Let \(G\) be a finite group and \(\mathbb{Z}[G]\) its integral group ring. Let \(\mathcal{Z} = Z(\mathbb{Z}[G])\) be the center of \(\mathbb{Z}[G]\). For each conjugacy class \(C \subseteq G\), let \(P_C = \sum_{g \in C} g\) be the class sum.
- Show that the class sums \(\{P_C\}\) form a \(\mathbb{Z}\)-basis for \(\mathcal{Z}\). Hence \(\mathcal{Z} \cong \mathbb{Z}^d\) as an abelian group, where \(d\) is the number of conjugacy classes in \(G\).
- Show that if a ring \(R\) is isomorphic to \(\mathbb{Z}^d\) as an abelian group, then every element in \(R\) satisfies a monic integral polynomial (i.e. is integral over \(\mathbb{Z}\)).
- Let \(\pi: G \to \operatorname{GL}(V)\) be an irreducible complex representation of \(G\). Show that \(\pi(P_C)\) acts on \(V\) as multiplication by the scalar: \[\omega_\pi(C) = \frac{|C| \chi_\pi(C)}{\dim V}\] where \(\chi_\pi(C)\) is the value of the character \(\chi_\pi\) on any element of \(C\).
- Conclude that \(\frac{|C| \chi_\pi(C)}{\dim V}\) is an algebraic integer.
Solution
Goal: Prove that class sums form a \(\mathbb{Z}\)-basis of \(Z(\mathbb{Z}[G])\), that finite free \(\mathbb{Z}\)-algebras are integral, that \(\pi(P_C)\) acts by scalar \(\omega_\pi(C)\) via Schur’s Lemma, and that \(\omega_\pi(C)\) is an algebraic integer.
1. Part 1: Class Sums Form a \(\mathbb{Z}\)-Basis for \(\mathcal{Z} = Z(\mathbb{Z}[G])\):Proof:
1.1. Let \(x = \sum_{g \in G} a_g g \in \mathbb{Z}[G]\) with \(a_g \in \mathbb{Z}\).
1.2. The element \(x\) lies in the center \(\mathcal{Z}\) if and only if \(h x h^{-1} = x\) for all \(h \in G\):\[h x h^{-1} = \sum_{g \in G} a_g (h g h^{-1}) = \sum_{y \in G} a_{h^{-1} y h} y = \sum_{y \in G} a_y y.\]
1.3. Equating coefficients of basis elements \(y \in G\):\[a_{h^{-1} y h} = a_y \quad \text{for all } h, y \in G.\]
1.4. This means the coefficient function \(g \mapsto a_g\) is constant on each conjugacy class \(C\) of \(G\).
1.5. Let \(C_1, C_2, \dots, C_d\) be the distinct conjugacy classes of \(G\).
1.6. Then \(x\) can be written uniquely as:\[x = \sum_{i=1}^d c_i \left( \sum_{g \in C_i} g \right) = \sum_{i=1}^d c_i P_{C_i} \quad (c_i \in \mathbb{Z}).\]
1.7. Since the conjugacy classes \(C_i\) partition \(G\), the elements \(P_{C_1}, \dots, P_{C_d}\) are support-disjoint in \(G\), hence linearly independent over \(\mathbb{Z}\).
1.8. Thus \(\{P_C\}\) forms a \(\mathbb{Z}\)-basis of \(\mathcal{Z}\), and \(\mathcal{Z} \cong \mathbb{Z}^d\) as an abelian group.
2. Part 2: Rings with \(R \cong \mathbb{Z}^d\) are Integral over \(\mathbb{Z}\):Proof:
2.1. Let \(R\) be a ring whose additive group is free abelian of finite rank \(d\), with \(\mathbb{Z}\)-basis \(\{v_1, v_2, \dots, v_d\}\).
2.2. For any fixed element \(r \in R\), multiplication by \(r\) defines an additive group endomorphism \(L_r: R \to R\) (\(L_r(x) = r x\)).
