The group that will not come apart

There is a formula for the roots of a quadratic, and making a square out of it shows where it comes from. There is one for the cubic, found in the 1530s, and one for the quartic shortly after. For degree five there is none, and the reason is not that nobody has looked hard enough.

The lattice that runs the other way sets up the correspondence this rests on: the subgroups of a polynomial’s symmetry group correspond, order-reversed, to the fields between the rationals and the field containing all its roots. That essay’s closing section names the insolubility of the quintic as the theorem the correspondence exists to prove. This is that theorem.

What solving by radicals means #

A formula in radicals builds the roots from the coefficients using the four arithmetic operations and the extraction of $n$-th roots. Each root extraction adds a new number to the field being worked in, and the field grows by a step.

So a formula is a tower: start at the rationals, adjoin an $n$-th root, adjoin another, and after finitely many steps arrive at a field containing all the roots. Each step is an extension generated by an $n$-th root of something already present.

By the correspondence, a tower of fields is a chain of subgroups running the other way — from the whole symmetry group down to the trivial one, each contained in the last. And a step that adjoins an $n$-th root is, once enough roots of unity are available, an extension whose own symmetry group is cyclic, hence abelian.

Putting the two together: a formula in radicals exists exactly when the polynomial’s symmetry group has a chain of subgroups

$$G = G_0 \supseteq G_1 \supseteq \cdots \supseteq G_k = \{e\}$$

with each $G_{i+1}$ normal in $G_i$ and each quotient $G_i/G_{i+1}$ abelian. A group with such a chain is called solvable, and the word means exactly what it looks like it means.

The blocks a subgroup cuts out builds the cosets and the quotient this depends on; a quotient exists only when the subgroup is normal, which is why normality is in the definition and not an afterthought.

The chain, when there is one #

For four letters the chain exists and is short enough to write out. The symmetric group on four letters has twenty-four elements; its even half, the alternating group, has twelve; inside that sits the four-element group of double transpositions; and inside that, the identity.

$$S_4 \;\rhd\; A_4 \;\rhd\; V_4 \;\rhd\; \{e\}$$

with quotients of sizes 2, 3 and 4 — all abelian, since groups of prime order are cyclic and the four-element quotient here is the Klein group. So $S_4$ is solvable, and the quartic has a formula.

Restricting the search to unions of conjugacy classes is not a shortcut that might miss something. A normal subgroup is carried into itself by every conjugation, so it is a union of classes; and it contains the identity. So the sixteen candidates are the complete list, and testing sixteen sets replaces testing all $2^{24}$ subsets.

Where the chain fails #

For five letters the same search returns a different answer.

That is the obstruction, and it is a finite object. A chain from $A_5$ down to the identity would need a normal subgroup of $A_5$ to step through, and there is none but the two trivial ones. The one available step, from $A_5$ straight to $\{e\}$, has quotient $A_5$ itself, which is not abelian — the group contains two five-cycles that do not commute, and a single such pair is enough.

So $A_5$ is not solvable. And $S_5$ contains it, with the derived series stalling exactly there: the commutators of $S_5$ generate $A_5$, and the commutators of $A_5$ generate $A_5$ again.

Reading the derived series #

The derived series is the cleanest test for solvability and the one the figures compute. At each step, replace the group by the subgroup generated by all its commutators $aba^{-1}b^{-1}$.

That subgroup is the smallest one whose quotient is abelian — the quotient is abelian precisely when every commutator dies, so killing exactly the commutators is the minimal way to abelianise. So if any chain with abelian quotients exists, the derived series is one, and it descends at least as fast as any other. A group is solvable exactly when its derived series reaches the identity.

For $S_3$: 6, 3, 1. For $S_4$: 24, 12, 4, 1. For $S_5$: 120, 60, 60, 60, … The series stops, and stopping above the identity is the whole statement.

The computation asserts what it needs at each step: that the commutators really do generate a subgroup of the group they came from, that the subgroup is carried into itself by every conjugation, and that each step’s size divides the last. None of those is assumed from the theory.

The cubic’s formula, read as a chain #

The chain for $S_3$ is short enough to match against the formula it corresponds to, and doing so is the fastest way to see that the correspondence is not an analogy.

$$S_3 \rhd A_3 \rhd \{e\}$$

with quotients of size 2 and 3. Reading it upside down as a tower of fields: first an extension of degree 2, then one of degree 3.

Cardano’s method does exactly that. Given $x^3 + px + q$, it first forms the discriminant and takes its square root — that is the degree-2 step, and it is the step that decides whether the cubic has one real root or three. Then it takes a cube root of an expression built from that square root — the degree-3 step. Two radicals, of orders 2 and 3, in that order, matching the two quotients.

The quartic’s chain has three steps of sizes 2, 3 and 4, and Ferrari’s method has three corresponding extractions — a resolvent cubic solved first, which is the size-3 step, sandwiched between two square roots. The formula’s shape is the chain’s shape, and neither was designed to match the other.

For the quintic there is no chain, so there is no shape for a formula to have. That is a stronger statement than “nobody has found one”: it says that any expression built by successive root extractions generates a tower whose corresponding chain has abelian quotients, and $S_5$ admits no such chain, so no such expression can reach the roots.

What is actually being claimed #

Two statements are easy to run together and only one is true.

True: there is no formula in radicals giving the roots of the general quintic in terms of its coefficients — no expression built from the coefficients by arithmetic and root extraction that works for every quintic.

