What is the reason that A5 is simple and A4 is not?

Question

I am fascinated by the proof that the general quintic equation isn't solvable in radicals. I have a basic understanding that it eventually boils down to showing that A5 is a simple group. I know that this can be proved by considering the sizes of its conjugacy classes but something about this is so unsatisfying to me.

So I guess my question is, why did the universe (or God, if you believe in one) decide that A5 would be simple, and not A4, or A3? What is the fundamental difference between A5 and A4? Or is this just a pointless question?

Answer 1

Since we're being heuristic anyway, it might be easier to think about $S_n$. The reason that $S_n$ doesn't want to have a lot of normal subgroups is that conjugation in $S_n$ moves stuff around violently and only preserves the thing it's forced to preserve: cycle decomposition.

So a subgroup is going to be normal iff it is a disjoint union of the set of all elements of a given cycle type. Heuristically, this is just not going to happen very much, since multiplication doesn't play well with cycle type decomposition and as soon as you get a transposition in your normal subgroup you have all of $S_n$.

The fact that even permutations multiply together to make even permutations forces a normal subgroup to exist, namely $A_n$. Otherwise you're going to need some coincidence that depends on $n$. For large $n$ there are no coincidences, but for $n=4$ we have the weird result that (products of two disjoint transpositions) are closed under mutiplication, something that manifestly stops happening when $n > 4$.

Answer 2

In general, $A_{n+1}$ is the group of rotations of the $n$-simplex; explicitly, $A_{n+1}$ acts by permutation matrices on

$$\Delta_n = \{ (x_0, \dots x_n) \in \mathbb{R}^{n+1}_{\ge 0} : \sum_{i=0}^{n+1} x_i = 1 \}.$$

This is most familiar when $n = 2$, where it just says that $A_3 \cong C_3$, the cyclic group of order $3$, is the group of rotations of the triangle. When $n = 3$ we get that $A_4$ is the group of rotations of the tetrahedron.

We can use this to show conceptually that $A_4$ is not simple: the action of $A_4$ on the tetrahedron induces an action on the set of pairs of opposite edges of the tetrahedron, of which there are $3$. This gives a nontrivial homomorphism $A_4 \to S_3$, whose kernel must therefore be a nontrivial normal subgroup. So $A_4$ is not simple. If we allow reflections then we get the full symmetric group $S_4$ and a surjective homomorphism $S_4 \to S_3$, and this turns out to be responsible for the existence of the resolvent cubic of a quartic.

This argument doesn't work in general because we just can't find a corresponding set of features of the $n$-simplex which $A_{n+1}$ permutes and which is still small enough to get a homomorphism to a smaller group (or rather, there are no obvious choices, and simplicity guarantees that there aren't any non-obvious choices either). For example, for $n = 4$ the set of pairs of non-adjacent edges has cardinality $\frac{5!}{2! 2! 2!} = 15$ and it just gets worse from there.

As Arturo says in the comments, this sort of "law of small numbers" phenomenon happens all the time in mathematics; we have a general argument that works most of the time but in some small cases it doesn't have enough "room" to work, so those small cases end up being exceptions to the general pattern. Sometimes one can find interesting explanations of the exceptions.