Generalized characters whose values on non-identity elements are

roots of unity Geoffrey R. Robinson, Institute of Mathematics, University of Aberdeen, Aberdeen AB24 3UE

Introduction: Let H be a finite group. We are concerned here with the question of when the values taken by a generalized character θ of H on non-identity elements are all roots of unity. By elementary Galois theory and algebraic number theory, this is equivalent to the requirement that |θ(h)|^{2} = 1 for all h ∈ H^{#}. Note that the generalized characters to be considered form an Abelian group under pointwise multiplication, which we done by U(H). We will completely classify all such generalized characters θ as long as the Sylow 2-subgroup of H does not have a cyclic subgroup of index 2 (in particular, this allows the possibility that H has odd order). It will turn out that (even if a Sylow 2subgroup of H does have a cyclic subgroup of index 2), then θ always factors as the product of a linear character λ of H and a generalized character ψ such that ψ(h) ∈ {1, −1} for all h ∈ H^{#}. When H = O^{2}(H), this implies that U(H) is the direct product of H/H′ with an elementary Abelian 2-group. The main theorem of this note shows that (when H = O^{2}(H) and a Sylow 2-subgroup of H does not have a cyclic subgroup of index 2), the rank of the elementary Abelian 2-group can be calculated directly from the character table of H.

One of the motivations for our work is to attempt a better understanding of some aspects of Glauberman correspondence. Glauberman correspondence is concerned with the irreducible characters of a semi-direct product L = GA, where A is group of automorphism of the finite group G with gcd(|G|, |A|) = 1. If χ is an A-stable irreducible character of G, then there is a unique irreducible extension χ^{(A)} of χ to an irreducible character of L such that Res_{A}^{L} is afforded by a unimodular representation. Any extension of χ to an irreducible character of L differs from χ^{(A)} only by multiplication by a linear character with G in its kernel. For each a ∈ A^{#}, we have Res_{G⟨a⟩}^{L} = χ^{⟨a⟩}. Moreover, there is a corresponding irreducible character, μ_{⟨a⟩} of C_{G}(a), known as the Glauberman correspondent of χ, and a sign ε_{⟨a⟩}, such that χ^{(A)}(ag) = ε_{⟨a⟩}μ(g) for each g ∈ C_{G}(a). It would be desirable to understand the relationship between the signs ε_{⟨a⟩} which appear as a ranges through elements of A. If each μ_{⟨a⟩} is linear, then the values taken by Res_{A}^{L}(χ) on A^{#} are all roots of unity, so that our main theorem below may be applied.

As with many questions relating to Glauberman correspondence, and elsewhere in representation theory, determinants of characters play an important role in our considerations here. Given a complex character χ of H, there is a well-defined linear character (detχ) which is defined by (detχ)[h] = det(hσ), where σ is any matrix representation affording character χ. We will also make use of the augmentation map τ : RH → R (for any commutative ring R with 1), defined by

( ∑_{h∈H} r_{h}h)τ = ∑_{h∈H} r_{h}.

By a slight abuse, we let ΩR(H) denote the kernel of the augmentation map. When R is an integral domain of characteristic zero, the character afforded by ΩR(H) takes the value −1 on all non-identity elements of H and takes value |H| − 1 on 1_{H}. For ΩR(H) ∩ R(∑_{h∈H} h) = {0} and H acts trivially on R(∑_{h∈H} h).

We will adopt some other standard conventions: π(H) denotes the set of prime divisors of the order of the group H. If π is a set of primes and n is a positive integer, we let n_{π} denote the largest divisor of n which is only divisible by primes in π. If x is an element of a finite group X and we write (uniquely) x = ab with a, b ∈ ⟨x⟩ and with |⟨a⟩| = |⟨x⟩|_{π}, and |⟨b⟩| = |⟨x⟩|_{π′}, we denote the element a by x_{π}, and call x_{π} the π-part of x.

