# Exercise 6.3.16 from Scott, Group Theory

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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This requires quite a bit of theory, so you need to indicate what background you are assuming. Have you seen the transfer homomorphism, for example? – Geoff Robinson Dec 19 '11 at 19:20
@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? – user641 Dec 19 '11 at 20:34
Presumably, 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. – Thomas Andrews Dec 19 '11 at 21:56
@SteveD, there's a simpler argument than F-T to show that there are no odd-order non-abelian simple groups? – Mariano Suárez-Alvarez Dec 19 '11 at 22:23
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. – Geoff Robinson Dec 19 '11 at 23:04