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\begin{document}
\title{Games with Galois}%
\author{Daniel Mathews}
\maketitle
\section{Mathematics and Games}
All the world's a stage, they say, and all the people merely
players. Well, we might also say, all the world's a game, and all
the people merely play them.
Mathematics is clearly home to a great many games and amusements.
But the fruits of such excellent entertainment may extend to
matters of even greater import than a closely-fought victory.
There is no greater example than John Conway's celebrated \emph{On
Numbers and Games}, in which a full, rigorous development of
numbers is presented starting from the concept of games. Conway
produces the integers, rationals, infinitesimals, transfinite
numbers and much more. So, occasionally games may be of use in
other areas of mathematics, which are traditionally treated with
more gravity: the present article is about another example.
\'{E}variste Galois was no game theorist, but his infamous
downfall arose from participation in an extraordinarily risky game
--- a duel with pistols at 25 paces. His tragic story is one of
rejection, strife, passion and, as E. T. Bell put it, ``massed
stupidity aligned against him". His short but eventful life, swept
up in the political currents of early nineteenth-century France,
met with continual disaster. He was demoted at school; he was
rejected from the esteemed Polytechnique twice (once, so the
legend goes, putting the blackboard duster to better use as a
missile against an imbecile examiner); his papers submitted to the
Academy were lost, and on one occasion, the secretary responsible
for his paper died; he was expelled from school for standing up
for his democratic ideals; he joined the artillery of the National
Guard to fight for the Republican cause; he was tried and
acquitted for threatening the king; and then jailed simply for
being ``dangerous", later convicted on a trumped-up charge. The
precise circumstances of his duel are not known: ostensibly it was
over a woman, but it may also have involved his political rivals.
Today it is beyond doubt that Galois made epochal contributions to
mathematics, despite his work being dismissed, neglected, or
described as ``incomprehensible" during his lifetime. By his death
at the age of 20, young \'{E}variste had established the branch of
mathematics that today bears his name.
Galois theory is often regarded as a difficult area of
mathematics. It's certainly not easy. But in a certain sense, as
we shall see, it's all really just a game.
\section{Solving as a learning game}
The most immediately important question addressed by Galois theory
is the solubility of polynomial equations. Our first player,
valiantly attempting to solve polynomials, we will name Sol ---
she's very bright! The aim of any game, for her, is to solve
polynomials.
Sol must convince me that a polynomial is soluble. While I
understand algebra and general mathematical argument (sometimes),
I am not very knowledgeable, and I only know the rational numbers.
I want to be able to write down the solutions using the following
symbols only:
$$
+, -, \times, \div, \sqrt[n].
$$
That is, Sol has to show me how to find the roots and write them
as a radical expression.
\noindent {\bf Game 1.} $x^2 + 1 = 0$. The roots, you should know,
are $\pm i$. But I don't know that. How can Sol explain them to
me? She could give me a lecture course about complex numbers, but
that's asking a bit much. Rather, she could introduce a number
$\alpha$ such that $\alpha^2 = -1$. Then the other root would be
$-\alpha$. That is, she can introduce numbers \emph{purely in
terms of their algebraic properties}. Provided I am happy that,
manipulating the end result by the rules, it satisfies the
polynomial, that's not a problem. I can write the roots as $\pm
\alpha$ or $\pm \sqrt{-1}$.
Here's an interesting question: is the number $\alpha$ supposed to
be $i$ or $-i$? Sol could again take the ``complex numbers lecture
course" approach, or think of other methods. But, if she sticks to
purely algebraic properties of numbers, she \emph{must fail},
because really $i$ and $-i$ have identical algebraic properties:
they both just sit there, until they're squared, at which point
they become $-1$. Thus Sol's aim is an impossible one, and the
numbers $i$ and $-i$ are too symmetric for me to tell them apart.
We must settle with simply writing down $\pm \alpha$ or $\pm
\sqrt{-1}$ without really knowing what the individual numbers are
in ``reality". This may be a little disconcerting, but since I'm
only worried about being able to write down roots as radical
expressions, it's perfectly adequate.
In fact, for the same reasons, I can't ever distinguish between
$\sqrt{5}$ and $-\sqrt{5}$, in fact any set of numbers satisfying
the same irreducible rational polynomial. Such numbers are called
\emph{Galois conjugate}.
