the same conjugacy class as 1, the orthogonality relation for the columns of the group's character table gives: Now the χi(g) are algebraic integers, because they are sums of roots of unity. If all the nontrivial irreducible characters which don't vanish at g take a value divisible by q at 1, we deduce that is an algebraic integer (since it is a sum of integer multiples of algebraic integers), which is absurd. This proves the statement. The complex number qdχ(g)/n is an algebraic integer. The set of integer-valued class functions on G, Z([G]), is a commutative ring, finitely generated over . All of its elements are thus integral over , in particular the mapping u which takes the value 1 on the conjugacy class of g and 0 elsewhere. The mapping which sends a class function f to is a ring homomorphism. Because ρ(s)−1A(u)ρ(s) = A(u) for all s, Schur's lemma implies that A(u) is a homothety λIn. Its trace nλ is equal to Because the homothety λIn is the homomorphic image of an integral element, this proves that the complex number λ = qdχ(g)/n is an algebraic integer. The complex number χ(g)/n is an algebraic integer. Since q is relatively prime to n, by Bézout's identity there are two integers x and y such that: Because a linear combination with integer coefficients of algebraic integers is again an algebraic integer, this proves the statement. The image of g, under the representation ρ, is a homothety. Let ζ be the complex number χ(g)/n. It is an algebraic integer, so its norm N(ζ) (i.e. the product of its conjugates, that is the roots of its minimal polynomial over ) is a nonzero integer. Now ζ is the average of roots of unity (the eigenvalues of ρ(g)), hence so are its conjugates, so they all have an absolute value less than or equal to 1. Because the absolute value of their product N(ζ) is greater than or equal to 1, their absolute value must all be 1, in particular ζ, which means that the eigenvalues of ρ(g) are all equal, so ρ(g) is a homothety. Conclusion Let N be the kernel of ρ. The homothety ρ(g) is central in Im(ρ) (which is canonically isomorphic to G/N), whereas g is not central in G. Consequently, the normal subgroup N of the simple group G is nontrivial, hence it is equal to G, which