By S. Zaidman

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**Example text**

J 4. homC (M, N ) ∼ = N ⊗ M ∗ so this follows from the previous two parts. 3. Let G act on the finite set Ω and let CΩ be the corresponding permutation module and χΩ its character. For g ∈ G let fixΩ (g) = {ω ∈ Ω : g · ω = ω}. Then for any g ∈ G we have χΩ (g) = | fixΩ (g)|. Version of Sunday 29th March, 2015 at 20:00 40 Proof. Consider the matrix for the action of g with respect to the basis Ω of CΩ. The column of this matrix corresponding to ω ∈ Ω has a 1 at position g · ω and zeros elsewhere, so it contributes 1 to the trace if g · ω = ω and zero otherwise.

Let m1 , . . be a basis of M such that mi is an eigenvector for the action of g with eigenvalue µi . Then dim M = |χ(g)| = | µj | ≤ j |µj | = dim M j since µj is a root of unity and therefore has absolute value 1. The triangle inequality for complex numbers says |z + w| ≤ |z| + |w| with equality if and only if z and w have the same argument. Using this we get that each µj has the same argument, and so since they all have absolute value 1, they are all equal, say to µ. Then χ(g) = (dim M )µ, but χ(g) = dim M , so µ = 1.

Ns of N such that mi and nj are an eigenvectors for the action of g with eigenvalues µi νj respectively. Thus χM (g) = i µi and χN (g) = j νj . 1. Let ρM : G → GLr (C) and ρN : G → GLs (C) be the matrix representations corresponding to our bases, so that χM (g) = Tr ρM (g) and χN (g) = Tr ρN (g). The module M ⊕ N has a basis consisting of the (mi , 0) and (0, nj ), and the matrix representation ρ corresponding to this basis is ρM (g) 0 ρ(g) = . 0 ρN (g) Thus χM ⊕N (g) = Tr ρ(g) = Tr ρM (g) + Tr ρN (g) = χM (g) + χN (g).