Our main result is:

Theorem: Let H be a finite group whose Sylow 2-subgroup does not have a cyclic subgroup of index 2. Let θ be a generalized character of H such that |θ(h)|^{2} = 1 for each non-identity element of H. Then θ = λ.ψ, where λ is a linear character of H and ψ is a generalized character of H whose values on non-identity elements are all in {1, −1}. If ψ takes only one value on H^{#}, then there is a sign ε such that θ(h) = ελ(h) for all h ∈ H^{#}. Otherwise, it is possible to partition the prime divisors of H into two non-empty disjoint sets π_{0} and π_{1} so that for δ ∈ {0, 1} we have ψ(x) = (−1)^{δ} for each non-identity π_{δ}-element x ∈ H. Furthermore, in that case, if |H| has odd order, or if 2 ∈ π_{δ}, then C_{H}(x) has a Hall π_{δ}-subgroup of order dividing 2 for each non-identity π_{1−δ}-element x ∈ H. Moreover, in that case, if x is an element of even order of H, then we have ψ(x) = ψ(x_{2}).

Conversely, if a Sylow 2-subgroup of H does not have a cyclic subgroup of index 2 and it is possible to partition the prime divisors of |H| into two nonempty disjoint sets π and τ such that C_{H}(x) has a Hall π-subgroup of order dividing 2 for each non-identity τ-element x ∈ H, then H has a generalized character θ such that θ(x) = 1 if x is a non-identity π-element and θ(x) = −1 if x is a non-identity τ-element and such that, furthermore, θ(x) = θ(x_{2}) for all x of even order in H.

Remarks: The hypothesis that a Sylow 2-subgroup of H should not have a cyclic subgroup of index 2 is necessary. For example, when H ≅ PSL(2, 7), then using Brauer’s characterization of characters, we see that there is a generalized character θ of H such that θ(1) = 43, θ(x) = 1 if x ∈ H^{#} is not an involution, and θ(y) = −1 if y ∈ H^{#} is an involution. Thus π(H) can’t be partitioned as in the conclusion of the above Theorem.

An application of the theorem to Glauberman correspondence is given by:

Corollary: Let A be a (not necessarily Abelian) finite group which acts as a coprime group of automorphisms of another finite group G, and let L denote the semidirect product GA. Suppose that a Sylow 2-subgroup of A has no cyclic subgroup of index 2 (in particular, this allows the possibility that A has odd order). Let χ be an A-stable irreducible character of G, and let χ̃ be an extension of χ to an irreducible character of L. Suppose that for each a ∈ A^{#} the Glauberman correspondent of χ for C_{G}(a) is linear. Then one of the following occurs:

i) There is a linear character λ of A and a sign ε such that χ̃(a) = ελ(a) for all a ∈ A^{#}. ii) For some linear character γ of A, it is possible to partition the prime divisors of A into two non-empty subsets π_{0} and π_{1} such that if |H| is odd, or 2 ∈ π_{δ}, then whenever x is a non-identity π_{1−δ}-element of A, C_{A}(x) has a Hall π_{δ}-subgroup of order dividing 2, and:

a) χ̃(x) = (−1)^{β}γ(x), whenever x is a π_{β}-element for β ∈ {0, 1}.

b) χ̃(x) = (−1)^{δ}γ(x) whenever x ∈ A has even order.

1 Local results

Now let R be a complete discrete valuation ring of characteristic 0 such that R/J(R) is algebraically closed of finite characteristic p > 0. Let 𝕂 be the field of fractions of R, and suppose that 𝕂 contains sufficiently many p-power order roots of unity to be a splitting field for any finite group considered here.

Our first result is well-known:

Lemma 1.1: Let ⟨x⟩ be a finite cyclic p-group, and let R denote the trivial R⟨x⟩-module. Then ΩR(⟨x⟩) affords a unimodular representation when p is odd, and x acts with determinant −1 on ΩR(⟨x⟩) when p = 2. In either case the determinant of the representation of ⟨x⟩ on ΩR(⟨x⟩) is the same as the determinant of the regular representation of ⟨x⟩, so that ΩR(⟨x⟩) ⊕ R⟨x⟩ affords a unimodular representation of ⟨x⟩.

