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.