\section{Symmetry strategy}
Solving equations like the previous one really just amounts to
taking square roots. Things get a bit more difficult when the
polynomial is more complicated. Here we need to try something more
interesting.
Let us first recall a few properties of polynomials. Consider a polynomial
\[
p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0,
\]
and let the roots be $\alpha_1, \ldots, \alpha_n$. The roots allow us to factorise the polynomial:
\[
p(x) = a_n (x - \alpha_1) (x - \alpha_2) \cdots (x - \alpha_n).
\]
Let's now assume our polynomial is \emph{monic}, i.e. that the leading coefficient $a_n = 1$. We can equate our two descriptions of $p(x)$
\[
x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 = (x - \alpha_1) (x - \alpha_2) \cdots (x - \alpha_n),
\]
and then expanding out the brackets gives a rather complicated (but extremely useful!) equality.
\begin{eqnarray*}
x^n + a_{n-1} x^{n-1} + \cdots +
a_0 &=& x^n - \left( \alpha_1 + \cdots + \alpha_n \right)
x^{n-1} + \left( \alpha_1 \alpha_2 + \cdots + \alpha_{n-1}
\alpha_n \right) x^{n-1} \\
&& + \cdots + (-1)^{n} \alpha_1 \cdots
\alpha_n
\end{eqnarray*}
Here, the expressions in brackets are very interesting. The coefficient of $x^{n-1}$ is $-(\alpha_1 + \cdots + \cdots \alpha_n)$, the (negative) sum of the roots. The coefficient of $x^{n-2}$, written out a little more fully, is:
$$
\begin{array}{ccccccccc}
\alpha_1 \alpha_2 &+& \alpha_1 \alpha_3 &+& \alpha_1 \alpha_4 &+& \cdots &+& \alpha_1 \alpha_n \\
&+& \alpha_2 \alpha_3 &+& \alpha_2 \alpha_4 &+& \cdots &+& \alpha_2 \alpha_n \\
& & &+& \alpha_3 \alpha_4 &+& \cdots &+& \alpha_3 \alpha_n \\
& & & & &+& \ddots &+& \vdots \\
& & & & & & &+& \alpha_{n-1} \alpha_n
\end{array}
$$
which is the \emph{sum of all the possible products of two distinct roots}. It is itself a polynomial of the roots.
In general, $a_{n-k}$ is (plus or minus) the \emph{sum of all possible products of roots taken $k$ at a time}. These expressions involving sums of all possible products of the roots are polynomials of the roots called \emph{symmetric polynomials}. For instance, if $n=3$, the symmetric polynomials are
\[
\alpha_1 + \alpha_2 + \alpha_3, \quad \alpha_1 \alpha_2 + \alpha_1 \alpha_3 + \alpha_2 \alpha_3, \quad
\alpha_1 \alpha_2 \alpha_3.
\]
This gives us the following key fact; a nice relationship between the coefficients of a polynomial and the roots.
\noindent {\bf Fact 1.}
For a monic polynomial,
\begin{eqnarray*}
a_{n-1} &=& - \left( \alpha_1 + \cdots + \alpha_n \right) \\
a_{n-k} &=& (-1)^k \; \left\{ \text{ sum of all products of roots
taken $k$ at a time } \right\} \\
a_0 &=& (-1)^n \; \alpha_1 \alpha_2 \cdots \alpha_n
\end{eqnarray*}
Thus, the coefficients of a polynomial are (up to sign) the symmetric polynomials of the roots.
We will also take as a fact the following deeper theorem.
\noindent {\bf Fact 2.} (Gauss' theorem on symmetric functions.) A
polynomial in $\alpha_1, \cdots, \alpha_n$ which is symmetric in
all the variables can be expressed in terms of the symmetric
polynomials above, i.e. in terms of $a_0, a_1, \ldots, a_{n-1}$.
