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I know from questions such as this one or this one that Galois theory has proven there exists no "closed formula" (ie. via radicals) for solving quintic equations.

My question is more of a meta-mathematical question, which is distinct from the two questions linked above. Is there any particular reason why this occurs at $n=5$ instead of at any other number? Is there anything special that happens to the Galois groups when $n=5$?

From one of the answers linked above,

It happens that for the general polynomial of degree $n$ (i.e. with indeterminate coefficients) the Galois group is $S_n$, and that the alternating group $A_n$ is simple for $n\geq 5$, and of course, not abelian. Hence $S_n$ is not solvable.

My question could perhaps be made more specific by asking: is there any higher-level reason why $A_n$ is simple when $n\geq 5$ instead of this occuring at $n=4$, or say $n=3000$?

pyridoxal_trigeminus
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    Can you provide an analogy of a higher-level reason? Without an analogy, my impulse is to say that the reason is the proof that the alternating groups are simple in the aforementioned instances. I don't really understand what you're looking for. – Git Gud May 05 '22 at 22:03
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    The reason is buried in the deep mysteries of Math and might be the same reason for the existence of the monster group, or perhaps for the existence of only 5 regular polytopes! – Ruy May 05 '22 at 22:21
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    @Ruy - You mean 5 regular 3-polytopes. There are 6 regular 4-polytopes, and then only 3 regular $n$-polytopes for any $n\geq5$.... – mr_e_man May 05 '22 at 22:24
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    Not the reason, but a plausible corroboration: $,n=5,$ is the lowest degree for which the Lagrange resolvent has degree higher than $,n,$, so it cannot be used to reduce the equation to one of lower degree. – dxiv May 05 '22 at 22:30
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    @mr_e_man, yes, of course! But we shouldn't forget the exceptional Lie Algebras, or the exotic differentiable structures of $\mathbb R^4$. – Ruy May 05 '22 at 22:31
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    @Ruy After doing some research (I had not heard of the Monster group), I realize that my question may not even be answerable, and that it is, in fact, directly analogous to asking why the monster group exists. I suppose it has to do with the very foundations of our particular system of math and the definitions we have employed, but where exactly in the logical chain necessitates the existence of these things is anyone's guess... – pyridoxal_trigeminus May 06 '22 at 19:39
  • Why $5$ and not a bigger number for simplicity of $A_n$? This has been tried to answer also here. It is crucial that a non-trivial normal subgroup contains a $3$-cycle, and this is for $n\ge 5$. – Dietrich Burde May 07 '22 at 13:49

1 Answers1

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Teaser: In one of the proofs that $S_n$ isn't solvable, we have to pick $5$ different elements from the set on which $S_n$ acts. This is only possible if there are at least $5$ elements there to begin with.

~

There is a theorem that a group $G$ is solvable if and only if there is an iterated commutator of $G$ which is trivial. We need to unpack this statement:

For $a,b\in G$ we define the commutator $[a,b]:=aba^{-1}b^{-1}$. This basically measures how much $a$ and $b$ don't commute, since $[a,b]=e$ if and only if $a$ and $b$ commute. Now we define the commutator $[G,G]$ as the subgroup generated by all commutators of elements of $G$. This subgroup essentially measures how non-Abelian $G$ is, since intuitively speaking, the more non-commuting elements there are in $G$, the more elements there are in $[G,G]$, and $[G,G]$ is the trivial group iff $G$ is Abelian.

Now an iterated commutator $D^n G$ of $G$ is defined inductively as a commutator of a commutator of a ...: $D^0G:=G,~D^{n+1}G:=[D^nG,D^nG]$. The theorem from the beginning now says that $G$ is solvable if and only if one of the iterated commutators $D^n G$ is trivial.

Now the fact that $S_n$ is not solvable for $n\geq5$ hinges on the fact that for $n\geq 5$ we have $D^kS_n=A_n$, for all $k\geq1$. The reason for this is that in this case $A_n$ is its own commutator, that is, applying the commutator to $S_n$ yields $A_n$, and then applying the commutator again just returns $A_n$ again, so we're stuck at $A_n$ and will never get the trivial group as an iterated commutator. So the question is: why is $A_n$ its own commutator starting at $n\geq5$?

To answer this question, we rely on the fact that the alternating group $A_n$ is generated by the $3$-cycles in $S_n$. Now if we can show that every $3$-cycle is a commutator of two other $3$-cycles, then the generators of $[A_n,A_n]$ contain all generators of $A_n$, meaning that $A_n\subseteq[A_n,A_n]$, but since $[A_n,A_n]$ is a subgroup of $A_n$, this would make $[A_n,A_n]=A_n$.

Now to proving this: Take any $3$-cycle $(x_1~x_2~x_3)$ with $x_1,x_2,x_3\in\{1,2,\dots,n\}$, all pairwise different. We can now take another two elements $X_4,x_5\in\{1,2,\dots,n\}$, such that $x_1,\dots,x_5$ are still pairwise different. Then we have:

$$(x_1~x_2~x_3)=(x_1~x_2~x_4)(x_1~x_3~x_5)(x_1~x_2~x_4)^{-1}(x_1~x_3~x_5)^{-1}.$$

You can verify this by hand if you want. But we can now see that every $3$-cycle is a commutator of $3$-cycles, and thus $A_n=[A_n,A_n]$, making $A_n$ (and also $S_n$) not solvable.

The whole thing hinges on the decomposition of $(x_1~x_2~x_3)$ into the four cycles hinted at above. But this decomposition only works if we can pick another two elements $x_4$ and $x_5$ from the set $\{1,\dots,n\}$ we are acting on. This is of course only possible if it contains at least $5$ elements, otherwise we can't pick 5 different ones. So that's the smallest number where this proof starts to work.

You can read about the surrounding details (like the proof for the theorem stated at the beginning) in Bosch: Algebra.

Vercassivelaunos
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