# Sylow theorem and icosahedral group

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?

• Why do the presence of subgroups of order 4 contradict simplicity? Commented Apr 14, 2012 at 2:50
• I believe the icosahedral group is isomorphic to $A_5$ that is simple.
– user38268
Commented Apr 14, 2012 at 2:53
• 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
– user39074
Commented Aug 30, 2012 at 12:36

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.

• 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 ${1} and G$ Commented Apr 14, 2012 at 3:13
• 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 $G$. Commented Apr 14, 2012 at 3:23
• @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. Commented Apr 14, 2012 at 3:52
• thanks , get the idea now. Commented Apr 14, 2012 at 3:52