Why are only the first four alternating groups are non-simple? I know asking for intuition in math is a generally flawed approach, but can anyone give any reason why only the first four alternating groups are non-simple?
 A: Actually $A_0,A_1,A_2$ are trivial and $A_3\cong C_3$ is simple, so you're really just asking about $A_4$.
Let $V_n$ be the set of all permutations on $n$ letters of the form $(2)(2)$ - i.e. all products of $2$-cycles.  We can quickly see that $V_n$ is stable under conjugation because cycle conjugation preserves cycle type.  Thus $\langle V_n \rangle$ is always a normal subgroup of $A_n$. (Note. For a broader way to see that $V_n$ is normal, indeed characteristic, in $A_n$, see the EDIT of my answer here.)
There are exactly three elements of type $(2)(2)$ which can be made from $4$ letters: $$V_4 = \{(12)(34),(13)(24),(14)(23)\}.$$  We can easily verify that $V_4$ (along with the identity) is closed, so $\langle V_4 \rangle$ is a normal subgroup of $A_4$ of order $4$.
For $n\geq 5$, however, this doesn't work because $V_n$ generates $A_n$ and thus is not proper.  This alone is not enough to prove that $A_n$ is not simple, but it is enough to suggest it intuitively.  Sylow theory finishes the job.
A: I think the best exercise for you would be to go carefully examine a proof of the fact that $A_n$ is simple for $n\geq5$.
In a nutshell, you will find that 5 symbols is the minimum necessary to pull tricks to show that $A_n$ is simple. For $A_3$, there is simply no room for proper normal subgroups. For $A_4$, there are just enough symbols to have a subgroup, but too few symbols available to keep it from being normal. So, $A_4$ just happens to have the Klein 4-group as a normal subgroup. 
Because of the added variety in the 5+ symbol groups, it is always possible to show that a normal subgroup contains a 3-cycle, and hence (by a lemma you will probably see in your book) is the entire alternating subgroup.
