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I think this might be a stupid question.

The icosahedral group $A_5$ with order $60$ is a simple group $60=2^2\times 3\times5$ but according to Sylow theorem $A_5$ must have subgroups of order $4$. But that's contradictory to that $A_5$ is a simple group right ?

Is $A_5$ simple?

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    I observe 3 cathegories within all non normal subgroups of icosahedral group A_5. That is also wh$y$ they cannot be bound by (only) one "p-cathegory".Pecik2012-08-30

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You seem to be recalling the definition of a simple group incorrectly. A simple group is a group $G$ whose only normal subgroups are the trivial group and $G$ itself. None of the order $4$ subgroups of the icosahedral group (or any of the nontrivial proper subgroups, for that matter) are normal. $I$ is isomorphic to $A_5$, which is a well-known example of a simple group.

In fact, the only nontrivial groups with no nontrivial proper subgroups at all are $\mathbb{Z}/p\mathbb{Z}$, where $p$ is a prime.

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    thanks , get the idea now.2012-04-14