2.3. Expressing \(L_r(v_i)\) in the basis:\[r v_i = \sum_{j=1}^d a_{ij} v_j \quad \text{where } A = (a_{ij}) \in M_d(\mathbb{Z}).\]
2.4. In matrix-vector notation: \((r I_d - A) \mathbf{v} = \mathbf{0}\), where \(\mathbf{v} = [v_1, \dots, v_d]^T\).
2.5. Multiplying by the adjugate matrix \(\operatorname{adj}(r I_d - A)\):\[\det(r I_d - A) \cdot \mathbf{v} = \mathbf{0} \implies \det(r I_d - A) v_i = 0 \quad \text{for all } i = 1, \dots, d.\]
2.6. Since \(\{v_1, \dots, v_d\}\) is a \(\mathbb{Z}\)-basis of \(R\), the identity element \(1 = \sum c_i v_i \in R\) satisfies \(\det(r I_d - A) \cdot 1 = 0\), so:\[\det(r I_d - A) = 0 \in R.\]
2.7. The characteristic polynomial \(p_A(t) = \det(t I_d - A) \in \mathbb{Z}[t]\) is a monic polynomial with integer coefficients of degree \(d\).
2.8. Since \(p_A(r) = 0\), every element \(r \in R\) is an algebraic integer (integral over \(\mathbb{Z}\)).
3. Part 3: Action of \(\pi(P_C)\) via Schur’s Lemma:Proof:
3.1. Let \((\pi, V)\) be an irreducible complex representation of \(G\), and let \(n = \dim_\mathbb{C}(V)\).
3.2. Since \(P_C \in \mathcal{Z}\), for every \(h \in G\), \(\pi(h) \pi(P_C) = \pi(h P_C) = \pi(P_C h) = \pi(P_C) \pi(h)\).
3.3. By Schur’s Lemma, any \(G\)-equivariant endomorphism of an irreducible complex representation is a scalar multiple of the identity:\[\pi(P_C) = \lambda_C \operatorname{id}_V \quad \text{for some scalar } \lambda_C \in \mathbb{C}.\]
3.4. We compute the trace of both sides:\[\operatorname{Tr}(\pi(P_C)) = \operatorname{Tr}(\lambda_C \operatorname{id}_V) = \lambda_C \dim(V) = n \lambda_C.\]
3.5. On the other hand, by linearity of trace and definition of character \(\chi_\pi\):\[\operatorname{Tr}(\pi(P_C)) = \sum_{g \in C} \operatorname{Tr}(\pi(g)) = \sum_{g \in C} \chi_\pi(g) = |C| \chi_\pi(C)\]since \(\chi_\pi\) is constant on the conjugacy class \(C\).
3.6. Equating the two trace computations:\[n \lambda_C = |C| \chi_\pi(C) \implies \lambda_C = \frac{|C| \chi_\pi(C)}{\dim V}.\]
4. Part 4: \(\frac{|C|\chi_\pi(C)}{\dim V}\) is an Algebraic Integer:Proof:
4.1. The algebra homomorphism \(\pi: \mathbb{Z}[G] \to \operatorname{End}_\mathbb{C}(V)\) restricts on the center to a ring homomorphism:\[\pi|_\mathcal{Z}: \mathcal{Z} \longrightarrow \mathbb{C} \cdot \operatorname{id}_V \cong \mathbb{C}, \qquad P_C \longmapsto \lambda_C = \frac{|C|\chi_\pi(C)}{\dim V}.\]
4.2. By Part 1, \(\mathcal{Z} \cong \mathbb{Z}^d\) as an abelian group.
4.3. By Part 2, the class sum \(P_C \in \mathcal{Z}\) satisfies a monic polynomial \(p(t) \in \mathbb{Z}[t]\) with \(p(P_C) = 0\).
4.4. Applying the ring homomorphism \(\pi\):\[p(\lambda_C) = p\left(\frac{|C|\chi_\pi(C)}{\dim V}\right) = 0.\]
4.5. Since \(\lambda_C\) is a root of a monic polynomial with integer coefficients, \(\frac{|C|\chi_\pi(C)}{\dim V}\) is an algebraic integer.
5. Conclusion:Class sums form a \(\mathbb{Z}\)-basis of \(\mathcal{Z} \cong \mathbb{Z}^d\), whose elements are integral over \(\mathbb{Z}\); by Schur’s Lemma \(\pi(P_C) = \omega_\pi(C) \operatorname{id}_V\), so \(\omega_\pi(C) = \frac{|C|\chi_\pi(C)}{\dim V}\) is an algebraic integer. Q.E.D.