Brian Bi
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Remark The proof of theorem 5.4.4 relies on the result that the matrices of a finite group representation over \(\mathbb{C}\) are diagonalizable. This useful fact was never explicitly proven in the text. The proof is as follows: suppose \(G\) is a finite group, \(g \in G\), and \(V\) is a complex finite-dimensional representation of \(G\). Then the matrix \(M = \rho(g)\) satisfies \(M^k = I\) for some positive integer \(k\), so that \(M^k - I = 0\). The minimal polynomial of \(M\) divides \(x^k - 1\), so it factors as \((x - \zeta_1) \cdot \ldots \cdot (x - \zeta_m)\) for some distinct roots of unity \(\zeta_i\). Therefore \(M\) is diagonalizable.

Remark In the proof of Burnside's theorem, it is asserted that if \(G\) is the smallest nonsolvable group of order \(p^a q^b\), then \(G\) is simple. To see why, suppose \(G\) were not simple. Then it would have some nontrivial proper normal subgroup \(H\), and both of the groups \(H\) and \(G/H\), being smaller than \(G\), must be solvable. But \(H\) and \(G/H\) being both solvable would imply \(G\) is solvable as well, a contradiction.