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We know that:

Theorem: If a simple group $G$ has a proper subgroup $H$ such that $[G:H]=n$ then $G\hookrightarrow A_n$.

This fact can help us to prove that any group of $G$ of order $120$ is not simple. In fact, since $n_5(G)=6$ then $[G:N_G(P)]=6$ where $P\in Syl_5(G)$ and so $A_6$ has a subgroup of order $120$ which is impossible. My question is:

Can we prove that $G$ of order $120$ is not simple without employing the theorem? Thanks.

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Just curious: Are you refering to some specific approaches when asking the question? Thanks. –  awllower Aug 6 '12 at 3:00
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Well, you could classify all groups of order 120. Why do you want to avoid the theorem? –  Kevin Carlson Aug 6 '12 at 4:54
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It does seem to me that this probably is the easiest way to prove that there is no simple group of order $120$. There are other ways to do it, of course, but the ones I can think of are more complicated than that. –  Geoff Robinson Aug 6 '12 at 7:02
    
@GeoffRobinson: Yes, it does, but I am just curious if there is other way helping us for such this kinds of problems. Thanks for the time. Thanks for all comments. –  B. S. Aug 6 '12 at 7:23
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1 Answer 1

up vote 2 down vote accepted

Well, you can obtain a contradiction to the simplicity of a finite group $G$ of order $120$ by showing that a Sylow $2$-subgroup $S$ of $G$ can't be a maximal subgroup of $G,$ for example (I won't give the details, but they require somewhat more background than the theorem you want to avoid). Hence $G$ has a subgroup of index $3$ or $5$, but then you are using the embedding in a symmetric group to obtain a contradiction in any case. Or you can do a complicated fusion and transfer analysis with the prime $2,$ but there is a perfect group of order $120$, so that is not straightforward either (the perfect group of order $120$ has a center of order $2$).

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Thanks Geoff for your time. I think, I'd better use the theorem instead than the other way. Thanks for your great answer and hints. :) –  B. S. Aug 6 '12 at 12:31
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