2-elements.

In our first result, we will prove the following.

Theorem A

Let G be a finite group. Suppose that the simple group L₂(3^2f+ 1) is not involved in G for any integer f ≥ 1 and that all rational irreducible characters of G have odd degree. Then all rational elements of G are 2-elements.

In order to prove Theorem A, we need the following solvability result.

Theorem B

Let G be a finite group. Suppose that no simple group L₂(3^2f+ 1) with f ≥ 1 is involved in G and that all rational irreducible characters of G have odd degree. Then G is solvable.

The proof of Theorem B will depend on the following result on simple groups.

Theorem C

Let S be a normal non-abelian simple subgroup of a finite group G and C_G(S) = 1. If S is not isomorphic to L₂(3^2f+ 1) then there exists $\chi \in \operatorname {Irr}_{\mathbb Q}(G)$ of even degree not containing S in its kernel.

Apparently, the non-existence of rational elements of certain orders might have stronger impact on the group structures (see, e.g., [4, Theorem C] and Theorem 3.1).

As mentioned earlier, [17, Theorem A] states that a finite group G has exactly two rational irreducible characters if and only if it has two rational classes. The non-trivial rational class of such groups consists of involutions and hence our conjecture clearly holds in this case. Extending this one step further, it turns out that there is no finite group having exactly three rational irreducible characters whose degrees are all odd.

Theorem D

Let G be a finite group. Suppose that G has exactly three rational irreducible characters. Then G has a rational irreducible character of even degree.

In fact, it was conjectured by Navarro and Tiep that a finite group G has exactly three rational irreducible characters if and only if it has three rational conjugacy classes. One direction of this conjecture was proved by Rossi in [19]. For finite solvable groups G of even order, we can show that the number of odd degree rational irreducible characters of G is even (see Corollary 5.2). However, this is not true in general. The simple Janko group J₄ has exactly 13 rational, irreducible characters of odd degree.

Finally, we mention that the converse of our conjecture does not hold. As a counterexample, all rational elements of the dihedral group D₈ are of order 1 and 2 but D₈ has a rational irreducible character of degree 2. It would be an interesting problem to obtain a group-theoretical characterization of finite groups whose all rational irreducible characters have odd degree. Motivated by [16, Theorem A], where it is shown that if all rational irreducible characters of G are linear, then all rational irreducible characters of some Sylow 2-subgroup of G are linear, one might ask whether a similar conclusion holds under the hypothesis of our conjecture. Unfortunately, as pointed out to us by the reviewer, this is also not true. For example, one can take $G:=\textrm {SmallGroup(160, 234)}\cong {C_{2}^{4}} \rtimes (C_{5} \rtimes C_{2})$. Then all rational irreducible characters of G have odd degree but a Sylow 2-subgroup of G has rational irreducible characters of degree 2.

Our notation is standard. We follow [11] for the character theory of finite group and [2] for the notation of non-abelian simple groups.

2 Preliminaries

We collect some useful results on rational characters and rational elements in this section.

Lemma 2.1

Let G be a finite group and let N be a normal subgroup of G. If χ is a real irreducible character of G of odd degree and 𝜃 is an irreducible constituent of χ_N, the restriction of χ to N, then 𝜃 is real of odd degree.

Proof

Assume that χ ∈ Irr(G) is real with χ(1) odd. Let 𝜃 ∈ Irr(N) be a constituent of χ_N. By Clifford’s theorem, χ_N = e(𝜃₁ + 𝜃₂ + ⋯ + 𝜃_t), where all $\theta _{i}^{\prime }s$ are conjugate to 𝜃 = 𝜃₁ and integers e,t ≥ 1. Since χ(1) = et𝜃(1) is odd, t is odd. As χ is real, we have $\overline {(\chi _{N})}=(\overline {\chi })_{N}=\chi _{N}$, and so

$$ e\left( \overline{\theta_{1}}+\overline{\theta_{2}}+\cdots+\overline{\theta_{t}}\right)=e\left( \theta_{1}+\theta_{2}+\cdots+\theta_{t}\right). $$

It follows that the G-orbit of all irreducible constituents of χ_N is closed under taking complex conjugate. Since t is odd, $\overline {\theta _{j}}=\theta _{j}$ for some j with 1 ≤ j ≤ t. As 𝜃_j is G-conjugate to 𝜃, 𝜃 is real. □

Lemma 2.2

Let G be a finite group and let $N\unlhd G.$ Let 𝜃 ∈ Irr(N) be rational. Then there exists a rational character χ ∈ Irr(G|𝜃) if either |G/N| is odd or 𝜃(1) is odd and o(𝜃) = 1.

Proof

These follow from [17, Corollaries 2.2 and 2.4]. □

Lemma 2.3

Let S be a finite non-abelian simple group. Then all rational irreducible characters of S have odd degree if and only if S≅L₂(3^2f+ 1) for some integer f ≥ 1. Moreover, the only non-trivial rational irreducible character of L₂(3^2f+ 1) with f ≥ 1 is the Steinberg character of degree 3^2f+ 1.

