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Prove that a group of order 150 is not simple.

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closed as off-topic by Jonas Meyer, LutzL, Claude Leibovici, draks ..., daw Mar 25 at 8:11

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(Please) Don't give orders. –  Rasmus Dec 12 '12 at 16:46
Is this homework? What have you done? Where are you stuck? Try to write your question in a nicer language. –  DonAntonio Dec 12 '12 at 16:49
@hmmmm A problem in S.-T. Yau College Student Mathematics Contests. The book I used to learn abstract algebra does not consider group operation it only use sylow-theorem to slove problem like group of order 36 15 100, a fellow student said I'd better consider group operation when discuss the p-sylow group,is it a must? –  Jebei Dec 12 '12 at 17:05
You really mean you want to consider a group without an operation ??!? And what that would mean? –  Andrea Mori Dec 12 '12 at 18:11
Ok, for "action" instead of "operation", but it should be observed that the usual proof of the Sylow Theorem makes use of certain group actions. –  Andrea Mori Dec 12 '12 at 19:54

2 Answers 2

up vote 0 down vote accepted

If we have the following:

For any finite group $G$ and subgroup $H$ we have $|G:H|=n$ then there exists a normal subgroup $K$ of $G$ contained in $H$ and $|G:K|$ divides $n!$.

So suppose that we have that $G$ is not simple and so by Sylow theory we have $6$ subgroups of order $25$, denote one $X$. We also have that $K=\{e\}$ above.

Then consider $N_{G}(X)$

Now we have that $|G:N_{G}(X)|=6$

Then as $|K|=1$ we have $|G|$ divides $6!$ but $150$ does not so this is a contradiction and so $G$ is simple.

Although looking at it I think we need an argument using group actions to prove the proposition that I used so it probably doesn't really help. (It's basically the argument in the other answer but using my proposition to cover up the group action I think.)

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Why did somebody down vote? I'm not saying there is nothing wrong with my answer just if there is I would like to know what?-Thanks very much –  hmmmm Dec 12 '12 at 21:57

A little different approach: if the 5-Sylow subgroup is not normal (and thus not unique), then there are 6 such subgroups, which means that the normalizer of any of them has index 6.

But if the group were simple, then making it act on the left cosets of that normalizer we'd get an embedding of the group into $\,S_6\,$ , something that is absurd as $\,150\nmid6!$

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