While \(\chi(g)/\chi(1)\) is not, in general, an algebraic integer, if we multiply it by \(|g^G|\) then it becomes one. To prove statements like this we generally use either conjugacy class sums or central idempotents.

Recall from Section 1.3 and Section 1.5 that \(\mathbb{C}G\) is a direct sum of matrix algebras, and we may write \(1\) as a sum of central idempotents \(e\), one for each matrix algebra. The central idempotents \(e\) can be written as a sum \(\sum_g a_gg\), and the coefficients \(a_g\) are equal for conjugate elements \(g\). Our first result is to give a formula for \(a_g\).

Proposition 3.8. Let \(\chi\in\mathrm{Irr}(G)\), and let \(e_\chi\) denote the associated central idempotent. Then\[ e_\chi=\frac{\chi(1)}{|G|}\sum_{g\in G} \chi(g^{-1})g.\]
Proof: Write \(e_\chi=\sum_g a_gg\), so we claim that \(a_g=(\chi(1)/|G|)\cdot \chi(g^{-1})\). Let \(\rho\) denote the regular representation of \(G\), and extend it by linearity to a representation of \(\mathbb{C}G\). Consider \((e_\chi g^{-1})\rho\), and in particular its trace. On the one hand, \(\rho\) has trace \(0\) for all non-trivial elements of \(G\) and \(|G|\) for the identity, so the trace is \(|G|a_g\). On the other hand, \(\rho\) is the sum of \(\deg\sigma\) copies of each irreducible representation \(\sigma\) of \(G\). This has trace\[\sum_{\psi\in \mathrm{Irr}(G)} \psi(1)\psi(e_\chi g^{-1}).\]If \(\pi\) is the representation associated to \(\psi\in\mathrm{Irr}(G)\), then \((e_\chi g^{-1})\pi=(e_\chi\pi)(g^{-1}\pi)\). If \(\psi\neq \chi\) (i.e., \(\pi\) does not have character \(\chi\)) then \(e_\chi\pi=0\) (because \(e_\chi\) annihilates all simple modules other than the one with character \(\chi\)), and if \(\pi\) has character \(\chi\) then \(e_\chi\pi\) is the identity matrix. Thus the trace of \((e_\chi g^{-1})\pi\) is simply \(\chi(g^{-1})\). Thus the terms in the sum \(\sum_{\psi\in \mathrm{Irr}(G)} \psi(1)\psi(e_\chi g^{-1})\) are zero unless \(\psi=\chi\), and for \(\psi=\chi\) we have contribution \(\chi(1)\chi(g^{-1})\).

Thus \(a_g=(\chi(1)/|G|)\cdot \chi(g^{-1})\), as claimed. □

Next, we consider an irreducible representation \(\rho\), and its action on a class sum (i.e., a sum of all elements in a conjugacy class of \(G\)). Since class sums under \(\rho\) are central in \(\mathrm{GL}(V)\) (where \(\rho:G\to\mathrm{GL}(V)\)) the image is a scalar matrix.

Proposition 3.9. Let \(G\) be a finite group, let \(C\) be a conjugacy class of \(G\), \(g\in C\), and let \(\rho\) be an irreducible representation with character \(\chi\). If \(\bar C\) denotes the sum of all elements of \(C\), then \(\bar C\rho=\lambda I\), where \(I\) is the identity matrix and\[ \lambda=|C|\cdot \frac{\chi(g)}{\chi(1)}.\]Furthermore, \(\lambda\) is an algebraic integer.

Proof: We take traces. First, if \(g\in C\) then \(g\rho\) has trace \(\chi(g)\), so the trace of \(\bar C\rho\) is \(|C|\cdot\chi(g)\). On the other hand, the trace is \(\lambda\cdot\chi(1)\) for some \(\lambda\), whence we obtain the result.

To prove that \(\lambda\) is an algebraic integer, we find \(\lambda\) as the eigenvalue of some integer-valued matrix. The (right) regular representation \(\pi\) is clearly integer valued, and so the matrix action of \(\bar C\) on it is also integer valued, hence the eigenvalues for any element of \(\mathbb{C}G\) are algebraic integers. On the other hand, \(\rho\) is equivalent to a subrepresentation of \(\pi\), so the eigenvalues of \(\bar C\rho\) are eigenvalues of \(\bar C\pi\).