Proof

The first statement is [5, Theorem 2.7] and the second follows from [17, Lemma 9.4]. □

Let G be a finite group. Recall that an element x ∈ G is rational (in G) if whenever 〈y〉 = 〈x〉, then y is G-conjugate to x. Also x ∈ G is real (in G) if x^g = x^− 1 for some g ∈ G. Clearly, every rational element is real. Moreover, if x ∈ G is an element of order 3, then x is real if and only if x is rational. We call a class x^G is rational if x is rational.

Lemma 2.4

Let G be a finite group and let $N\unlhd G.$

(a) If x ∈ G is rational, then xN is rational in G/N.(b) If g ∈ H ≤ G and g is rational in H, then g is rational in G.(c) Assume that x ∈ G has prime order p. Then, x is rational in G if and only if there exists a $p^{\prime }$-element g ∈ G such that x^g = x^t, where t (mod p) is any generator of $\mathbb Z_{p}^{\times }.$(d) If x ∈ G is rational, then every power of x is also rational; moreover, x_π is rational for every set of primes π.(e) If x ∈ G, $\gcd (o(x),|N|)=1$ and xN is rational in G/N, then x is rational in G.(f) If G/N has a rational element of prime order p, then G also has a rational element of order p.

Proof

These statements can be found in [17, Lemmas 5.1 and 5.2]. □

Lemma 2.5

Let S be a finite non-abelian simple group. Then, either S contains a rational element of order 3, or $S={}^{2}\mathrm {B}_{2}(2^{2f+1})$ and S contains a rational element of order 5, or S≅L₂(3^2f+ 1), where f ≥ 1 is an integer.

Proof

This is [17, Theorem 11.1]. □

3 Finite Groups with No Even Rational Character Degrees

Let G be a finite group. If $U\unlhd V$ are subgroups of G, then we call V/U a section of G. We say that a finite group T is involved in G if T is isomorphic to some section of G and G is said to be T-free if none of the section of G is isomorphic to T.

We first prove Theorem B, assuming Theorem C.

Proof of Theorem B

Assume that every rational, irreducible character of G has odd degree and that L₂(3^2f+ 1) is not involved in G for any integer f ≥ 1. We prove by induction on |G| that G is solvable. If $1<N\unlhd G$, then $\operatorname {Irr}_{\mathbb Q}(G/N)\subseteq \operatorname {Irr}_{\mathbb Q}(G)$ and L₂(3^2f+ 1) is not involved in G/N, by induction, G/N is solvable. It follows that G has a unique minimal normal subgroup, say M. If M is solvable, then since G/M is solvable by the claim above, G is solvable. Thus, we assume that M is non-solvable.

Write $M=S_{1}\times S_{2}\times \dots \times S_{n},$ where each S_i is non-abelian simple. Let S = S₁, H = N_G(S) and C = C_G(S). Since L₂(3^2f+ 1) is not involved in G, S≇L₂(3^2f+ 1) for all integers f ≥ 1. Let $\overline {H}=H/C$. Then, $\overline {S}\unlhd \overline {H}$ and $\mathbf {C}_{\overline {H}}(\overline {S})=1$. By Theorem C, there exists a rational, irreducible character $\delta \in \operatorname {Irr}_{\mathbb Q}(\overline {H})$ of even degree such that $[\delta _{\overline {S}},1_{\overline {S}}]=0$. Inflate δ to H, we still have that [δ_S,1_S] = 0. Hence, δ_S has an irreducible constituent 1≠𝜃 ∈ Irr(S). Let $\phi =\theta \times 1_{S}\times \dots \times 1_{S}\in \operatorname {Irr}(M)$. Let I be the inertia group of ϕ in G. Observe that δ lies over ϕ and I ≤ H. Let ψ be the Clifford correspondence of δ over ϕ. Then, ψ^H = δ and ψ^G ∈ Irr(G). Thus, δ^G = ψ^G ∈ Irr(G) and since δ is rational, $\psi ^{G}\in \operatorname {Irr}_{\mathbb Q}(G)$. Furthermore, δ(1) is even and hence δ^G is a rational irreducible character of even degree of G, which is a contradiction. □

Using Theorem B, we can now prove Theorem A. Recall that for a prime p and a finite p-solvable group G, B_p(G) is a canonical subset of Irr(G) with values in $\mathbb Q_{|G|_{p}}$.

Proof of Theorem A

We proceed by induction on |G|. By Theorem B, we know that G is solvable. If $1<N\unlhd G,$ then $\operatorname {Irr}_{\mathbb Q}(G/N)\subseteq \operatorname {Irr}_{\mathbb Q}(G)$, by induction G/N has no rational element of odd order > 1.

