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Recall some elementary algebra. For n ∈ N and x ∈ K, an expression nx stands for an n-fold sum x + · · · + x. A relation nx = 1 exists in K if and only if either K has characteristic zero or the characteristic p > 0 of K does not divide n. In this case we say, n is invertible in K. We denote this inverse as usual by n−1 . 5) Proposition. If KG is semi-simple, then |G| is invertible in K. Proof. Suppose KG is semi-simple. Then the fixed set F = {λΣ | λ ∈ K, Σ = g∈G g} is a sub-representation. Hence there exists a projection p : KG → F .

There is another Z-basis which consists of the permutation representations Q(Cm /Cn ) = Pm/n . The representation Pm/n contains the irreducible representations which have Cn in its kernel. The kernel of Vd is Cm/d . Hence Pk = obius-inversion one obtains d|k Vd . By M¨ Vk = d|k µ(k/d)Pd . 5) Example. Suppose the characteristic of K does not divide the order of G. We describe R(Cm ; K). Let L = K(ε) be the field extension, ε a primitive m-th root of unity. 3), R(Cm ; L) ∼ = Z[ρ]/(ρm − 1) ∗ where ρ : Cm → L is given by ρ(a) = ε.

Since these units are rational representations, R(G; Q) has the same units of finite order as R(G; R). 5 Cyclic Induction Let F be a set of subgroups of G.

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