For instance, take $n=3$, and take a random polynomial in $\alpha_1, \alpha_2, \alpha_3$ which is symmetric in all the variables, say $\alpha_1^3 + \alpha_2^3 + \alpha_3^3$. Then Gauss' theorem says that we can express this polynomial in terms of the standard symmetric polynomials, namely $\alpha_1 + \alpha_2 + \alpha_3$, $\alpha_1 \alpha_2 + \alpha_1 \alpha_3 + \alpha_2 \alpha_3$ and $\alpha_1 \alpha_2 \alpha_3$. Indeed, if we look hard enough, we see that this is possible!
\begin{align*}
\alpha_1^3 + \alpha_2^3 + \alpha_3^3 &= (\alpha_1 + \alpha_2 + \alpha_3)^3 - 3 (\alpha_1 + \alpha_2 + \alpha_3) (\alpha_1 \alpha_2 + \alpha_1 \alpha_3 + \alpha_2 \alpha_3) + 3 \alpha_1 \alpha_2 \alpha_3 \\
&= -a_2^3 + 3 a_1 a_2 - 3 a_0
\end{align*}
That took some effort, but Gauss' theorem tells us that if we seek long enough, we shall find!
With that background behind us, it's time for another game.
\noindent {\bf Game 2.} $x^2 - 6x + 7 = 0$. Let the roots be
$\alpha, \beta$. By the above formulae
$$
\alpha + \beta = 6, \qquad \alpha \beta = 7.
$$
\begin{verse}
S: Consider the expression $(\alpha - \beta)^2$. This is symmetric
in $\alpha$ and $\beta$: if we swap them, we get $(\beta -
\alpha)^2$ which is obviously the same.
D: Right! By Gauss' theorem, then, we can express it in terms of
the symmetric polynomials $\alpha+\beta=6$ and $\alpha \beta = 7$.
S: Precisely. We get
$$
(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4 \alpha \beta =
6^2 - 4 \times 7 = 8.
$$
So I'm going to introduce to you this number $\sqrt{8}$, which has
the property that its square is 8. If you play around with the
algebra you'll see that the square of $\sqrt{8}/2$ is $2$. So you
can write it as $2 \sqrt{2}$ if you like.
D: OK, I'm happy with that, I know algebra.
S: Now we've got $\alpha + \beta = 6$ and $\alpha - \beta = 2
\sqrt{2}$. Solving the simultaneous equations easily gives
$$
\alpha = 3 + \sqrt{2}, \quad \beta = 3 - \sqrt{2}
$$
\end{verse}
Using this method we can solve quadratics in general.
\noindent {\bf Game 3.} $x^3 - 2$. (As you know, the roots are
$$
\sqrt[3]{2}, \quad \sqrt[3]{2} \left( -\frac{1}{2} +
\frac{\sqrt{3}}{2} i \right), \quad \sqrt[3]{2} \left( - \frac{1}{2}
- \frac{\sqrt{3}}{2} i \right)
$$
but I don't know that.)
\begin{verse}
S: Well, there's this number $\sqrt[3]{2}$, it's a number $\alpha$
described by the property that $\alpha^3 = 2$. I won't even bother
to distinguish it from the other roots, because they're all Galois
conjugates.
D: OK, great, $\alpha = \sqrt[3]{2}$ is a root. I now know the
rational numbers and this number $\alpha$. But what are the other
two?
\end{verse}
At this point, Sol can try to describe to me the other two roots
by using rational numbers and $\alpha$. But without introducing
any new numbers, this is an impossible task. Using $\sqrt[3]{2}$
and rational numbers, she can't possibly describe complex numbers
to me, only real ones. She must introduce a new number.
\begin{verse}
S: Well, there's this number $\sqrt[3]{2}\left( -\frac{1}{2} +
\frac{\sqrt{3}}{2} \right)$. It's a number $\beta$ so that
$\beta^3 - 2 = 0$.
D: Hmmm... sounds like $\alpha$.
S: Well it's one of the roots other than $\alpha$.
