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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?

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The first thing to say is that this is not true: $A_0, A_1$ and $A_2$ are all the trivial group, which is not simple. – Chris Eagle Jan 10 '13 at 19:19
The first half of your sentence lost me. Most of math is intuition driven in the sense that it revolves around ideas. That said, after the ideas take on a life of their own, things can get very unintuitive. Still, it is a weird mistake to give up on intuition in mathematics :P – rschwieb Jan 10 '13 at 19:21
@ChrisEagle Without trying to sound like I'm preaching: it's probably pretty frustrating for inexperienced students to be smacked down with a niggle. Rather than posting "your question is wrong because of edge cases," could you consider converting that comment to "You probably want to edit that to exclude these edge cases". – rschwieb Jan 10 '13 at 19:29
Okay, so left me see if I understand. Because the first are trivial, they are not simple? Then all greater than three still are? I edited the question. – Pax Kivimae Jan 10 '13 at 19:31
It's too simple to be simple. – Zhen Lin Jan 10 '13 at 20:08
up vote 10 down vote accepted

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.

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Dear Rschwieb, In you fourth line, you probably mean "proper subgroups" (since having no proper normal subgroups is the hallmark of being simple). Regards, – Matt E Jan 11 '13 at 2:56
@MattE Yes, it's a hallmark that $A_3$ has, so I don't see where the sentence is problematic. The goal was to explain why $A_4$ is at a balance point where a proper normal subgroup slips in, but is not eliminated. I think we're reading the sentence with two different angles, that's all. – rschwieb Jan 11 '13 at 13:47
Dear rschwieb, I see; thanks! Cheers, – Matt E Jan 11 '13 at 21:27

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.

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