11. Let χ be an irreducible character of G. Prove that for every z∈Z(G) we have χ(z) = ζχ(1) for some root of unity ζ.
12. Let ψ be the character of some representation φ of G, and let g∈G. Prove (a) ψ(g) = ψ(1) if and only if g∈\text{ker }φ, and (b) |ψ(g)| = ψ(1) implies g ∈ Z(G) if φ is faithful.
In the exercises that follow, let F = \overline{ℚ} be the field of algebraic numbers.
14. Prove that a representation φ of G over F is irreducible over F if and only if it is irreducible when considered as a representation over ℂ. Deduce that the group characters of G when considered over F are the same as when considered over ℂ, and that every complex representation of G is equivalent to a representation over F.
16-17. Let σ be an automorphism of the field extension F/ℚ. If φ is a representation of G over F with character ψ, let φ^σ be the mapping given by applying the automorphism σ to each of the entries of φ. Prove that φ^σ is a representation with character ψ^σ = σ∘ψ.
18. If ψ is again a character of G, let ℚ(ψ) be the (finite) extension of ℚ by each value ψ(g)∈F. Prove that ℚ(ψ)⊆ℚ(ζ_n), where ζ_n is a primitive n^\text{th} root of unity. Deduce that ℚ(ψ) is a Galois extension of ℚ with abelian Galois group.
19-24. For each a relatively prime to n, let σ_a ∈ \text{Gal}(ℚ(ζ)/ℚ) ≅ (ℤ/nℤ)^× be the automorphism given by ζ_n ↦ ζ_n^a. Show that ψ^{σ_a}(g) = ψ(g^a). Prove that every character of a group has all integer values if and only if for all g∈G, one has g is conjugate to g^k for all k relatively prime to |g|.
Proof: (11) Let φ be the corresponding representation of G on V. Since z commutes with every element of G, it commutes with every element of ℂG. Therefore, multiplication by z is a ℂG-module endomorphism on V. Diagonalize V with respect to z, so that φ(z) is given by a matrix with roots of unity on the diagonal and zeros elsewhere. Let ζ be one such root. By Schur's lemma, z - ζ is either an automorphism or an annihilator of V. Since z - ζ annihilates at least one basis element of V (that from which ζ was chosen from φ(z)), we see it must be the latter, so that φ(z) = ζI and χ(z) = ζχ(1).
(12)(a-b) If g ∈ \text{ker }φ, then clearly ψ(g) = ψ(1). Conversely, if ψ(g) = ψ(1), then after diagonalization with respect to g, we may consider g as a matrix with ψ(1) roots of unity ζ_i on the diagonal. If ψ(g) = \sum ζ_i = ψ(1), then also |ψ(g)| = |\sum ζ_i| = \sum |ζ_i| = ψ(1), so by the triangle inequality each ζ_i is equal. From here we see (b), whereby φ(g) commutes with every matrix in φ(G)⊆\text{GL}(ℂ), then φ(g)∈Z(φ(G)) implying g∈Z(G) if φ is faithful. Given the stronger condition of (a) that ψ(g) = ψ(1), we in fact see ζ_i = 1, so g∈\text{ker }φ.
(14) Note that the computation of a character norm is independent of the field over which the matrices are considered, and irreducibility of a character/representation is equivalent to the norm being 1. Thus, a representation is irreducible over F if and only if it is irreducible over ℂ. Furthermore, there are r irreducible characters over F that are equivalent to their corresponding complex characters, where r is the number of conjugacy classes of G; this is also independent of the field considered, so this is the full set of irreducible complex characters of G. Thus, the F-characters and complex characters are the same. Finally, since every representation over ℂ will entail a character identical to that of some representation over F, this implies every representation over ℂ is equivalent to one over F.
(16-17) The action of σ on \text{GL}(F) is in fact an automorphism, so the composition φ^σ is a homomorphism from G into \text{GL}(V) corresponding to which we have clearly have the character ψ^σ.
(18) Examining Proposition 14 more closely, we see in fact that ψ(g)∈ℚ(ζ_k) for each g∈G with order |g|=k. Since necessarily k will divide n, and ℚ(ζ_k)⊆ℚ(ζ_n), we have ℚ(ψ)⊆ℚ(ζ_n), in fact ℚ(ψ)=ℚ(ζ_e) for the exponent e of φ(G). Thus it is a Galois extension with abelian Galois group.
(19-24) If we diagonalize with respect to g, we see that σ_a(ζ_n^k) = (ζ_n^k)^a for all k, hence \text{tr }φ^{σ_a}(g) = \text{tr }φ(g)^a = \text{tr }φ(g^a) = ψ(g^a). If g is conjugate to g^a for all a relatively prime to |g|, then ψ^{σ_a} = ψ as characters for all a relatively prime to n, so ψ(g) is within the fixed field of the Galois extension ℚ(ζ_n)/ℚ, that is, ℚ. By Proposition 14, character values are algebraic integers, so ψ actually takes values in ℤ.
Conversely, suppose g isn't conjugate to g^a for some a relatively prime to |g|. Let n = p_1^{α_1}p_2^{α_2}...p_k^{α_k} and |g| = p_1^{β_1}p_2^{β_2}...p_k^{β_k} be factorizations, necessarily with β_i ≤ α_i. Reordering the primes if necessary, let β_i = 0 for i > m. By the Chinese Remainder Theorem, let b be such that b≡a \mod p_i^{α_i} for i≤m, and b≡1 \mod p_i^{α_i} for i > m. Then b ≡ a \mod |g| (so g^a = g^b), and b is relatively prime to n. Then since the irreducible characters of G form a basis for the space of class functions on G, there is some irreducible character χ such that χ(g)≠χ(g^b)=χ^{σ_b}(g)=σ_b(χ(g)). Thus χ(g) is not within the fixed field of σ_b, hence not an integer.~\square