Assume first that O₂(G) > 1. Then, G/O₂(G) has no non-trivial rational element of odd order. Assume by contradiction that G has a non-trivial rational element x of odd order. By Lemma 2.4 (a), xO₂(G) is a non-trivial rational element of G/O₂(G) of odd order, which is a contradiction. Thus, we assume that O₂(G) = 1. It follows that $\mathbf {O}_{2^{\prime }}(G)>1$ since G is solvable.

Let N be a minimal normal subgroup of G. Then, N is an elementary abelian p-group for some odd prime p. Note that G/N has no rational element of odd prime order. Again, assume that G has a rational element x of odd prime order. Since G/N has no rational element of odd order > 1, we deduce that x ∈ N and thus o(x) = p > 2.

We now use the argument as in the proof of [17, Theorem 7.2] to produce an irreducible rational character of G which does not contain N in its kernel. By Lemma 2.4 (c), there exists a $p^{\prime }$-element g ∈ G such that x^g = x^t, where t (mod p) is a generator for the multiplicative group $\mathbb Z_{p}^{\times }.$ Let $\sigma \in \operatorname {Gal}(\mathbb Q_{|G|}/\mathbb Q)$ be an element of order p − 1 fixing $p^{\prime }$-roots of unity and ξ^σ = ξ^t, where ξ is a primitive p-root of unity. By [17, Corollary 6.4], there exists 1_N≠ψ ∈B_p(N) such that ψ^σ = ψ^g and $\mathbb Q(\psi )\subseteq \mathbb Q_{p}$. By [17, Corollary 6.3], there exists a rational irreducible character χ of G lying over ψ.

From the hypothesis of the theorem, we know that χ(1) is odd. By Lemma 2.1, the constituent ψ of χ_N is real as χ is real. However, as |N| is odd, the only real irreducible character of N is the trivial character 1_N which implies that ψ = 1_N, which is a contradiction. □

Observe that if G is a finite group and x ∈ G is an element of order 3, then x is a real element of G if and only if x is a rational element of G as the only generators of the cyclic group 〈x〉 are x and x² = x^− 1.

Theorem 3.1

Let G be a finite group. Assume that L₂(3^2f+ 1) is not involved in G for any integers f ≥ 1. If G has no rational element of order 3 and 5, then G is solvable.

Proof

We proceed by induction on the order of G. From the definition, we observe that every subgroup and quotient of G are L₂(3^2f+ 1)-free for all integers f ≥ 1. Let π = {3,5}.

Assume first that G has a non-trivial proper normal subgroup N. By Lemma 2.4 (b), N has no rational element of order p with p ∈ π. If G/N has a rational element of order p for some p ∈ π, then by Lemma 2.4 (f), G has a rational element of order p, which is impossible. Thus, G/N has no rational element of order p for any p ∈ π. Since |G/N| < |G| and |N| < |G|, by induction, both N and G/N are solvable and hence G is solvable. Therefore, we can assume that G is a finite simple group.

If G is abelian, then G is solvable. Thus, we can assume that G is a non-abelian simple group. Since G is not isomorphic to L₂(3^2f+ 1) for any integer f ≥ 1, by Lemma 2.5, G always contains a rational element of order 3 or 5. This contradiction proves that G is solvable. □

We recall the following result which describes the structure of finite groups having no rational element of order 3 (and which also implies Theorem 3.1).

Theorem 3.2

[4, Theorem C] Let G be a finite non-solvable group. Assume that G has no rational element of order 3. Let $L=\mathbf {O}^{2^{\prime }}(G)$ and $M=\mathbf {O}_{3^{\prime }}(L)$. Then, L/M = S₁ × S₂ ×⋯ × S_n, where n ≥ 1 is an integer and $S_{i}\cong \operatorname {L}_{2}(3^{2f_{i}+1}),f_{i}\ge 1$ for all 1 ≤ i ≤ n.

4 Proof of Theorem C

This section is devoted to proving Theorem C.

4.1 Alternating Groups and Sporadic Groups

Suppose S = A_n, where n ≥ 7. Then, S ≤ G ≤ Aut(S)≅S_n. Consider the irreducible characters α = χ^(n− 2,2) and $\beta = \chi ^{(n-2,1^{2})}$ of S_n, labeled by partitions (n − 2,2) and (n − 2,1²), and of degree n(n − 3)/2 and n(n − 3)/2 + 1. Since both partitions are non-associate, α and β restrict to rational, irreducible characters of G, and one of them has even degree, yielding the desired character χ.

If S = A₅, A₆, ${}^{2}F_{4}(2)^{\prime }$, or one of the 26 sporadic simple groups, then one can check directly using [2] that G always contains a rational irreducible character of even degree.

4.2 Lie-type Groups in Characteristic 2

Suppose S is a Lie-type group in characteristic 2. By [8], the Steinberg character St of S extends to a rational irreducible character of even degree of G.

From now on, we will assume that S is a simple group of Lie type, defined over a field $\mathbb F_{q}$ of odd characteristic p, q = p^f. Our proof is largely based on [18, Lemma 5.1], which we reformulate for the case of rational characters.