Proof: Clearly, the character afforded by ΩR(⟨x⟩) is the sum of all non-trivial linear characters of ⟨x⟩, and the determinant of the representation of ⟨x⟩ on ΩR(⟨x⟩) is the product of all non-trivial linear characters of ⟨x⟩. When p is odd, the non-trivial linear characters of ⟨x⟩ occur in complex conjugate pairs, so their product takes value 1 on x (and all its powers). When p = 2, there are just two real-valued linear characters of ⟨x⟩. One takes the value 1 at x, the other takes the value −1 at x. The remaining linear characters of ⟨x⟩ pair off in complex conjugate pairs and have trivial product.

Remark: For future reference, we remark that when p = 2 and x has order 2^{e}, ΩR(⟨x⟩) ⊕ R⟨x⟩ is a unimodular representation of ⟨x⟩ of degree 2^{e+1} − 1, and remains unimodular if its direct sum with an even number of copies of the regular representation is taken, while R is a unimodular representation of ⟨x⟩ of degree 1, and remains unimodular if its direct sum with an even number of copies of the regular representation is taken.

In the former case, the character afforded by the module takes constant value −1 on ⟨x⟩^{#}, while in the latter case, the character afforded by the module takes constant value 1 on ⟨x⟩^{#}.

When p is odd, the action of ⟨x⟩ on ΩR(⟨x⟩) is unimodular, and the character afforded by ΩR(⟨x⟩) takes value −1 on ⟨x⟩^{#}. Addition of direct sums of copies of the regular module does not affect the unimodularity of the action, or the value taken by the character on non-identity elements. Clearly the action of ⟨x⟩ on the trivial module R is unimodular. This allows us to record:

Lemma 1.2: Let χ be a complex character of a finite cyclic p-group ⟨x⟩ such that (detχ) is the trivial character, and such that χ(x^{j}) is a root of unity for each integer j with 1 ≤ j < |⟨x⟩|. Then:

i) If p is odd, there is a sign ε such that χ(1) ≡ ε (mod |⟨x⟩|) and χ(x^{j}) = ε for 1 ≤ j < |⟨x⟩|.

ii) If p = 2, and there is a sign ε such that χ(1) ≡ ε (mod 2|⟨x⟩|), then χ(x^{j}) = ε for 1 ≤ j < |⟨x⟩|.

Proof: Suppose first that p is odd. Since the action of x on the regular representation is unimodular, we may suppose that ⟨χ, μ⟩ = 0 for some linear character μ of ⟨x⟩ (otherwise we can subtract the regular character from χ and still leave a character with trivial determinant). Notice that

0 = ∑_{y∈⟨x⟩} χ(y)μ(y^{−1}) ≥ χ(1) − (|⟨x⟩| − 1),

since χ(y) is a root of unity for each non-identity y ∈ ⟨x⟩. Hence χ(1) < |⟨x⟩|.

However, the character χ\overline{χ} takes the value 1 on all non-identity powers of x, so that χ(1)^{2} ≡ 1 (mod |⟨x⟩|). As p is odd, there is a sign ε such that χ(1) ≡ ε (mod |⟨x⟩|). Since χ(1) < |⟨x⟩|, we either have χ(1) = 1 or χ(1) = |⟨x⟩| − 1. If χ(1) = 1, then χ must be the trivial character by unimodularity. Suppose then that χ(1) = |⟨x⟩| − 1. Then for the linear character μ above, we must have χ(y) = −μ(y) for each non-identity y ∈ ⟨x⟩. Let θ be the character of ⟨x⟩ afforded by ΩR(⟨x⟩). Then χ = μθ and (detχ) = μ^{χ(1)}(detθ). But the action of ⟨x⟩ on ΩR(⟨x⟩) is unimodular, and χ(1) is not divisible by p, so μ must be the trivial character, as detχ is trivial.

