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How to demonstrate that a simple non-abelian group of odd order has order divisible by the cube of its smallest prime divisor?

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    In fact, I believe it is already an exercise in Burnside's book that the order of a finite simple group is divisible either by $12$ or by the cube of its smallest prime divisor. It is also interesting that a slighlt more careful analysis allowed Feit and Thompson to conclude that if a finite group $G$ of odd order does not contain an elementary Abelian subgroup of order $p^3$, where $p$ is the smallest prime divisor of its order, then $G$ is not simple. The existence of elementary Abelian subgroups of rank $3$ is an important issue in the odd order theorem proof.2011-12-19

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