Lemma 4.1

Suppose that $G\lhd A$. Let ρ be a rational character of A, not necessarily irreducible. Assume that ρ|_G contains an A-invariant rational irreducible constituent α such that [α,ρ|_G] = 1. Then, there is a rational χ ∈ Irr(A) such that χ|_G = α.

4.3 The Case S = L_n(q), n ≥ 2

In the subsequent treatment of L_n(q), it is convenient to adopt the labeling of irreducible $\mathbb C \operatorname {GL}_{n}(q)$-modules as given in [12], which uses Harish-Chandra induction ∘. Each such a module is labeled as S(s₁,λ₁) ∘… ∘ S(s_m,λ_m), where $s_{i} \in \overline {\mathbb F}_{q}^{\times }$ has degree d_i (over $\mathbb F_{q}$), λ_i is a partition of k_i, and ${\sum }^{m}_{i=1}k_{i}d_{i} = n$ (cf. [12, 13]).

Lemma 4.2

[18, Lemma 5.4] Let B = GL₂(q) and α ∈ Irr(B) be of form S(s,(1)) ∘ S(s^− 1,(1)) for some s of order 4 and degree 1, or S(t,(1)) for some t of degree 2 and order 4. Then, α is rational and invariant under any field automorphism of B.

We view S as L/Z(L), where $L = \operatorname {SL}_{n}(q) \lhd H = \operatorname {GL}_{n}(q)$. Consider the natural module $\langle e_{1}, {\ldots } e_{n}\rangle _{\mathbb F_{q}}$ for H, the subgroup

$$T = \operatorname{Stab}_{H}(\langle e_{1} \rangle_{\mathbb F_{q}}, \langle e_{2}, {\ldots} ,e_{n}\rangle_{\mathbb F_{q}}) \cong \operatorname{GL}_{1}(q) \times \operatorname{GL}_{n-1}(q),$$

and the induced character ρ_n = (1_T)^H. Then, ρ_n is a rational character of degree q^n− 1(qⁿ − 1)/(q − 1). Since T > Z(H), ρ_n can be viewed as a character of $\bar {H} = \operatorname {PGL}_{n}(q)$. Recall that A = Aut(L_n(q)) is a semidirect product $\bar {H} \rtimes F$, where F is generated by a field automorphism σ, and also the transpose-inverse τ if n > 2. We can define σ and τ such that they stabilize T. It follows that ρ_n extends to the rational A-character $(1_{\bar {T} \rtimes F})^{A}$, where $\bar {T} = \mathbf Z(G)T/\mathbf Z(G)$.

If λ ⊩ n, then let χ^λ denote the unipotent character of G labeled by λ (cf. [1]). As shown in the proof of [18, Proposition 5.5], if n ≥ 2 then ρ_n has the following decomposition into distinct irreducible constituents:

$$ \begin{array}{@{}rcl@{}} \rho_{n} & = & \chi^{(n)} + 2\chi^{(n-1,1)} + \chi^{(n-2,1^{2})} + S(-1,(1^{2})) \circ S(1,(n-2))\\ & & + \sum\limits_{a \in \mathbb F_{q}^{\times},~a \neq \pm 1}S(a,(1)) \circ S(a^{-1},(1)) \circ S(1,(n-2))\\ & & + \sum\limits_{b \in \mathbb F_{q^{2}}^{\times},~b^{q+1} = 1,~b \neq \pm 1}S(b,(1)) \circ S(1,(n-2)), \end{array} $$

(1)

where in ${\sum }_{a}$, there is one summands for the pair {a,a^− 1}, and similarly for ${\sum }_{b}$; also, the summand $\chi ^{(n-2,1^{2})}$ occurs only for n ≥ 3.

First consider the case n = 2 and $2 \nmid f$, whence $\operatorname {Aut}(S) = K \rtimes C_{2}$ with $K = S \rtimes C_{f}$. By assumption, p≠ 3. Hence, by Lemma 2.3, S admits a rational irreducible character 𝜃 of even degree, and 𝜃 extends to a rational character of H by Lemma 2.2. In particular, we are done if G ≤ K. Suppose G≦̸K. Note that $\tilde {K}:=\operatorname {SL}_{2}(q) \rtimes C_{f} \leq \operatorname {Sp}_{2nf}(p)$, and Sp_2nf(p) admits irreducible Weil characters ξ₁,ξ₂ of degree (qⁿ + 1)/2 and η₁,η₂ of degree (qⁿ − 1)/2, all remaining irreducible upon restriction to $\tilde H$ and SL₂(q). Moreover, ξ₁ + ξ₂ and η₁ + η₂ are rational-valued. Furthermore, the outer diagonal automorphisms of S, and so any element of $G \smallsetminus K$, fuse (ξ₁)|_S with (ξ₂)|_S, and (η₁)|_S with (η₂)|_S. Choosing the one, say η₁, that is trivial at $\mathbf Z(\tilde K)$, we obtain an irreducible character ψ of G ∩ K. Setting χ := ψ^G, we see that χ = 0 on $G \smallsetminus K$, and χ(g) = (η₁ + η₂)(g) for g ∈ G ∩ K. Thus, χ is an even-degree rational irreducible character of G.