If p = 2, we may suppose that there is a linear character μ of ⟨x⟩ such that ⟨χ, μ⟩ ∈ {0, 1}, otherwise we may subtract an even integer multiple of the regular character from χ and still have a character with trivial determinant (without affecting values on non-identity elements, or the fact that χ(1) ≡ ε (mod 2|⟨x⟩|)). Then [χ(1) − (|⟨x⟩| − 1)] ≤ |⟨x⟩|, so that χ(1) < 2|⟨x⟩|. Hence either χ(1) = 1 (if ε = 1) or χ(1) = 2|⟨x⟩| − 1 (if ε = −1). If χ(1) = 1, we are done, since (detχ) is the trivial character. If χ(1) = 2|⟨x⟩| − 1, then for the linear character μ above, we must have ⟨χ, μ⟩ = 1 and χ(y) = −μ(y) for each non-identity element y of ⟨x⟩. Let θ be the character afforded by ΩR(⟨x⟩)⊕R⟨x⟩.

Then (detθ) is the trivial character, and χ = μθ, so that (detχ) = μ^{χ(1)}(detθ) and μ is the trivial character, since χ(1) is odd and (detχ) and (detθ) are both trivial.

2 From local to global

Theorem 2.1: Let A be a finite Abelian group. Let χ be a complex character of A such that |χ(a)|^{2} = 1 for each a ∈ A^{#}. Then there is a linear character λ of A and a sign ε such that χ(a) = ελ(a) for all a ∈ A^{#}. If O_{2}(A) is not cyclic, then we have (detχ) = λ^{χ(1)}. In particular, λ is trivial if O_{2}(A) is not cyclic and χ is afforded by a unimodular representation of A.

Proof: We may subtract multiples of the regular character from χ without changing its values on A^{#}. Furthermore, this does not affect (detχ) when O_{2}(A) is not cyclic. For if a ∈ A^{#}, the regular character of A restricts to [A : ⟨a⟩] times the regular character of ⟨a⟩. Now [O_{2}(A) : O_{2}(⟨a⟩)] is even when O_{2}(A) is not cyclic. The determinant of the regular character of ⟨a⟩ is the trivial character if a has odd order, and is the linear character λ with λ(a) = −1 if a has even order. Hence the determinant of the regular character of A restricted to ⟨a⟩ is trivial when O_{2}(A) is not cyclic. Thus we may indeed suppose that ⟨χ, λ⟩ = 0 for some linear character λ of A. Then χ(1) − (|A| − 1) ≤ 0, since |χ(a)|^{2} = 1 for each a ∈ A^{#}. Hence χ(1) < |A|.

Now write χ = ∑_{i=1}^{n} m_{i}λ_{i}, where the m_{i} are non-negative integers and λ_{1}, …, λ_{n} are the distinct linear characters of A. We have

|A| − 1 = ∑_{a∈A^{#}} |χ(a)|^{2} = |A| ∑_{i=1}^{n} m_{i}^{2} − χ(1)^{2} = ∑_{i<j}(m_{i} − m_{j})^{2}.

Let us label so that m_{1} = … = m_{k} = 0 and 0 < m_{k+1} ≤ … ≤ m_{n}. We have assumed that at least one m_{i} = 0.

Then (n − 1) = ∑_{i<j}(m_{i} − m_{j})^{2} ≥ k ∑_{j=k+1}^{n} m_{j}^{2} ≥ k(n − k). Hence k = 1 or k = n − 1. Furthermore, if k = 1 we must have m_{i} = 1 for 2 ≤ i ≤ n, while if k = n − 1, we must have m_{n} = 1. Hence we either have χ(a) = −λ_{1}(a) for all a ∈ A^{#} or else we have χ(a) = λ_{n}(a) for all a ∈ A^{#}. In the former case, we have χ(1) = |A| − 1, and in the latter case, we have χ(1) = 1 (but we recall that we have subtracted multiples of the regular character from χ so, in general, we have χ(1) ≡ ±1 (mod |A|)).

