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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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    $\begingroup$ Why do the presence of subgroups of order 4 contradict simplicity? $\endgroup$ Commented Apr 14, 2012 at 2:50
  • $\begingroup$ I believe the icosahedral group is isomorphic to $A_5$ that is simple. $\endgroup$
    – user38268
    Commented Apr 14, 2012 at 2:53
  • $\begingroup$ I observe 3 cathegories within all non normal subgroups of icosahedral group A_5. That is also why they cannot be bound by (only) one "p-cathegory".Pecik $\endgroup$
    – user39074
    Commented Aug 30, 2012 at 12:36

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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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  • $\begingroup$ I think I still didn't fully understand the definition of simple group, I am kind of comparing the definition of simple group to prime number, which seems to be incorrect. what does "normal subgroups are the trivial group and $G$ itself" means ? I explain it as has only subgroup ${1} and G$ $\endgroup$
    – zinking
    Commented Apr 14, 2012 at 3:13
  • $\begingroup$ By definition, a simple group has no nontrivial proper normal subgroups. A simple group $G$ can have a nontrivial proper subgroup $H$, but if $H$ is normal in $G$ then it must be that $H = 1$ or $H = G$. In other words, the only normal subgroups of a simple group $G$ are $1$ and $G$. This says nothing about the possibility of non-normal subgroups of $G$. $\endgroup$ Commented Apr 14, 2012 at 3:23
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    $\begingroup$ @zinking: If you want to draw the analogy to prime numbers, note that the only subgroups by which you can "divide" a group (form a quotient) are normal subgroups. So if you want to define "simple group" as "has no nontrivial 'divisors'", then the "divisors" will refer to normal subgroups, not arbitrary subgroups. $\endgroup$ Commented Apr 14, 2012 at 3:52
  • $\begingroup$ thanks , get the idea now. $\endgroup$
    – zinking
    Commented Apr 14, 2012 at 3:52

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