Assume now that n = 2 and 2|f. Then, 8|(q − 1), and we can find $a \in \mathbb F_{q}^{\times }$ of order 4. By Lemma 4.2, α := S(a,(1)) ∘ S(a^− 1,(1)) is rational and σ-invariant, of even degree q + 1. It is easy to check α, viewed as PGL₂-character, is irreducible upon restriction to S. By Lemma 4.1, α extends to a rational character φ of A, and now we can take χ := φ|_G.

Next we consider the case n ≥ 3. If q ≡ 1 (mod 4), choose $a \in \mathbb F_{q}^{\times }$ of order 4 and take α := S(a,(1)) ∘ S(a^− 1,(1)) ∘ S(1,(n − 2)) of even degree

$$(q+1)\frac{(q^{n}-1)(q^{n-1}-1)}{(q^{2}-1)(q-1)}.$$

If q ≡ 3 (mod 4), choose $b \in \mathbb F_{q^{2}}^{\times }$ of order 4 and take α := S(b,(1)) ∘ S(1,(n − 2)) of even degree

$$(q-1)\frac{(q^{n}-1)(q^{n-1}-1)}{(q^{2}-1)(q-1)}.$$

Arguing as in the proof of [18, Proposition 5.5], using [12, (7.33)] and [13, Lemmas 3.2, 4.1], we see that α is A-invariant, rational-valued, and irreducible upon restriction to S. By Lemma 4.1, α extends to a rational character φ of A, and now we can take χ := φ|_G.

4.4 The Case S = U₃(q), q ≥ 3

We will view S = L/Z(L) with L = SU₃(q), and use the notation for irreducible characters of L as given in [9]. Also, recall that $\operatorname {Aut}(S) = H \rtimes \langle \sigma \rangle$, with H = PGU₃(q), and σ is an outer automorphism of order 2f induced by the field automorphism x↦x^p. Note that S has a unique (unipotent) character of even degree q(q − 1), so rational. Hence, we are done by Lemma 2.2 if $2 \nmid |G/S|$. It remains to consider the case |G/S| is even.

First assume that q ≡ 3(mod 4); in particular, f is odd. Then, L has a pair of Weil characters $\chi ^{(u)}_{q^{2}-q+1}$ of degree q² − q + 1, with u = (q + 1)/4 and 3(q + 1)/4, which are dual to each other, and fused by σ^2f, and trivial at Z(L). In fact, it is straightforward to check that the semisimple characters of GU₃(q) labeled by the elements diag(b,b,− 1) with $b \in \mathbb F_{q^{2}}^{\times }$ of order 4 are trivial at Z(GU₃(q)), so can be viewed as H-characters 𝜃 and $\overline \theta$, restrict to the previous two Weil characters of S, dual to each other, with $\mathbb Q(\sqrt {-1})$ as field of values, and fused by σ. It follows that $\psi :=\theta ^{H \rtimes \langle \sigma ^{f} \rangle }$ is a rational irreducible character of $H \rtimes \langle \sigma ^{f} \rangle$, which is of odd index f in Aut(S). By Lemma 2.2, ψ extends to a rational character φ of Aut(S). The construction of φ shows that

$$\varphi|_{S} = \chi^{((q+1)/4)}_{q^{2}-q+1} + \chi^{(3(q+1)/4)}_{q^{2}-q+1};$$

hence, the inertia subgroup of $\chi ^{((q+1)/4)}_{q^{2}-q+1}$ in Aut(S) is precisely $H \rtimes \langle \sigma ^{f} \rangle$. As |G/S| is even, $\chi ^{((q+1)/4)}_{q^{2}-q+1}$ is not G-invariant. Hence, we can take χ := φ|_G.

Assume now that q ≡ 1(mod4). Then, L has a rational character $\chi ^{((q^{2}-1)/4)}_{q^{3}+1}$ of degree q³ + 1, which is trivial at Z(L). This character extends to the rational character $\alpha :=\chi ^{(0,(q^{2}-1)/4)}_{q^{3}+1}$ of M := GU₃(q), in the notation of [7]. Direct calculations using the character table of GU₂(q), also given in [7], show that [α|_N,1_N] = 1, where N := GU₁(q) × GU₂(q) embedded naturally in M. It follows that α is a multiplicity-one constituent of (1_N)^M. Note that σ can be defined to fix both N and M, so (1_N)^M extends to the permutation character $(1_{N \rtimes \langle \sigma \rangle })^{M \rtimes \langle \sigma \rangle }$, and moreover α is σ-invariant. By Lemma 4.1, α extends to a rational irreducible character ψ of $M \rtimes \langle \sigma \rangle$. Note that α is trivial at Z(M), so, after modding out by Z(M), ψ yields a rational irreducible character φ of $(M \rtimes \langle \sigma \rangle )/\mathbf Z(M) \cong \operatorname {Aut}(S)$, which is irreducible over S. Now we can take χ := φ|_G.