Now there is a sign ε and a linear character λ of A such that χ(a) = ελ(a) for each a ∈ A^{#} and χ(1) ≡ ε (mod |A|). Furthermore, we have reduced to the case that χ(1) ∈ {1, |A| − 1}. To prove the last claim, under the assumption that O_{2}(A) is not cyclic, we need to prove that (det\overline{λ}χ) is the trivial character in that case. This is clear when χ(1) = 1, so suppose that χ(1) = |A| − 1 and ε = −1. It suffices to prove that (det\overline{λ}χ) restricts trivially to O_{p}(A) for each prime p dividing |A|.

Choose such a prime p, and assume that R/J(R) has characteristic p. Then \overline{λ}χ is the character of O_{p}(A) afforded by ΩR(O_{p}(A)) ⊕ tRO_{p}(A), where t = [A : O_{p}(A)] − 1. The action of O_{p}(A) on this module is unimodular if p is odd. If p = 2, then as O_{p}(A) is not cyclic, each 2-element of A has unimodular action on RO_{p}(A). Since O_{2}(A) is not cyclic, for any a ∈ O_{2}(A)^{#}, ΩR(A) decomposes as R⟨a⟩-module as the direct sum of ΩR(⟨a⟩) with an odd number of copies of R⟨a⟩. Hence the action of a on ΩR(A) is unimodular. Thus (det\overline{λ}χ) is the trivial character, as required.

Theorem 2.2: Let H be a finite group which has a complex character χ such that |χ(h)|^{2} = 1 for each h ∈ H^{#}. Suppose also that the Sylow 2-subgroup of H does not have a cyclic subgroup of index 2. Then χ(1) ≡ ±1(mod|S|) for each Sylow subgroup S of H. Furthermore, let π_{δ} denote the set of prime divisors p of |H| such that χ(1) ≡ (−1)^{δ} (mod |H|_{p}) for δ ∈ {0, 1}. Then there is a linear character λ of H such that (detχ) = λ^{χ(1)} and we have χ(h) = (−1)^{δ}λ(h) whenever h is a π_{δ}-element of H^{#}. Furthermore, if 2 ∈ π_{δ}, then C_{H}(x) has a Hall π_{δ}-subgroup of order dividing 2 whenever x ∈ H is a non-identity π_{(1−δ)}- element of H.

Proof: We have χ(1)^{2} ≡ 1 (mod |H|). We may choose integers a and b such that a|H| + bχ(1) = 1. Let μ = (detχ). Then the determinant of μ^{−b}χ is the trivial character and (detχ) = μ^{bχ(1)}. Hence it suffices to consider the case that (detχ) is trivial.

Choose a prime divisor p of the order of H and let P be a Sylow p-subgroup of H. If p is odd, then χ(1) ≡ ε (mod |P |) for some sign ε. If p is odd, then by the previous result, χ(x) = ε for each x ∈ P^{#} since (detχ) = 1. If p = 2, we have χ(1) ≡ ε (mod \frac{|P |}{2}) for some sign ε. Since P has no cyclic subgroup of index 2, we have χ(x) = ε for each x ∈ P^{#} by Lemma 1.2. Hence we in fact have χ(1) ≡ ε (mod |P |) in this case too.

Let a be a non-identity π_{0}-element of H. We want to prove that χ(a) = 1. This is the case if a has p-power order, so suppose otherwise. Then by Theorem 2.1, there is a sign δ and there is a linear character μ of ⟨a⟩ such that χ(u) = δμ(u) for each non-identity element u ∈ ⟨a⟩^{#}. We have χ(a_{p}) = 1 for each prime divisor p of the order of a. Hence we have μ(a_{p}) = δ for each such prime. Hence must have δ = 1 since at least one of these primes is odd. Then μ contains each a_{p} in its kernel, so is the trivial character and χ(a) = 1.