False: no quintic can be solved by radicals. Plenty can. $x^5 - 2$ has roots $\sqrt[5]{2}$ times fifth roots of unity, and its Galois group has order 20 and is solvable. $x^5 - x - 1$ has Galois group $S_5$ and cannot be solved by radicals; $x^5 - 5x + 12$ has Galois group the dihedral group of order 10 and can.

The theorem is about the group of a particular polynomial, and it says a polynomial is soluble by radicals exactly when that group is. The general quintic has group $S_5$ — the largest possible — and $S_5$ is not solvable, so the general formula does not exist.

Why non-commutativity is the whole obstruction #

Every quotient in the chain has to be abelian, and it is worth seeing where the failure of commutativity actually bites.

A group is abelian exactly when every commutator is the identity. So the derived subgroup is a measurement of how far a group is from commuting, and the derived series is that measurement applied repeatedly — measure the failure, then measure the failure of the failure, and so on.

For $S_3$ the failure is small: one round produces the rotations, and the rotations commute, so a second round kills everything. For $S_4$ it takes three rounds. For $S_5$ the process reaches $A_5$ and finds that the failure of $A_5$ to commute is $A_5$ itself — the group is perfect, equal to its own derived subgroup, so measuring the failure returns the whole thing and no progress is ever made.

Perfect groups are exactly the ones on which this method stalls, and a simple non-abelian group is automatically perfect: its derived subgroup is normal, so it is the identity or the whole group, and it cannot be the identity because the group is not abelian.

Where the fifth degree is different #

It is worth asking why five and not four, since nothing in the statement of the problem changes.

The answer is that $A_n$ is simple for every $n \ge 5$, and is not for $n = 4$. The proof of simplicity turns on being able to move three chosen letters while leaving the rest alone: for $n \ge 5$ there are at least two letters left over, and that slack is enough to conjugate any 3-cycle to any other and to force a normal subgroup containing one 3-cycle to contain all of them.

At $n = 4$ the slack runs out. The double transpositions $\{e, (12)(34), (13)(24), (14)(23)\}$ form a normal subgroup precisely because there is no room to break the pattern up, and it is exactly that accident that makes the quartic soluble.

There is a way of seeing the slack that makes it concrete. In $A_5$, take any two distinct 3-cycles; there are twenty of them, and they fall into a single conjugacy class, so any one can be turned into any other by relabelling. A normal subgroup containing one 3-cycle is closed under relabelling and therefore contains all twenty — and twenty 3-cycles generate the whole of $A_5$. The same argument in $A_4$ produces the eight 3-cycles, which generate $A_4$ and not the smaller normal subgroup; the double transpositions are reached by a different route, and they are a class of their own with nothing to conjugate them into.

So the failure at five is that a class is too big to be contained, and the success at four is that a class is small enough to close up. That is visible in the class sizes the two searches report: $S_4$ has classes 1, 3, 6, 6, 8, and the subgroup of size 4 is $1 + 3$; $A_5$ has classes 1, 12, 12, 15, 20, and no sub-collection containing 1 adds to a divisor of 60 that is also closed.

So the general formula fails at five for a reason that is combinatorial rather than analytic. The crossings that will not come out even builds the even permutations and the parity that separates them; this is the next question about the same object, and it turns out to be the last question with a positive answer.

What was known before, and what changed #

Ruffini published an argument in 1799, six hundred pages long, which was essentially correct and which almost nobody read or believed. Abel gave a complete proof in 1824, short and correct, and it settles the question: there is no general formula.

Galois, between 1829 and 1832, did something else. Abel’s theorem says the general quintic has no formula; Galois’s theory says exactly which polynomials do, by attaching a group to each and reducing the question to a property of that group. The difference is between an impossibility and a criterion.

That is the pattern worth carrying. The cube that will not double settles one impossibility by a degree count, and the degree count answers that question and no other. The correspondence answers a class of questions at once, and answers them by computing in a finite object — which is why the figures here are tables of subgroups rather than arguments about fields.

What the pictures cannot show #

The searches here run on groups of order 6, 24, 60 and 120, which is where exhaustion over conjugacy-class unions is cheap. The theorem is about $A_n$ for every $n \ge 5$, and no computation covers that — the simplicity proof does, and it is an argument rather than a search.

The derived series table reports sizes and not structures. That $S_5$’s series stalls at 60 is visible; that the 60-element group it stalls at is $A_5$, and that $A_5$ is the same group for every route into it, is not something a column of numbers says.

And nothing here draws a quintic or its roots. The whole argument happens in a group of permutations of five symbols, and the connection back to the equation — that the group of the general quintic is the full symmetric group on its roots — is the correspondence, quoted rather than computed. That is the step the lattice that runs the other way establishes and this essay uses.

The ladder from here #

Below: the lattice that runs the other way, the correspondence between subgroups and intermediate fields, and the blocks a subgroup cuts out, where normality and quotients are built. Sideways: the cube that will not double, an impossibility settled by a degree rather than a structure, and the crossings that will not come out even, which builds the alternating group. Above: the simplicity of $A_n$ for all $n \ge 5$, the classification of solvable quintics, the resolvent cubic and where the quartic’s formula comes from, and Klein’s icosahedral solution, which solves the quintic with functions that are not radicals.

What is worth carrying away #

An impossibility proof is usually a demonstration that something would have a property it cannot have. Here the property is a chain, the something is a group of sixty elements, and the demonstration is a search over sixteen candidate subsets that returns two.

The remarkable part is the reduction, not the search. A question about formulas — about symbols on a page, over an infinite field, for infinitely many polynomials at once — has become a question about whether a particular finite group has a normal subgroup. The whole value of the Galois correspondence is that it moves a question into a place where the objects can be listed, and listing them is then a triviality that a table can carry.