4.5 Other Classical Groups

For the remaining simple classical groups $S \not \cong \operatorname {P{\Omega }}^{+}_{8}(q)$, we can follow the proof of [5, Theorem 2.1], which produces two irreducible constituents α,β of a rank 3 permutation character of S (see, e.g., [20]), one of even degree and another of odd degree, which extend to rational-valued irreducible characters of Aut(S). One of this extensions, say φ, has even degree, so we can take χ := φ|_G.

Suppose $S = \operatorname {P{\Omega }}^{+}_{8}(q)$. We consider the parabolic subgroups S_i with i = 1,2 in the notation of [3, Lemma (6.4)], and let $\pi _{i}:= (1_{S_{i}})^{S}$. By [3, Lemma (6.4)], π₁ and π₂ are both multiplicity-free, [π₁,π₁] = 3, [π₂,π₂] = 6, and π₂ contains π₁. It is well-known (see, e.g., [20, Table 1]) that the rank 3 permutation character π₁ = 1_G + α + β, with α(1) = q(q² + 1)² and β(1) = q²(q⁴ + q² + 1). It follows that α is a multiplicity-one constituent of π₂, and it is well-known that α is Aut(S)-invariant and rational (as the unique unipotent character of this degree). Since S₂ corresponds to the branching node of the Dynkin diagram D₄ of S, π₂ extends to a permutation character of Aut(S). Hence, α extends to a rational character of G ≤ Aut(S) by Lemma 4.1.

4.6 Exceptional Groups

Suppose that S =²G₂(q) with q = 3^f ≥ 27. Here |Aut(S)/S| = f is odd. By Lemma 2.3, S has an even-degree rational character 𝜃, which then extends to a rational character of G by Lemma 2.2.

Suppose that S = G₂(q) with q ≥ 3, or ³D₄(q). Let B denote the Borel subgroup of G. By [3, Proposition (7.22) (iv)], (1_B)^G is the sum of 5 distinct irreducible characters, among which one constituent, say α, is the unique character ϕ_2,1 (in the notation of [1, Section 13.9]) of S of even degree (1/6)q(q + 1)²(q² + q + 1), respectively (1/2)q³(q³ + 1)², hence rational and Aut(S)-invariant. As B can be chosen to be invariant under (suitable) outer automorphisms of S, α extends to a rational character of G ≤ Aut(S) by Lemma 4.1.

Suppose that S = E₆(q) or ²E₆(q). The proof of [5, Theorem 2.1] shows that S has a rank 5 permutation character, which is the sum of 5 distinct irreducible characters, among which three are of even degree, and all extendible to rational characters of Aut(S).

Suppose that S = F₄(q), respectively E₈(q). As shown in [3, Proposition (4.2)], S has a parabolic subgroup S_i, with i = 1, respectively 8, such that the permutation character $(1_{S_{i}})^{S}$ is the sum of 5 distinct irreducible characters, among which one constituent, say α, is the so-called reflection character, of even degree

$$\frac{1}{2}q(q^{3}+1)^{2}(q^{4}+1), \text{ respectively }q(q^{10}+1)\frac{q^{24}-1}{q^{6}-1}.$$

Now, α is Aut(S)-invariant (as it is the unique unipotent character of this degree), and S_i can be chosen to be invariant under (suitable) outer automorphisms of S. Hence, α extends to a rational character of G ≤ Aut(S) by Lemma 4.1.

Finally, suppose that S = E₇(q). We consider the parabolic subgroups S₁ and S₇, in the notation of [3, Propositions (4.3) and (5.2)]. By these propositions of [3], [π₇,π₇] = 4 and [π₁,π₇] = 3, for $\pi _{i} := (1_{S_{i}})^{S}$, and

$$\pi_{1} = 1_{S} + \phi_{7,1} + \chi_{1} +\chi_{2} + \chi_{3},$$

a sum of 5 distinct irreducible characters, with ϕ_7,1 (in the notation of [1, Section 13.9]) being the reflection character, but now of odd degree. Next,

$$\pi_{7}(1) = (q^{5}+1)(q^{9}+1)(q^{14}-1)/(q-1),$$

and π₇ is the sum of 4 unipotent characters of the principal series. Checking the degrees of the latter as given in [1, Section 13.9], we see that π₇ = 1_S + ϕ_7,1 + ϕ_27,2 + ϕ_21,3. Again using the unipotent degrees listed in [1, Section 13.9] and [π₁,π₇] = 3, we obtain

$$\pi_{1}=1_{S} + \phi_{7,1} + \phi_{27,2} + \phi_{35,4}+ \phi_{56,3},$$

with α := ϕ_56,3 rational, Aut(S)-invariant, and of even degree. Since S₁ can be chosen to be invariant under (suitable) outer automorphisms of S, α extends to a rational character of G ≤ Aut(S) by Lemma 4.1, completing the proof of Theorem C.