This proves that \(\lambda\) is an algebraic integer. □

While \(\chi(g)/\chi(1)\) need not be an algebraic integer in general, we see from the above that \(|g^G|\cdot \chi(g)/\chi(1)\) is an algebraic integer. This is enough to prove the following important result.

Corollary 3.10. If \(\chi\) is an irreducible character of the finite group \(G\), then \(\chi(1)\) is a divisor of \(|G|\).

Proof: Take the row orthogonality relation between \(\chi\) and itself. We have\[ \sum_{i=1}^r |g_i^G|\chi(g_i)\overline{\chi(g_i)}=|G|,\]where \(g_1,\dots,g_r\) are conjugacy class representatives for \(G\). We may divide both sides by \(\chi(1)\) to obtain\[ \sum_{i=1}^r\left(|g_i^G|\frac{\chi(g_i)}{\chi(1)}\right)\overline{\chi(g_i)}=\frac{|G|}{\chi(1)}.\]The left-hand side is a sum of products of algebraic integers, with the important one being Proposition 3.9, so the right-hand side must be an algebraic integer, hence an integer as it is a rational number. This proves the result. □

Having proved this, we have one more result in this section that has a number of applications. We start with a definition.

Definition 3.11. Let \(G\) be a finite group, and let \(C_1,\dots,C_r\) be the conjugacy classes of \(G\), with class sums \(\bar C_i\). Viewed as elements of \(\mathbb{C}G\), we have that \(\bar C_i\bar C_j\) is a linear combination of class sums. Define \(a_{ijk}\) by\[ \bar C_i\bar C_j=\sum_{k=1}^r a_{ijk}\bar C_k.\]The \(a_{ijk}\) are the class structure constants.

Another definition of \(a_{ijk}\) is it is the number of solutions to \(xy=z\), where \(x\in C_i\), \(y\in C_j\) and \(z\in C_k\). A character-theoretic formula for \(a_{ijk}\) is easy to prove, using our previous results.

Theorem 3.12. Let \(G\) be a finite group, and let \(C_1,\dots,C_r\) be the conjugacy classes of \(G\), with class sums \(\bar C_i\). Let \(a_{ijk}\) denote the class structure constants. Letting \(x_i\) be any element of \(C_i\), we have\[ a_{ijk}=\frac{|C_i|\cdot |C_j|}{|G|}\sum_{\chi\in\mathrm{Irr}(G)} \frac{\chi(x_i)\chi(x_j)\overline{\chi(x_k)}}{\chi(1)}.\]
Proof: Let \(\rho\) be an irreducible representation, with character \(\chi\). On the one hand, via Proposition 3.9,\[ (\bar C_i\bar C_j)\rho=|C_i|\cdot |C_j|\cdot \frac{\chi(x_i)\chi(x_j)}{\chi(1)^2}I.\]On the other, we obtain\[ \sum_{k=1}^r a_{ijl}\bar C_l\rho=\sum_{l=1}^r a_{ijl} |C_l|\frac{\chi(g)}{\chi(1)}I=\frac{1}{\chi(1)}\left(\sum_{l=1}^r a_{ijl} |C_l|\chi(g)\right)I.\]Thus \(|C_i|\cdot|C_j|\cdot\chi(x_i)\chi(x_j)/\chi(1)=\sum_l a_{ijl}|C_l|\cdot\chi(x_l)\). Multiply both sides by \(\overline{\chi(x_k)}\) for some \(1\leq k\leq r\), and sum over all irreducible characters \(\chi\) to obtain\[|C_i|\cdot|C_j|\sum_{\chi\in\mathrm{Irr}(G)}\frac{\chi(x_i)\chi(x_j)\overline{\chi(x_k)}}{\chi(1)}=\sum_{l=1}^r a_{ijl}|C_l|\sum_{\chi\in\mathrm{Irr}(G)}\chi(x_l)\overline{\chi(x_k)}.\]The summation term on the right-hand side is \(0\) unless \(l=k\), and \(|C_G(x_k)|=|G|/|C_k|\) if \(l=k\), by the column orthogonality relation. Substituting this yields\[|C_i|\cdot|C_j|\sum_{\chi\in\mathrm{Irr}(G)}\frac{\chi(x_i)\chi(x_j)\overline{\chi(x_k)}}{\chi(1)}=a_{ijk}|C_k|\frac{|G|}{|C_k|}.\]From this we clearly obtain the desired result. □

