A Dubiously Simple Proof of the Insolubility of Symmetric Groups

June 2023

Abel's Theorem in Problems and Solutions by V.B. Alekseev is a great problem-oriented textbook that leads the reader to naturally rediscover the titular theorem for himself. And as all roads to the solvability of equations passes through symmetric groups, even Vladimir Arnold—whose lectures the book is based on—is forced to make the pilgrimage.

The important detail about symmetric groups $S_n$ is that they are insoluble for all $n \geq 5.$ Alekseev spends quite a few pages—about four—proving this fact. What's interesting is that it all can be skipped by some simple programming.

One approach to proving this theorem is by simple induction. If we show that $S_5$ is insoluble, and also show that the insolubility of $S_n$ implies the insolubility of $S_{n+1},$ then we have shown that $S_n$ is insoluble for all $n \geq 5.$

Look, I'm not going to go through all the basics of group-theory here, all you need to know is that the commutant $K(G) := \{aba^{-1}b^{-1}\,:\,\forall a, b\in G\}$ of a group $G$ is a normal subgroup of it. The other thing you need to know is that a group is soluble if $K(K(\dots K(G)\dots))$ is the trivial group after some finite number of iterations of $K.$ We denote the $n$th iteration as $K_n(G).$ So $G$ is soluble if there exists an $r \in \N$ such that $K_r(G)$ is the trivial group.

Well, actually, one more thing you need to know is that subgroups of a soluble group are all soluble. So suppose $G$ is soluble and $H \subseteq G.$ Notice that $K(H) \subseteq K(G),$ and similarly $K_n(H) \subseteq K_n(G).$ These facts are easy to verify from the definitions. As such, if $K_r(G)$ is the trivial group, then $K_r(H)$—being a group—must also be the trivial group. Therefore $H\subseteq G$ is a soluble group.

The contrapositive of this statement—that the insolubility of $H \subseteq G$ implies the insolubility of $G$—gives us the inductive step of our proof. We wish to show that the insolubility of $S_n$ implies the insolubility of $S_{n+1}.$ If we can show that $S_n$ is isomorphic to some subgroup of $S_{n+1},$ then we have shown that $S_{n+1}$ is insoluble.

It can easily be shown, for instance, that $S_3$ is isomorphic to a subset of $S_4.$ Just take a good look at $$S_3 = \{123, 132, 213, 231, 312, 321\}$$ and then look at $$H = \{ 1234, 1324, 2134, 2314, 3124, 3214 \} \subseteq S_4$$ Can you convince yourself that $H\cong S_3?$ If not, consider this map from $S_3$ to $H$ where $a,b,c$ are digits $\phi(abc) = abc4.$ This idea can easily be generalized to arbitrary $n.$ The symmetric group $S_n$ is isomorphic to a subset of $S_{n+1}$ via $\phi(i_1i_2\dots i_n) = i_1i_2\dots i_n(n+1).$

All that's left for us to do now is to prove our base case: that $S_5$ is insoluble. Contrary to popular belief, base cases are actually much harder than inductive cases, and this is the exception that proves the rule. Because the usual proof is quite tedious, while the proof I present here is incredibly easy to do, though it's not really math in the traditional sense. Usually, one would take quite a long detour through cyclic permutations and alternating groups and whatnot to prove that $S_5$ is insoluble. But, remember that $S_5$ is just some finite set with a binary operation—something that can be represented as a computational object. What I'm trying to say is that brute force is really fun.

So after acquiring your group-theory software of choice (I wrote one for myself but then threw it away after I realized that sage does group theory), implement a function that gives you the commutant of a group. Sage already has a method for this task so mine is pretty simple:

sage: K = lambda x: x.commutator()

Then just play around a little bit. First I loaded $S_5$ onto the system:

sage: S = SymmetricGroup(5)

My instinct at this point was to play around with the sizes of this group and its commutant. The idea was that if the order of the commutants stops decreasing at some point then that means that the group is insoluble. And, as it happens, the order of the commutants stops decreasing instantly.

sage: len(K(S)) == len(K(K(S))) True

Well, why stop at the sizes of the commutants? If we can show—sorry, I meant compute—that $K(S_5) = K(K(S_5)),$ then it's obvious that $K_n(S_5) = K(S_5)$ which means that $S_5$ is not soluble. So I tried it out:

sage: K(S) == K(K(S)) True

Bam! Just what we wanted. And as we would expect, $K(K(S_4)) \neq K(S_4)$ which allows $S_4$ to be soluble. In fact we can see how long it takes for $S_4$ to get to the trivial group:

sage: S = SymmetricGroup(4) sage: K(K(K(S))) Permutation Group with generators [()]

That is only $3$ iterations!

Well, there you have it dear readers. That's the show for tonight. If only Abel had an m1 macbook air. He could've skipped four pages!