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\title{Lecture 4-8: Values of characters}
\date{April 8, 2024}
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We now explore the values $\chi(g)$ of characters at elements of $g$ in more detail.  We begin with a brief digression into ring theory.  Given a commutative ring $B$ with a subring $A$, we say that $b\in B$ is \alert{integral over $A$} if there is a {\sl monic} polynomial $p\in A[x]$ such that $p(b) = b^n+\sum\limits_{i=0}^{n-1}a_ib^i = 0$ (so that $a_i\in A$ for all $i$).  Equivalently, \alert{the subring $B'=A[b]$ of $B$ generated by $A$ and $b$ is a finitely generated $A$-module $\sum Ab_i$ for some $b_i\in B'$}; for then if we write each $b_i$ as a polynomial in $b$ of degree $d_i$ with coefficients in $A$ and choose $d>d_i$ for all $i$, then $b^d$, as a combination of $b_i$, must be an $A$-linear combination of lower powers of $b$.  
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Moreover, if $b$ is integral over $A$, then so are all elements of the subring $A[b]$.  To see this, let $b_1,\ldots,b_n$ be a set of generators for $B'=A[b]$ as an $A$-module.  For $c\in B'$ write each product $cb_i$ as a combination $\sum_{j=1}^n c_{ji}b_j$, where the $c_{ji}$ lie in $A$, and let $C$ be the $n\times n$ matrix over $A$ with $ij$ entry $c_{ij}$.  Given a combination $d=\sum a_ib_i\in B'$ with $a_i\in A$ the matrix product $C(a_1,\ldots,a_n)^t$ is a column vector $(a_1',\ldots,a_n')^t$; the definition of $C$ shows that $cd=\sum a_i'b_i$.  If we now make $B'$ into a module over the polynomial ring $A[t]$ by decreeing that $t$ act by multiplication by $c$, then we see that the matrix $C-tI$ over $A[t]$ acts by 0 on $B'$.  Multiplying this matrix on the left by the transpose of its cofactor matrix, we get the scalar matrix $p(t)I$, where $p(t)$ is the characteristic polynomial of $C$; it too acts by 0 on $B'$, whence $p(c)=0$ and $c$ is integral over $A$.  More generally, \alert{given any ring extension $A\subset B$, the set of elements of $B$ integral over $A$ is a subring of $B$ containing $A$}.
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In particular, the \alert{algebraic integers} in $\mathbb C$, i.e. the complex numbers integral over $\mathbb Z\subset\mathbb C$ in this sense, form a subring of $\mathbb C$. .
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But the algebraic integers, unlike the algebraic numbers, do not form a subfield of $\mathbb C$.  On the contrary, if the rational number $\frac r s$ in lowest terms is an algebraic integer, then we have a relation $(\frac r s)^n+\sum\limits_{i=0}^{n-1} z_i(\frac r s)^i = 0, r^n+\sum\limits_{i=0}^{n-1} z_ir^is^{n-i}=0$.  If $s$ has any prime divisor $p$, then $p$ cannot divide any power of $r$, since $\frac r s$ is in lowest terms, whence $p$ does not divide the left side of this last equation, a contradiction.  We conclude that \alert{the integers are the only rational numbers that are algebraic integers}.  Indeed, ordinary integers are sometimes called {\sl rational integers} for emphasis, so as to distinguish them from algebraic integers.
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We now return to character theory.  Any character value $\chi(g)$ is the sum of the eigenvalues of a matrix, each of which is a root of unity, so $\chi(g)$ is an algebraic integer.  In particular, $\chi(g)\in\mathbb Q$ if and only if $\chi(g)\in\mathbb Z$:  \alert{there are no fractions in character tables}.  Now let $V$ be an irreducible representation of $G$ with character $\chi$.  Let $C_1,\ldots C_m$ be the conjugacy classes in $G$, with $C_i$ of size $c_i$.  For each $i$ denote the common value of $\chi$ on all elements of $C_i$ by $d_i$.
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\begin{alertblock}{Proposition 4, DF p. 887}
The element $x_i=\sum_{g\in C_i}g\in\mathbb CG$ acts by the scalar $e_i=\frac {c_id_i} {\chi(1)}$ on $V$ and $e_i$ is an algebraic integer.
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\begin{proof}
We know that $x_i$ is central in $\mathbb CG$, whence it acts by a scalar on $V$ by Schur's Lemma; taking traces, we see that this scalar is indeed $e_i$.  For any $i,j$, the product $x_ix_j$ is also central in $\mathbb CG$, whence it is a nonnegative integral combination of $x_k$.  It follows that the $\mathbb Z$-submodule of $\mathbb C$ generated by the $e_i$ is also a subring, whence the $e_i$ are algebraic integers, as claimed.
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\begin{alertblock}{Corollary 5, DF p. 888}
The degree $n_i=\chi_i(1)$ of any irreducible character $\chi_i$ of $G$ divides the order $n$ of $G$.
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\begin{proof}
Let $d_j$ be the common value of $\chi_i$ on the elements of $C_j$.  By Schur orthogonality, the sum $\sum\limits_{j=1}^m e_j\overline{d_j} = \frac n {\chi_i(1)}$; but this sum is generated by algebraic integers and so is an algebraic integer, forcing $\frac n {\chi_i(1)}$ to be an integer, as claimed.