Example 3.13. Let \(G\) be the simple group \(\mathrm{GL}_3(2)=\mathrm{PSL}_2(7)\). The conjugacy classes are labelled by \(g_1,\dots,g_6\), of orders \(1,2,3,4,7,7\), and sizes 1, 21, 56, 42, 24, 24. Thus \(168\) is the sum of six squares, one of which is \(1\). There are five such possibilities (from a computer check!), but Exercise 3.1 shows that \(2\) is not a character degree, so the degrees are at least \(3\). This eliminates three of the five, leaving\[ 168=1^2+3^2+3^2+6^2+7^2+8^2=1^2+3^2+4^2+5^2+6^2+9^2.\]The second of these cannot work as \(9\not\mid |G|\), so this furnishes us with the character degrees: 1, 3, 3, 6, 7, 8. Label these \(\chi_1\) to \(\chi_6\).

(It is possible to work this out by hand. Since \(\sqrt{168}\lt 13\), the character degrees are at most \(12\). Certainly \(12\) isn't possible as the other degrees are at least \(3\), and since \(\chi(1)\mid 168\), the other characters have degree at most \(8\). Take congruences modulo \(8\): \(1^2\), \(3^2\), \(7^2\) all are \(1\) modulo \(8\), \(4\) and \(8\) square to zero modulo \(8\), and \(6\) squares to \(4\) modulo \(8\). Thus to get six squares to be \(0\) modulo \(8\) we need four from \(1\), \(3\) and \(7\), one \(6\) and one \(4\) or \(8\). Working modulo \(16\) we have \(1^2\equiv 7^2\equiv 1\), \(3^2\equiv 9\), \(4^2\equiv 8^2\equiv 0\) and \(6^2\equiv 4\), and our degrees must sum to \(8\) modulo \(16\), with a guaranteed \(1\) and \(6\), and one \(4\) or \(8\). We see we need \(1\), \(3\), \(3\), \(7\), \(6\) and either \(4\) or \(8\). We must get \(8\) as they have to sum square to \(168\).)

Since \(G\cong \mathrm{PSL}_2(7)\), it acts transitively on the projective line, which has size \(|\mathbb{F}_q|+1\). Thus \(G\) embeds in \(A_8\). (In fact \(S_8\), but \(G\) has no subgroup of index \(2\), so \(G\leq A_8\).) There is a single trivial composition factor in this transitive representation, and the remainder has degree \(7\). Thus this is \(\chi_5\). Clearly an element of order \(7\) in this action has a single fixed point, so \(\chi_5(x_5)=\chi_5(x_6)=0\). Elements of order \(4\) must act as a \(4\)-cycle and a transposition or two \(4\)-cycles, as they are even permutations, so \(\chi_5(x_4)=\pm 1\). For \(x_2\), it is either a double transposition or quadruple transposition, so \(\chi_5(x_2)\in\{-1,3\}\). Finally, \(\chi_5(x_3)\in\{1,4\}\) since it acts as \((1,2,3)\) or \((1,2,3)(4,5,6)\), hence fixes two or five points. So far we have the following table.

	\(1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(x_6\)
\(\chi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\chi_2\)	\(3\)
\(\chi_3\)	\(3\)
\(\chi_4\)	\(6\)
\(\chi_5\)	\(7\)	\(a\)	\(b\)	\(c\)	\(0\)	\(0\)
\(\chi_6\)	\(8\)

The inner product \(\langle\chi_1,\chi_5\rangle\) is \(0\), so we must have that \(7+21a+56b+42c=0\), where \(a=-1,3\), \(b=1,4\), \(c=1,-1\). Thus \(a=c=-1\), \(b=1\).

	\(1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(x_6\)
\(\chi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\chi_2\)	\(3\)
\(\chi_3\)	\(3\)
\(\chi_4\)	\(6\)
\(\chi_5\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(0\)
\(\chi_6\)	\(8\)

Also \(G\cong \mathrm{GL}_3(2)\), so it acts on the seven non-zero vectors in an \(\mathbb{F}_2\)-vector space. This yields an embedding of \(G\) into \(A_7\). Again we can compute this permutation character and it must have values \(7,3,b,1,0,0\), where \(b=4,1\). (The values on \(x_2\) and \(x_4\) are now fixed since they cannot move eight points.) The norm of this character is
\[ \frac1{168} (7^2+21\cdot 3^2+56b^2+42\cdot 1^2)=\frac{280+56b^2}{168}.\]If \(b=1\) then we get norm \(2\), and if \(b=4\) then we get norm \(8/3\), so \(b=1\) and this permutation character is \(\chi_1+\chi_4\).