D: That's a bit silly, why don't you factorise out $(x -
\alpha)$ then?!
\begin{eqnarray*}
x^3 - 2 &=& \left( x - \sqrt[3]{2} \right) \left( x^2 +
\sqrt[3]{2} x + \sqrt[3]{4} \right) \\
&=& (x - \alpha) (x^2 + \alpha x + \alpha^2)
\end{eqnarray*}
D: You should have said, ``there's this number $\beta$ that
satisfies $\beta^2 + \alpha \beta + \alpha^2 = 0$". It would have
been more efficient. That's fine, I can solve the quadratic and
obtain the two roots as
$$
\alpha \left( \frac{ -1 + \sqrt{-3} }{2} \right), \qquad
\alpha \left( \frac{ -1 - \sqrt{-3} }{2} \right).
$$
\end{verse}
\section{Multi-stage learning and partial symmetry}
As we just saw, with cubics I don't learn everything in one fell
swoop. I need to learn in several stages. With a more difficult
cubic, the problems become much worse. As we'll see, we need to
consider equations which are only partially symmetric as we go.
\noindent {\bf Game 4.} $x^3 - 6x - 2 = 0.$ Let the roots be
$\alpha, \beta, \gamma$ so that
$$
\alpha + \beta + \gamma = 0, \quad \alpha \beta + \beta \gamma
+ \gamma \alpha = -6, \quad \alpha \beta \gamma = 2.
$$
\begin{verse}
S: First, there's this number $\omega$, a cube root of unity. It
satisfies $\omega^2 + \omega + 1 = 0$ and we'll write it as
$$
\omega = \frac{ -1 + \sqrt{-3}}{2}.
$$
D: Hmmm... yes, we can solve quadratic equations now. But I can't really see the point of this.
S: Give me a minute. Do you agree that $\omega$ is a cube root of unity, $\omega^3 = 1$?
D: Well, I know algebra, but all I know is that $\omega^2 + \omega + 1 = 0$. Let me see:
\begin{eqnarray*}
\omega^3 &=& \omega^2 \cdot \omega \\
&=& (- \omega - 1) \cdot \omega \\
&=& - \omega^2 - \omega \\
&=& 1.
\end{eqnarray*}
OK! I'm with you now.
S: Great. Introduce the two numbers $y$ and $z$, which
are defined in terms of the roots of the cubic.
$$
y = \alpha + \omega \beta + \omega^2 \gamma, \quad z = \alpha
+ \omega^2 \beta + \omega \gamma.
$$
D: I see, you're trying to find a symmetric function in $\alpha,
\beta, \gamma$, just like in the quadratic case. But hang on a
minute, these aren't symmetric! If you swap $\alpha$ and $\beta$
around, for instance, both the expressions for $y$ and $z$ change!
S: Yes, you're right. But if you \emph{cycle} $\alpha, \beta,
\gamma$ around in $y$, then $y$ becomes $y \omega$ or $y
\omega^2$... I'll show you.
\begin{eqnarray*}
\alpha \beta \gamma &\rightarrow& \beta \gamma \alpha, \\
y = \alpha + \omega \beta + \omega^2 \gamma &\rightarrow&
\beta + \omega \gamma + \omega^2 \alpha \\
&=& \omega^3 \beta +
\omega^4 \gamma + \omega^2 \alpha \\
&=& \omega^2 \left( \alpha +
\omega \beta + \omega^2 \gamma \right) \\
&=& \omega^2 y
\end{eqnarray*}
Similarly $z$ becomes $\omega z$ or $\omega^2 z$. So if we
consider
$$
y^3 = \left( \alpha + \omega \beta + \omega^2 \gamma
\right)^3,
\quad z^3 = \left( \alpha + \omega^2 \beta + \omega \gamma
\right)^3
$$
then both $y^3$ and $z^3$ are invariant under cycles.
D: That's tricky! You are very bright, Sol! But neither expression is symmetric!
S: Correct. But they are \emph{partially} symmetric --- if we
content ourselves with only cycling $\alpha, \beta, \gamma$
around. On the other hand, if you do a swap like $\alpha
\leftrightarrow \beta$ then you change $y^3$ into $z^3$!
\begin{eqnarray*}
\alpha \beta \gamma &\rightarrow& \alpha \gamma \beta \\
y^3 = \left( \alpha + \omega \beta + \omega^2 \gamma \right)^3 &\rightarrow&
\left( \beta + \omega \alpha + \omega^2 \gamma \right)^3 \\
&=& \left( \omega^3 \beta + \omega \alpha + \omega^2
\gamma \right)^3 \\
&=& \omega^3 \left( \alpha + \omega^2 \beta + \omega \gamma
\right)^3 = z^3
\end{eqnarray*}
D: You're quite the algebra whiz!