We will need the following result for the proof of Theorem D.

Lemma 4.3

Let G be a finite group and let $S\unlhd G$ with C_G(S) = 1. Assume that S≅L₂(3^2f+ 1) for some integer f ≥ 1. If $|\operatorname {Irr}_{\mathbb Q}(G)|\ge 3$, then G has a rational irreducible character of even degree.

Proof

Assume that $|\operatorname {Irr}_{\mathbb Q}(G)|\ge 3$. Suppose by contradiction that all rational irreducible characters of G have odd degree. We have that $S\unlhd G\leq \operatorname {Aut}(S)$. Note that Out(S) is cyclic of order 2(2f + 1). Assume that |G/S| is even. Then, G has a normal subgroup G₁ such that G₁≅PGL₂(3^2f+ 1). In this case, G₁ has a rational, irreducible character α of degree q − 1 (lying over the characters labeled by η₁,η₂ as in [6, Theorem 38.1]). As |G/G₁| is odd, there exists $\chi \in \operatorname {Irr}_{\mathbb Q}(G)$ lying over α and has even degree.

Assume that G/S is odd. Let $\chi \in \operatorname {Irr}_{\mathbb Q}(G)$ be such that χ is non-trivial and is not the extension of the Steinberg character St_S of S to G. Clearly [χ,1_S] = 0 as G/S is of odd order. Let 1_S≠𝜃 ∈ Irr(S) be an irreducible constituent of χ_S. By Lemma 2.1, $\theta \in \operatorname {Irr}_{\mathbb R}(S)$. Since 𝜃∉{1_S,St_S}, by [6, Theorem 38.1] 𝜃(1) ∈{3^2f+ 1 ± 1} and so 𝜃(1) is even and hence χ(1) is even. This completes the proof. □

5 The Number of Rational Irreducible Characters of Odd Degrees

We first record the following result which should be well-known.

Lemma 5.1

Let G be a finite group of even order. Then, the number of irreducible real characters of G of odd degree must be even.

Proof

Let G be a finite group of even order. Clearly, if χ is not real, then $\chi \neq \overline {\chi }$. Since $|G|={\sum }_{\chi \in \operatorname {Irr}(G)}\chi (1)^{2}$ is even, we deduce easily that the number of real characters in Irr(G) of odd degree must be even. □

The next result will be needed in the proof of Theorem D.

Corollary 5.2

Let G be a finite solvable group of even order. Then, the number of rational irreducible characters of G of odd degree must be even.

Proof

By the main result in [10], every real irreducible character of odd degree of G is rational. Hence, the number of rational irreducible characters of G of odd degree coincides with the number of irreducible real characters of G of odd degree. Now, the corollary follows from the previous lemma. □

Proof

Let G be a counterexample to the theorem with minimal order. Then, $|\operatorname {Irr}_{\mathbb Q}(G)|=3$ and all rational irreducible characters of G have odd degree. Clearly |G| is even and so by Corollary 5.2, G is non-solvable. It follows from Lemma 2.1 that $\mathbf {O}_{2^{\prime }}(G)$ lies in the kernel of all rational irreducible characters of odd degrees of G. Thus, $|\operatorname {Irr}_{\mathbb Q}(G/\mathbf {O}_{2^{\prime }}(G))|=3$. By the minimality of |G|, we can assume that $\mathbf {O}_{2^{\prime }}(G)=1$.

Let N be a minimal normal subgroup of G. Since $\operatorname {Irr}_{\mathbb Q}(G/N)\subseteq \operatorname {Irr}_{\mathbb Q}(G)$ and |G/N| < |G|, by the minimality of G, $|\operatorname {Irr}_{\mathbb Q}(G/N)|\leq 2.$ We consider the following cases.

(a) N is non-solvable. Write $N=S_{1}\times S_{2}\times \dots \times S_{n}$, where each S_i is conjugate in G to S = S₁, where S is a non-abelian simple group. By Theorem B and its proof, we deduce that S≅L₂(3^2f+ 1) for some integer f ≥ 1. Let 𝜃 be the Steinberg character of S. We know that 𝜃 is rational and o(𝜃) = 1 since N is perfect.

Assume that n ≥ 3. Let ψ₁ = 𝜃 × 1 ×⋯ × 1, ψ₂ = 𝜃 × 𝜃 ×⋯ × 1, and ψ₃ = 𝜃 × 𝜃 ×⋯ × 𝜃. Then, ψ_i ∈ Irr(N) are all rational irreducible characters of odd degree and o(ψ_i) = 1 for all i. By Lemma 2.2, for each i, there exists $\chi _{i}\in \operatorname {Irr}_{\mathbb Q}(G)$ lying above ψ_i. Since ψ_i,1 ≤ i ≤ 3, lie in different G-orbits of irreducible characters of N, all $\chi _{i}^{\prime }s$ are pairwise distinct and thus $|\operatorname {Irr}_{\mathbb Q}(G)|\ge 4,$ a contradiction.

