Suppose that $G$ is a finite group such that $(|G|, 15)=1$. Why $G$ is solvable group?
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1$\begingroup$ Is there any reason to think that this question will have an answer simpler than the proof of the odd order theorem, which is the same claim with $2$ in place of $15$? $\endgroup$– David E SpeyerSep 11, 2012 at 14:46
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1$\begingroup$ @David Speyer: I have studied in a paper " it is not hard to prove by the classification that if $(|G|,15)=1$, then $G$ is solvable". So I thought that it has an answer simpler. $\endgroup$– A ESep 11, 2012 at 15:04
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$\begingroup$ If you allow the classification, the odd order theorem is proved the same way: Look at the table of finite noncyclic simple groups; notice they all have even order. $\endgroup$– David E SpeyerSep 11, 2012 at 18:00
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1$\begingroup$ I think the odd order theorem is an explicit step in the Classification, while the "$(G,15)$ is not. $\endgroup$– YCorSep 11, 2012 at 22:47
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1$\begingroup$ en.wikipedia.org/wiki/N-group_(finite_group_theory) $\endgroup$– Ian AgolSep 11, 2012 at 23:58
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2 Answers
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We note that the Suzuki groups are only non-Abelian simple groups of order prime to $3$ and $5$ is a prime divisor of the Suzuki groups. Let $G$ be unsolvable group, then $G$ has the following normal series: $1\unlhd K\lhd M\unlhd G$ such that $M/K$ is a non-Abelian simple group (and or $M/K\cong S\times $ $ S\cdot \cdot \cdot \times S$ where $S$ is non-Abelian simple group). As $ 3\nmid |G|$, then $M/K$ is a Suzuki group ( and or $M/K\cong S\times $ $ S\cdot \cdot \cdot \times S$ where $S$ is a Suzuki group). On the other hand $5\nmid |G|$, then $M/K$ is not isomorphic to a Suzuki group (and or $ M/K\ncong S\times $ $S\cdot \cdot \cdot \times S$ where $S$ is a Suzuki group), a contradiction.
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Because as you go through the list of finite simple noncyclic groups, you can observe that their orders are divisible by either 3 or 5. In fact, only a few of them, for instance, $\,^2B_2(8)$, will have orders not divisible by 3...
Off course, this argument depends on the classification... Maybe, there is a trick to show it directly...
answered Sep 11, 2012 at 13:27
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2$\begingroup$ Certainly, one should be able to get away with Thompson's classification of minimal non-solvable groups, which predates the classification. $\endgroup$ Sep 11, 2012 at 13:52