S: Therefore, if we add $y^3$ and $z^3$ together, then they're
invariant under any permutation! The same applies if we multiply
them to get $y^3 z^3$!
\begin{eqnarray*}
y^3 + z^3 &=& \left( \alpha + \omega \beta + \omega^2 \gamma
\right)^3 + \left( \alpha + \omega^2 \beta + \omega \gamma
\right)^3, \\
y^3 z^3 &=& \left( \alpha + \omega \beta + \omega^2 \gamma
\right)^3 \left( \alpha + \omega^2 \beta + \omega \gamma
\right)^3
\end{eqnarray*}
D: That's very cool.
S: No, Galois is cool. Thus, $y^3 + z^3$ and $y^3 z^3$ are totally
symmetric. By Gauss' theorem on symmetric functions, they can be
written in terms of $\alpha + \beta + \gamma$, $\alpha \beta +
\beta \gamma + \gamma \alpha$ and $\alpha \beta \gamma$ alone. In
fact I did some algebra earlier and figured out that
\begin{eqnarray*}
y^3 + z^3 &=& 2 \left( \alpha + \beta + \gamma \right)^3 - 9
(\alpha + \beta + \gamma)(\alpha \beta + \beta \gamma + \gamma
\alpha) + 27 \alpha \beta \gamma, \\
y^3 z^3 &=& \left( \alpha + \beta + \gamma \right)^6 - 9 \left(
\alpha + \beta + \gamma \right)^2 (\alpha \beta + \beta \gamma
+ \gamma \alpha)^2 - 27 (\alpha \beta + \beta \gamma + \gamma
\alpha)^3.
\end{eqnarray*}
For our polynomial, we know that $\alpha + \beta + \gamma = 0$,
$\alpha \beta + \beta \gamma + \gamma \alpha = -6$, $\alpha \beta
\gamma = 2$. Therefore we have
$$
y^3 + z^3 = 54, \quad y^3 z^3 = 18^3,
$$
and $y^3$ and $z^3$ are the roots of the quadratic
$$
t^2 - 54 t + 18^3 = 0.
$$
D: We solved quadratics earlier. I'll solve this one.
$$
y^3 = 27 + 27 \sqrt{-7}, \qquad z^3 = 27 - 27 \sqrt{-7}.
$$
S: Now I'll teach you the number $y$, which satisfies the equation
that its cube is $y^3$.
D: Der...!
S: Yes, but I had to teach you, otherwise you wouldn't know.
D: OK. I guess so.
$$
y = \sqrt[3]{ 27 + 27 \sqrt{-7}} = 3 \sqrt[3]{1 + \sqrt{-7}},
\quad z = \sqrt[3]{27 - 27 \sqrt{-7}} = 3 \sqrt[3]{1 -
\sqrt{-7}}
$$
S: Now we're nearly done! Remember we had $y = \alpha + \omega
\beta + \omega^2 \gamma$ and $z = \alpha + \omega^2 \beta + \omega
\gamma$. Now we know that $\omega + \omega^2 = -1$, so
$$
y + z = 2 \alpha - \beta - \gamma
$$
and hence
$$
y + z + (\alpha + \beta + \gamma) = 3 \alpha.
$$
We know $y$ and $z$, and we know $\alpha + \beta + \gamma = 0$.
Therefore
$$
\alpha = \frac{y+z}{3} = \sqrt[3]{1 + \sqrt{-7}} +
\sqrt[3]{1 - \sqrt{-7}}.
$$
The other roots can be found out similarly. I don't have to teach
you any new numbers.
\end{verse}
\section{Galois theory}
We'll need to know something about permutations. Recall that a
\emph{permutation} of a set of objects is just a rearrangement of
them. Speaking formally, a permutation is a bijective function
from a set of objects to itself. Normally, if there are $n$
objects, we just use the numbers $1, 2, \ldots, n$. We may write
permutations in different ways:
$$
\begin{array}{c}
12345 \rightarrow 23145, \qquad
\left( \begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 1 & 4
& 5 \end{array} \right) \\
1 \mapsto 2 \mapsto 3
\mapsto 1, \qquad 4 \mapsto 4,
\qquad 5 \mapsto 5 \\
(123)(4)(5) \\
(123).