Let a be a non-identity π_{1}-element of H. We want to prove that χ(a) = −1. This is the case if a has p-power order for some prime p, so suppose otherwise. Then by Theorem 2.1, there is a sign δ and there is a linear character μ of ⟨a⟩ such that χ(u) = δμ(u) for each non-identity element u ∈ ⟨a⟩^{#}. We have χ(a_{p}) = −1 for each prime divisor p of the order of a. Hence we have μ(a_{p}) = −δ for each such prime. Hence must have δ = −1 since at least one of these primes is odd. Then μ contains each a_{p} in its kernel, so is the trivial character and χ(a) = −1.

Now suppose that a_{0} is a non-identity π_{0}-element and a_{1} is a non-identity π_{1} element such that a_{0}a_{1} = a_{1}a_{0}. Set A = ⟨a_{0}a_{1}⟩. By Theorem 2.1, there is a sign γ and there is a linear character μ of A such that χ(u) = γμ(u) for all u ∈ A^{#}. Now χ(a_{0}) = 1 and χ(a_{1}) = −1. Hence μ(a_{0}) = γ and μ(a_{1}) = −γ. Suppose that γ = (−1)^{(1−β)}, where β ∈ {0, 1}. Then μ(a_{β}) = −1, so that a_{β} has even order and 2 ∈ π_{β}. This argument shows that if u is any non-identity π_{(1−β)}-element of A and v is any non-identity π_{β}-element of A, then μ(u) = 1 and μ(v) = −1. In particular, we must have v^{2} = 1 and especially a_{β}^{2} = 1. This shows that C_{H}(a_{(1−β)}) has an elementary Abelian Sylow 2-subgroup A_{β} which is a Hall π_{β}-subgroup of C_{H}(a_{(1−β)}). Let B = ⟨a_{(1−β)}⟩ × A_{β}.

There is a sign η and there is a linear character α of B such that χ(b) = ηα(b) for each b ∈ B^{#}. Now a_{(1−β)} has odd order, so that α(a_{(1−β)}) = 1 and η = χ(a_{(1−β)}) = (−1)^{(1−β)}. Thus (−1)^{β} = χ(c) = −η for all c ∈ A_{β}^{#}, so that α(c) = −1 for each c ∈ A_{β}^{#}. Since α is a linear character of B, it must be the case that A_{β} is cyclic.

This allows us to give a complete description, for any finite group H whose Sylow 2-subgroup does not have a cyclic subgroup of index 2, of generalized characters whose values on non-identity elements are always roots of unity.

Theorem 2.3: Let H be a finite group whose Sylow 2-subgroup does not have a cyclic subgroup of index 2. Let θ be a generalized character of H such that |θ(h)|^{2} = 1 for each non-identity element of H. Then θ = λ.ψ, where λ is a linear character of H and ψ is a generalized character of H whose values on non-identity elements are all in {1, −1}. It is possible to partition the prime divisors of H into two disjoint sets π_{0} and π_{1} so that for δ ∈ {0, 1} we have ψ(x) = (−1)^{δ} if x is a non-identity π_{δ}-element of H. Furthermore, if 2 ∈ π_{δ}, then C_{H}(x) has a Hall π_{δ}-subgroup of order dividing 2 for each non-identity π_{1−δ}-element. Moreover, if x is an element of H which is both π_{δ}-singular and π_{1−δ}-singular, then x has twice odd order, and we have ψ(x) = ψ(x_{2}) = (−1)^{δ}.

Conversely, if a Sylow 2-subgroup of H does not have a cyclic subgroup of index 2 and it is possible to partition the prime divisors of |H| into two disjoint sets π and τ such that no non-identity π-element of odd order commutes with a non-identity τ-element, and such that 2 ∈ π, and C_{H}(x) has a Sylow 2-subgroup of order at most 2 for each non-identity τ-element x, then H has a generalized character θ such that θ(x) = 1 if x is a non-identity π-element and θ(x) = −1 if x is a non-identity τ-element and such that, furthermore, θ(x) = θ(x_{2}) for all x of even order in H.

