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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?

Jyrki Lahtonen
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kelvin
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    The framing is pointless. one constructive proof that $A_n$ is simple for $n\geq 5$ relies on there being "enough room" to play with an arbitrary nontrivial even permutation and get a 3-cycle. The argument breaks down for smaller $n$ because you don't have enough room to choose indices. The non-simplicity of $A_4$ is really a consequence of the Strong Law of Small Numbers: there aren't enough small numbers to have all the properties you would like them to have. – Arturo Magidin Sep 17 '22 at 16:57
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    Added comment $A_3$ is simple! There it's so small that there isn't room for nontrivial subgroups at all, much less nontrivial normal subgroups. Unsolvability by radicals requires that $A_5$ (and higher $A_k$) are both simple and noncyclic. – Greg Martin Sep 17 '22 at 16:58
  • It shouldn't be that surprising that at small numbers there might be exceptions. $A_3$ is simple as well by the way, though for a different reason. I agree that the proof for $n\geq 5$ is not very intuitive, but that's math. There are even stranger results. – Mark Sep 17 '22 at 16:58
  • To add to what others have put, the order of $A_4$ has precisely two distinct prime factors. It is not cyclic of prime order and is not complicated enough to be a non-cyclic simple group. – Mark Bennet Sep 17 '22 at 17:05
  • Many approaches collected here. – Jyrki Lahtonen Sep 17 '22 at 18:28
  • @ArturoMagidin I see, this Law is new to me. I suppose it does make sense that there would be exceptions on the small scale. – kelvin Sep 17 '22 at 19:02

2 Answers2

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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.

Qiaochu Yuan
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  • I never knew A4 was the group of rotations on the tetrahedron. It's starting to make more sense intuitively for me now, thank you – kelvin Sep 17 '22 at 19:17
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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 $n > 4$.

hunter
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    Philosophizing: I generally don't think that heuristics and questions about "why facts are true" are a waste of time. There might even be some beautiful result about extra symmetry in four-dimensional "vector spaces over $\mathbb{F}_1$" that nobody has yet discovered. But I do agree with all the other commenters that perhaps the atheist view on this one is correct and there's not much to explain. – hunter Sep 17 '22 at 17:11
  • Thanks for your answer. So I guess it's just a funny coincidence – kelvin Sep 17 '22 at 19:10