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Under an extra hypothesis on the size $c_j$ of a conjugacy class $C_j$ and the degree $n_i$ of an irreducible representation $\pi_i:G\rightarrow GL(V_i)$ we get a severe constraint on the value $d_j$ of the corresponding character $\chi_i$ on the elements of $C_j$.  This is
\begin{alertblock}{Lemma 6, DF p. 889}
With notation as above, if $n_i=\chi_i(1)$ and $c_j$ are relatively prime, then either $d_j=0$ or $\pi_i(x)$ is a scalar matrix for all $x\in C_j$.
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\begin{proof}
Under the hypothesis both $\frac {c_jd_j} {n_i}$ and $d_j=\frac {n_id_j} {n_i}$ are algebraic integers, whence upon taking a suitable integral combination $f_j=\frac {d_j} {n_i}$ is an algebraic integer.  Now $d_j$ is the sum of $n_i$ roots of 1 in $\mathbb C$ (the eigenvalues of $\pi_i(x)$) divided by $n_i$, so its norm as a complex number is at most 1, with equality holding if and only if all the roots of 1 coincide, which in turn holds if and only if $\pi_i(x)$ is a scalar.  But the minimal polynomial of $f_j$ over $\mathbb Q$ (or $\mathbb Z$) must have integer coefficients, by Gauss's Lemma from last quarter.  The product of the Galois conjugates of these roots equals the constant term of this polynomial, up to sign, so must be an integer.  Yet each of these conjugates, like $f_j$ itself, is the sum of $n_i$ roots of 1 divided by $n_i$, so has absolute value at most 1.  Thus if none of the roots is 0, then $\pi_i(x)$ is a scalar, while otherwise all roots are 0 and $d_j=0$, as claimed.
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Thus it is no coincidence, for example, that the two-dimensional character $\chi_r$ of $S_3$ takes the value 0 on the conjugacy class of transpositions, since this class has size 3 and 2 and 3 are relatively prime.  
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Under an even stronger hypothesis on the size of a conjugacy class we get a consequence for the structure of $G$.
\begin{alertblock}{Burnside's Nonsimplicity Criterion:  DF p. 890}
Let $G$ be a nonabelian group such that some nonidentity conjugacy class $C$ has size a power of a prime $p$.  Then $G$ is not simple; that is, it has a proper normal subgroup.
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Suppose first that there is a nontrivial irreducible representation $\pi$ of $G$ with $\pi(g)$ a scalar matrix for some $g\in G,g\ne1$.  Then the set of all $h\in G$ with $\pi(h)$ a scalar matrix is a nontrivial normal subgroup of $G$; if it is all of $G$, then the kernel $K$ of $\pi$, consisting of all $k\in G$ with $\pi(k)=1$, is another normal subgroup of $G$.  If $K=1$, then $G$ is isomorphic to a subgroup of $\mathbb C^*$ and so is abelian, contrary to hypothesis; so in any event $G$ is not simple.  Now suppose that there is no such representation $\pi$.  Then every irreducible character $\chi_i$ of $G$ either has degree $n_i=\chi_1(1)$ a multiple of $p$ or has $\chi_i(g)=0$ for all $g\in C$.  Applying the second Schur orthogonality relation, we get $1+\sum_i \chi_i(1)\chi_i(g) = 0$, where the sum ranges over the nontrivial characters of $G$, for all $g\in C$.  Omitting the terms where $\chi_i(g)=0$ and dividing all the remaining $\chi_i(1)$ by $p$, we deduce that $-1/p$ is an algebraic integer, a contradiction.  Thus $G$ is not simple, as claimed.
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For example, Burnside's Criterion implies that the alternating group $G=A_4$ is not simple, since it has a conjugacy class $C$ of 3-cycles of size $4=2^2$.  Note however that while $A_r$ does have a nontrivial irreducible representation $\pi$ such that $\pi(x)$ is a scalar matrix for all $x\in C$, it is {\sl not} true that $C$ lies in a proper normal subgroup of $G$.  Instead the kernel of $\pi$ is the proper normal subgroup, consisting of another conjugacy class of $G$ together with the identity.  A further famous consequence of Burnside's Criterion is
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\begin{alertblock}{Burnside's Theorem:  Theorem 1, DF p. 886}
Any group $G$ whose order is the product $p^aq^b$ of two prime powers is solvable. 
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Indeed, given any such $G$, choose a nonidentity central element $g$ of a Sylow $p$-subgroup, which exists by the theory of $p$-groups.  Then its conjugacy class has order dividing $\frac {p^aq^b} {p^a} = q^b$, so this order is a power of $q$, whence $G$ is not simple by Burnside's Criterion.  Passing to a quotient $G/N$ and arguing by induction on the order of $G$, we may assume that $G$ has a solvable normal subgroup $N$ such that $G/N$ is also solvable, whence so is $G$.  On the other hand, a group whose order is the product of three prime powers need not be nonsimple, as the example of the alternating group $A_5$ shows.
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