Proof: We may add multiple of the regular character of H to θ and suppose that θ is a character. We may also suppose that (detθ) is trivial, as before. For the first part, it only remains to prove that θ(x) = θ(x_{2}) when x has even order. We prove this by induction on the number of odd prime divisors of the order of x, the result being clear when this number is zero. Write x = uvw, where u is a 2-element, v is a p′-element of odd order and w is a non-identity p-element, where p is an odd prime, and u, v, w all commute with each other. By induction, we may suppose that θ(uv) = θ(u), and we do so. Now we know that θ(uv) = ±1 and θ(uvw) = ±1. Since w is a p-element and θ(uv) and θ(uvw) are both integers, we know that θ(uv) ≡ θ(uvw) (mod p). Since p is odd, we must have θ(uvw) = θ(uv) = θ(u), as required.

For the converse, suppose that we can partition the prime divisors of |H| in the stated fashion. Choose an integer m such that m ≡ 1 (mod |H|_{π}) and m ≡ −1 (mod |H|_{τ}). Note that m is odd. Define a class function θ of H via θ(1) = m, θ(x) = 1 if x is a non-identity π-element, θ(x) = −1 if x is a non-identity τ-element, and θ(x) = θ(x_{2}) when x has even order.

By Brauer’s characterization of characters, it suffices to prove that Res_{E}^{H}(θ) is a generalized character whenever E is Brauer elementary. This is clear when E is a π-group or a τ-group. If |E| is divisible by primes both in τ and π, then O_{π}(E) has order 2. In that case, θ takes the constant value −1 on O_{τ} (E)^{#} and takes the constant value 1 on the coset xO_{τ} (E) when x is the unique involution of E. In fact, by Brauer’s characterization, we only need to check ⟨Res_{E}^{H}(θ), λ⟩ when λ is a linear character of E. If O_{τ} (E) is not contained in the kernel of λ, we obtain |E|⟨Res_{E}^{H}(θ), λ⟩ = m − (−1). This is an even integer and is divisible by |H|_{τ} , so is divisible by |E|. Hence the given inner product is an integer in this case.

If O_{τ} (E) is contained in the kernel of λ, we obtain

|E|⟨Res_{E}^{H}(θ), λ⟩ = \frac{λ(x)|E|}{2} + m − (\frac{|E|}{2} − 1) = m + 1 + \frac{|E|(λ(x) − 1)}{2},

again an integer multiple of |E|, since λ(x) ∈ {1, −1}. Hence θ is a generalized character in any case.

Remark: Let us expand a little on the situation when a Sylow 2-subgroup of H does have a cyclic subgroup of index 2. It suffices to consider a character θ of H whose values on non-identity elements of H are all roots of unity, and such that (detθ) is the trivial character. We want to prove in this case that the values taken by θ on H^{#} are still all in {1, −1}. We know that if A is any Abelian subgroup of H, there is a sign ε and a linear character λ of A such that θ(a) = ελ(a) for all a ∈ A^{#}. Now \overline{λ}θ takes the constant value ε on A^{#}. Hence \overline{λ}^{θ(1)} is either the trivial character of A or (detρ), where ρ is the regular character of A. In either case, the values taken are in {1, −1}. Since θ(1)^{2} ≡ 1 (mod |A|), there are integers s and t such that sθ(1) + t|A| = 1. Then λ = (λ)^{sθ(1)+t|A|} = λ^{sθ(1)}, and the last character takes all its values in {1, −1}.

Corollary 2.4: Let H be a finite group and let θ be a generalized character of H whose values on elements of H^{#} are all roots of unity. Then if θ is not rational valued, H/H′ is has exponent greater than 2.

We may suppose that θ is a character. We have seen that θ is expressible as a product λ.ψ, where λ is a linear character of H and ψ is a generalized character taking values in {1, −1}. If θ is not rational valued, then neither is λ, so that [H : kerλ] ≥ 3.