The Abel-Ruffini Theorem

July 2nd, 2017

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We all remember the quadratic equation.

$$x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$$

In general, any quadratic equation with integer coefficients is solvable by radicals, meaning that the answer can be computed in a finite number of steps using only the $+, -, \times, \div$ operations and also taking $n$th roots. Though the formulas get longer and longer, the same is true for cubic and quartic polynomials.

The quartic formula.

But something incredible happens when you hit $5$: it turns out that there is no such formula. Proving this result took literally hundreds of years and ended with the birth of modern algebra in the $1800$s. I’m going to be going through the proof in pretty rigorous detail, and hopefully you and I will both learn something at the end of the day. I will assume for the purposes of this proof that you know basic group theory and algebra (the first couple paragraphs on Wikipedia should do).

Solvable Groups

We first start with group theory. Though the two seem pretty unrelated at first, group theory is actually intimately connected with solving algebraic equations. We will start with proving a theorem about groups; namely, that for $n \ge 5$, the symmetric group $S_n$ is not solvable.

Definition: $S_{n}$ is the symmetric group on $n$ letters, and it describes the permutations you can do on $n$ elements. Multiplication in this group is simply composition of operations.

Definition: Suppose we have an element $x$ in some group $G$. Then we define the operation conjugation by another element $g \in G$ as the multiplication operation $g\ x\ g^{-1}$.

Definition: A subgroup $N$ of a group $G$ is said to be normal if it is preserved under conjugation by any member of $G$. In other words, for any $g \in G$, we have that $gNg^{-1} = N$; conjugation sends elements of $N$ to other elements of $N$.

The normal subgroups are very important in group theory, because we can “divide” only by normal subgroups to obtain a quotient group:

Definition: Suppose we have a group $G$ with a normal subgroup $N$. The quotient group $\frac{G}{N}$ refers to all cosets of the form $aN$ for $a \in G$. We can think of this with analogy to the “modulo” operation; elements in $G$ are sent to the same “bucket” in $\frac{G}{N}$ if they differ by an element $n \in N$. Multiplication is defined as follows: $(aN)(bN) = (abN)$.

Now, we define what it means for a group to be solvable:

Definition: Let $G$ be a group. We call $G$ solvable if and only if there exists a sequence of subgroups:

$$ G = H_0 \supset H_1 \supset H_2 \supset ... \supset H_m = {e} $$

such that $H_i$ is a normal subgroup of $H_{i-1}$, and the quotient group $\frac{H_{i-1} }{H_i}$ is abelian.

We will prove our theorem in two parts. First, we consider what conditions we need in order to make sure a quotient group is abelian.

Theorem:

Let $G$ be a group with normal subgroup $H$. Then $\frac{G}{H}$ is abelian if and only if $H$ contains all elements of $G$ of the form:

$$ xyx^{-1}y^{-1} $$

Proof:

First, we assume $\frac{G}{H}$ is abelian. Let us refer to $f$ as the “canonical” homomorphism; the mapping that sends an element in $G$ to its natural element in $\frac{G}{H}$. This mapping is a homomorphism due to the way that we defined multiplication in the quotient group. So we have:

$$ f(xyx^{-1}y^{-1}) = f(x)\ f(y)\ f(x^{-1})\ f(y^{-1}) $$

By hypothesis, $\frac{G}{H}$ is abelian; so multiplication can be done in any order. It is evident, then, that the right hand side is the identity; and so, the left hand side is an element of $H$.


Conversely, let us assume $H$ contains all elements of the form $xyx^{-1}y^{-1}$. Then, let’s take any two elements in $\frac{G}{H}$, called $a, b$. We must have $a = f(x)$ for some $x \in G$, and similarly $b = f(y)$ for some $y \in G$. By hypothesis, then, $H$ contains $xyx^{-1}y^{-1}$. We already know that:

$$ f(xyx^{-1}y^{-1}) = f(x)\ f(y)\ f(x^{-1})\ f(y^{-1}) = e $$

But this means that:

$$\begin{aligned} f(xyx^{-1}y^{-1}) &= a\ b\ a^{-1}\ b^{-1} = e \\ ab&=ba \end{aligned}$$

And so indeed, $\frac{G}{H}$ must be abelian.

(Before this point it may be helpful to review permutations and how to multiply cycles; my convention is multiplying right to left.)

Moving forward, we prove an intermediate lemma.

Lemma

Take $H,N$ subgroups of $S_n$, $n\ge 5$ with $N \subset H$ and $N$ normal in $H$, with $\frac{H}{N}$ abelian as we want here. Then suppose that $H$ contains every $3$-cycle; we claim that $N$ also contains every $3$-cycle.

To prove this, take an arbitrary pair of cycles $\sigma=(abc)$ and $\tau=(cdf)$, with $a,b,c,d,f$ as $5$ distinct letters. This construction will make sense in just a minute, and you could reverse engineer it from the following calculation. We know from earllier that if the quotient group is abelian, then $N$ must necessarily contain $\sigma \tau \sigma^{-1} \tau^{-1}$. Computing that element, we have:

$$\begin{aligned} \sigma \tau \sigma^{-1} \tau^{-1} &= (abc)(cdf)(cba)(fdc) \\ &= (cad) \end{aligned}$$

It is evident that if $H$ contains all possible $3$-cycles, then by a similar construction we can proof that $N$ contains all of the $3$-cycles as well, since our choice of $5$ letters was arbitrary.

Note that this is not the case if we only have 4 letters to choose from. In that case, the product becomes a transposition (2-cycle):

$$\begin{aligned} \sigma \tau \sigma^{-1} \tau^{-1} &= (abc)(bcd)(cba)(dcb) \\ &= (cb) \end{aligned}$$

And in the degenerate case of 3 letters to choose from:

$$\begin{aligned} \sigma \tau \sigma^{-1} \tau^{-1} &= (abc)(abc)(cba)(cba) \\ &= (e) \end{aligned}$$

Earlier we showed that if $\frac{H}{N}$ is abelian, then $N$ contains all the elements of the form $xyx^{-1}y^{-1}$, where $x,y \in H$. So, in the case where $H$ contains all the $3$-cycles, we know $N$ contains all the $3$-cycles as well.

Now, we can complete the proof with a simple argument. Since $S_n$ contains all $3$-cycles for $n\ge 5$, then $H_1$ must also contain all $3$-cycles, and so on. But then, the trivial group would contain all $3$-cycles; so we are done.


So we’ve proved that $S_n$ is not a solvable group for $n \ge 5$. But what does any of this have to do with algebraic formulas? We’ll find out in the next part!

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