Assume that n = 2. Then, $\psi =\theta \times 1\in \operatorname {Irr}_{\mathbb R}(N)$ is rational, irreducible character of odd degree > 1. Let $\chi \in \operatorname {Irr}_{\mathbb Q}(G)$ be lying over ψ. Observe that the inertia group of ψ lies inside N_G(S₁) and since |G : N_G(S₁)| = 2, from Clifford’s theory, χ(1) is even, which is a contradiction.

Assume that n = 1. Then, N = S≅L₂(3^2f+ 1). By a result in [8], 𝜃 extends to $\chi \in \operatorname {Irr}_{\mathbb Q}(G)$. If $|\operatorname {Irr}_{\mathbb Q}(G/S)|=2$, then by Gallagher’s theorem, G would have at least 4 distinct rational irreducible characters. Hence, G/S is of odd order. Let C = C_G(S). Then, |C| is odd and so $C\subseteq \mathbf {O}_{2^{\prime }}(G)=1$. Thus, $S\unlhd G$ and C_G(S) = 1. By Lemma 4.3, G has a rational irreducible character of even degree, which is a contradiction.

(b) Assume N is solvable. Since $\mathbf {O}_{2^{\prime }}(G)=1$, N is an elementary abelian subgroup of order 2ⁿ for some integer n ≥ 1. Moreover G/N is non-solvable and $|\operatorname {Irr}_{\mathbb Q}(G/N)|=2$. By [17, Theorem 10.2], there exist normal subgroups N ≤ K ≤ L such that $K/N=\mathbf {O}_{2^{\prime }}(G/N)$, $L/N=\mathbf {O}^{2^{\prime }}(G/N)$ is perfect and L/K≅S = L₂(3^2f+ 1) for some integer f ≥ 1. Write $\operatorname {Irr}_{\mathbb Q}(G/N)=\{1,\chi \}$, where χ is an extension of the Steinberg character of L/K which is rational of degree 3^2f+ 1. Let ψ be the remaining rational irreducible character of G. Note that [ψ_N,1_N] = 1. Since O₂(G/N) = O₂(L/N) = 1, N = O₂(G) is a unique minimal normal subgroup of G.

Since $N\cap L^{\prime }\unlhd G$ and N is a unique minimal normal subgroup of G, $N\leq L^{\prime }$ which implies that $L=L^{\prime }$ is perfect. Since G/L is of odd order, by Lemma 2.2, each rational irreducible character of L lies under some rational irreducible character of G and thus every rational irreducible character of L has odd degree.

Assume first that G has a component V, that is, $V=V^{\prime }$ is perfect and V/Z(V ) is a non-abelian simple group and V is subnormal in G. Then, the layer E(G) of G, which is the normal subgroup of G generated by all components of G, is non-trivial. We know that E(G)/Z(E(G)) is a direct product of non-abelian simple groups. It follows that E(G)/Z(E(G))≅L₂(3^2f+ 1). If Z(E(G)) > 1, then E(G)≅SL₂(3^2f+ 1). In this case, E(G) has a rational irreducible character μ of degree q − 1 (labelled by 𝜃_j with j = (q + 1)/4 as in [6, Theorem 38.1]) and o(μ) = 1 (since E(G) is perfect). By Lemma 2.2, G has a rational irreducible character lying over μ and so this character has even degree, a contradiction. Thus, Z(E(G)) = 1 and so E(G) is a minimal normal subgroup of G, contradicting the uniqueness of N. Hence, $\mathbf {F}^{*}(G)=\mathbf {F}(G)\mathbf {E}(G)=\mathbf {O}_{2}(G)=N$, where F^∗(G) is the generalized Fitting subgroup of G. By Bender’s theorem, we have C_G(N) = N.

Assume that K > N. Note that N is a normal Sylow 2-subgroup of K. It follows that O²(K) = K since K is not a 2-group and N is the unique minimal normal subgroup of G. Let $1\neq \lambda \in \operatorname {Irr}_{\mathbb Q}(K)$. Then, λ(1) is odd and o(λ) = 1. By Lemma 2.2, $\psi \in \operatorname {Irr}_{\mathbb Q}(G)$ lies over λ. Thus, all non-trivial rational irreducible characters of K are G-conjugate. Let V = Irr(N)≅N. Then, G/N acts transitively on V −{1_N} (note that C_G(N) = C_G(V ) = N). Thus, the semidirect product $V \rtimes G/N$ is a doubly transitive permutation group. However, this cannot occur by Hering’s theorem. (See [14, Appendix 1]).

Assume that K = N. As C_G(N) = N, C_L(N) = N. So V = Irr(N) is a non-trivial GF(2)-module for S. By [17, Theorem 10.1], there exists 1≠λ ∈ V such that T/N is of odd order, where T is the inertia group of λ in L. Let ν ∈ Irr(T) be the canonical extension of λ. Then, ν^L is a rational irreducible character of L of even degree as |L : T| is even, which is a contradiction. □