How to demonstrate that a simple non-abelian group of odd order has order divisible by the cube of its smallest prime divisor?
Exercise 6.3.16 from Scott, Group Theory
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group-theory
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6This requires quite a bit of theory, so you need to indicate what background you are assuming. Have you seen the transfer homomorphism, for example? – 2011-12-19
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1@GeoffRobinson: I just checked Scott's book; transfer has been covered. To the OP: What do you know about groups of order $p^2$ or order $p$? Why do you think the question uses the *smallest* prime divisor? Why do you think Geoff mentioned transfer? – 2011-12-19
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0Non-abelian simple group of *odd* order? Such a group does not exist ... Or am I misunderstanding the exercise? – 2011-12-19
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0Nicky is correct - it comes from Feit-Thomson. – 2011-12-19
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0There are much simpler arguments than Feit-Thompson here! – 2011-12-19
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1Presumably, since he said it was an exercise from a book, he doesn't want to know that it is true, but how he could prove it given what the book has taught him so far. – 2011-12-19
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1@SteveD, there's a simpler argument than F-T to show that there are no odd-order non-abelian simple groups? – 2011-12-19
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0@MarianoSuárez-Alvarez: Of course not. I meant this exercise can be solved without an appeal to F-T. – 2011-12-19
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0Ah :) ${}{}{}{}$ – 2011-12-19
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2In 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