\end{array}
$$
The last type of notation is called cycle notation and we say, for
instance, that $(123)$ is a \emph{3-cycle}. In the last line, we used the common notational convenience of omitting 1-cycles like $(4)$ and $(5)$, since they just stay where they are. The way in which the
cycles are grouped in a permutation is called its \emph{cycle
structure}. Thus $(123)(45)$ and $(124)(35)$ have the same cycle
structure --- they are $(2,3)$- or \emph{$(3,2)$-cycles} --- but $(123)(45)$ and $(123)$ do not. Permutations are
multiplied by composing the functions involved. We multiply from
left to right. For instance
$$
(123)(45) \cdot (345) = (1243)(5) = (1243).
$$
The do-nothing, or identity permutation is denoted $(1)$.
Permutations form a \emph{group}: when you multiply two
permutations, you get another permutation; each permutation has an
\emph{inverse} permutation which undoes it; and the multiplication
is \emph{associative}. Subsets of the set of permutations which have
these properties are called \emph{subgroups}.
Now it's worth summarising how we solved the cubic:
\begin{enumerate}
\item
$y^3 + z^3$ and $y^3 z^3$ are rational. The expressions
for them were totally symmetric. That is, they were
invariant under \emph{any} permutation of $\alpha, \beta,
\gamma$ --- the entire permutation group $S_3$.
$$
\begin{array}{cccccc}
\alpha \beta \gamma & \alpha \gamma \beta & \beta \alpha
\gamma & \beta \gamma \alpha & \gamma \alpha \beta &
\gamma \beta \alpha \\
(1) & (23) & (12) & (123) & (132) & (13)
\end{array}
$$
\item
Then we learnt $y^3$ and $z^3$ --- an intermediate level
of understanding --- as the roots of a quadratic. $y^3$
and $z^3$ were not totally symmetric, but they were
invariant under cycles of $\alpha, \beta, \gamma$. The
expressions for them were \emph{partially} symmetric ---
symmetric under cycling but not swaps. These cycles form
a subgroup of $S_3$ called $A_3$.
$$
\begin{array}{ccc}
\alpha \beta \gamma & \beta \gamma \alpha & \gamma
\alpha \beta \\
(1) & (123) & (132)
\end{array}
$$
\item
Then we learnt $y$ and $z$, by taking cube roots. They are
the roots of a (really easy) cubic. The expressions for
them are not symmetric at all! They were only invariant
under the trivial permutation.
\item
Understanding $y$ and $z$ is really the same level of understanding as
$\alpha, \beta, \gamma$, since we can then describe the
roots in terms of $y$ and $z$.
\end{enumerate}
Galois' amazing idea is that stages of learning, and types of
symmetry, are the same thing.
\begin{thm}[Fundamental theorem of Galois theory]
There is a one-to-one correspondence between levels of
understanding in solving a general degree-$n$ polynomial,
and subgroups of the
group of permutations of the $n$ roots.
\end{thm}
This statement is necessarily imprecise, since we have not
bothered to define anything properly. The correspondence is
order-reversing.
\begin{center}
\begin{tabular}{c|ccccc}
Knowledge & $\Q$ &$\subset$& $\Q, y^3, z^3$ & $\subset$ & $\Q, y,
z \Rightarrow \alpha, \beta, \gamma$ \\
\hline
Subgroup & $S_3$ & $\supset$ & $A_3$ & $\supset$ & (1)
\end{tabular}
\end{center}
Thus, the appropriate thing to do in solving polynomial equations
is not to perform algebraic torture, but to find the appropriate
middle levels of symmetry, and find expressions with that
symmetry. With higher degree polynomials we need more levels of
symmetry, because the middle levels of symmetry must be fairly
``close together". If I am to be able to write down the answer as
a radical expression, the only way to jump to a new level of
understanding is when the ``loss of symmetry" corresponds to
``taking an $n$'th root".
The technical condition is that we must have a nested sequence of
subgroups which have abelian quotients, down to the trivial group.