	\(1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(x_6\)
\(\chi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\chi_2\)	\(3\)
\(\chi_3\)	\(3\)
\(\chi_4\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)	\(-1\)
\(\chi_5\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(0\)
\(\chi_6\)	\(8\)

We claim that \(x_5^{-1}\) is conjugate to \(x_6\). This implies that \(\chi_3=\chi_2^*\) since there must be a non-real irreducible character. If there are two conjugacy classes of elements of order \(7\) then we can see which powers of \(x_5\) are conjugate to \(x_5\) by looking at a non-abelian group of order \(21\). This shows that \(x_5\), \(x_5^2\) and \(x_5^4\) are conjugate, and \(x_5^3\), \(x_5^5\) and \(x_5^6\) are conjugate. in particular, \(x_5\) and \(x_6=x_5^{-1}\) are not conjugate, so there are non-real irreducible characters. These must have degree \(3\).

Also, because \(G\) is simple, any homomorphism \(G\to \mathrm{GL}_3(\mathbb{C})\) must have image in \(\mathrm{SL}_3(\mathbb{C})\), so the determinant of the matrices involved is \(1\). This is the product of the eigenvalues, so the eigenvalues of \(x_2\) are \(-1,-1,+1\), and \(\chi_2(x_2)=\chi_3(x_2)=-1\).

	\(1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(x_6\)
\(\chi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\chi_2\)	\(3\)	\(-1\)	\(b\)	\(c\)	\(d\)	\(\bar d\)
\(\chi_3\)	\(3\)	\(-1\)	\(b\)	\(c\)	\(\bar d\)	\(d\)
\(\chi_4\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)	\(-1\)
\(\chi_5\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(0\)
\(\chi_6\)	\(8\)

Column orthogonality between columns 1 and 2 yields \(\chi_6(x_2)=0\). We can take \(\langle\chi_2,\chi_4\rangle\) to obtain\[ 0=|G|\langle\chi_2,\chi_4\rangle=18+21\cdot(-1)\cdot2+24\cdot(-1)\cdot(d+\bar d)=-24-24(d+\bar d).\]Thus \(d+\bar d=-1\). Hence \(d+\bar d\) is the sum of the six primitive seventh roots of unity, and since \(x_5\), \(x_5^3\) and \(x_5^4\) are conjugate, we have that the eigenvalues of the representation for \(x_5\) are \(\zeta\), \(\zeta^3\) and \(\zeta^4\), where \(\zeta^7=1\). This has an alternative expression as \(\alpha=(-1\pm 7\mathrm{i})/2\).

The inner products \(\langle\chi_2,\chi_1\rangle\) and \(\langle\chi_2,\chi_5\rangle\) are also both zero, and are (up to \(1/|G|\))\[ 3+21\cdot(-1)+56b+42c+24(d+\bar d)=-42+56b+42c=0,\]and\[ 21+21(-1)^2+56b-42c=42+56b-42c=0.\]Thus \(b=0\) and \(c=1\).

	\(1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(x_6\)
\(\chi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\chi_2\)	\(3\)	\(-1\)	\(0\)	\(1\)	\(\alpha\)	\(\bar \alpha\)
\(\chi_3\)	\(3\)	\(-1\)	\(0\)	\(1\)	\(\bar \alpha\)	\(\alpha\)
\(\chi_4\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)	\(-1\)
\(\chi_5\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(0\)
\(\chi_6\)	\(8\)	\(0\)

Now column orthogonality with the first column yields the row \(\chi_6\), and we have the complete table.

	\(1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(x_6\)
\(\chi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\chi_2\)	\(3\)	\(-1\)	\(0\)	\(1\)	\(\alpha\)	\(\bar \alpha\)
\(\chi_3\)	\(3\)	\(-1\)	\(0\)	\(1\)	\(\bar \alpha\)	\(\alpha\)
\(\chi_4\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)	\(-1\)
\(\chi_5\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(0\)
\(\chi_6\)	\(8\)	\(0\)	\(-1\)	\(0\)	\(1\)	\(1\)