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.$~\square$