A group which has such a sequence of subgroups is called
\emph{soluble}. The cubic polynomial is soluble precisely because
the group $S_3$ is soluble. It turns out that solubility is
equivalent to the derived series of the group eventually becoming
trivial. I won't define these concepts here. Suffice it to note
that the real game is about permutations; specifically, about
whether or not sufficient ``middle levels" of symmetry --- that
is, intermediate subgroups of permutations --- exist.
\section{The Game of Galois}
Here is a game about permutations then. Since the actual game
involved in determining whether a group is soluble is quite
complicated, we'll start with a simpler version. Sol has a friend
named Insol, who as the name suggests, insolently maintains that
many polynomials are insoluble.
\begin{defn}[$n$-Galois-lite]
Sol chooses a starting permutation $x_0$ of $\{1, 2, \ldots, n\}$,
which cannot be the
identity, and then takes no further part in the game.
Insol chooses a permutation $y_0$, which
\emph{must have the same cycle structure as $x_0$}. We then form
$x_0 y_0 = x_1$. Insol chooses another permutation
$y_1$ with the same cycle structure as $x_1$, and forms $x_1 y_1 =
x_2$. Play continues in this fashion. If Insol can keep away from
the identity permutation for arbitrarily long, she wins. If
she hits the identity, Sol wins.
\end{defn}
This sounds like a peculiar game. But the idea, in a more refined
version of the game, is as follows: in a soluble group of
permutations, Insol is forced to move down to lower levels of
symmetry, eventually to the identity.
\noindent {\bf Game 5.} $n = \infty$. We have $\infty$ objects,
say the natural numbers $1, 2, \ldots$, but the players are only
permitted to permute finitely many at a time.
Insol has a winning strategy in this game. Because there is an
infinite set of objects, Insol can always pick a permutation which
affects only objects which have been hitherto unused. When these
permutations are multiplied, we just write the cycles next to each
other! E.g.
$$
(12)(34) \cdot (56)(78) = (12)(34)(56)(78).
$$
\noindent {\bf Game 6.} $n=1$. This is a non-starter, literally.
There's only the identity permutation, so Sol cannot begin.
\noindent {\bf Game 7.} $n=2$. Sol can only choose $(12)$ to
start, and Insol can only respond with $(12)$. Then $(12)(12) =
(1)$ and Sol wins.
\noindent {\bf Game 8.} $n=3$. Sol has two types of permutations
to use, either the $(123)$ type or the $(12)$ type. Hopefully you
should be able to see that choosing different starting
permutations with the same cycle structure makes no difference to
the overall strategy.
Suppose Sol starts with $(123)$. Then Insol responds with $(123)$
and we form $(123)(123) = (132)$. Then Insol chooses $(132)$ and
we form $(132)(132) = (123)$. If Insol continues alternately
choosing $(123)$ and $(132)$, she can keep away from the identity
indefinitely and win. We'll call this the \emph{copy} tactic. Sol
must try something else, and starts with $(12)$. But, alas, it is
to no avail and Insol chooses $(13)$, giving $(12)(13) = (123)$
and from there can maintain a 3-cycle with the copy tactic. Thus
Insol has a winning strategy.
\noindent {\bf Game 9.} $n=4$. This is a little more complicated,
but no matter how Sol starts, Insol can get to a permutation of
the form $(123)$ or $(12)(34)$ and maintain it. The following
diagram shows the possible moves Insol can make between the
various cycle structures.
$$
\begin{array}{ccccc}
& & (123) & & \\
& \nearrow & \vline \; \downarrow \; \circlearrowleft & \nwarrow & \\
(1234) & \rightarrow & (12)(34) & \leftarrow & (12) \\
& \searrow & \downarrow \; \downarrow \; \circlearrowleft & \swarrow & \\
& & (1) & &
\end{array}
$$
[figure 1]
For instance, if Sol starts with $(1234)$ then Insol chooses
$(1342)$ and obtains $(1234)(1342) = (243)$, and can stay there by
the copy tactic. If Sol starts with $(12)(34)$ then Insol can
stick there, choosing $(13)(24)$ to get $(12)(34)(13)(24) =
(14)(23)$, maintaining a $(2,2)$-cycle. So Insol wins again.
In general, Insol has many strategies available. First, Insol can
remove any cycle she wants in a permutation, because each cycle
has an inverse, which is obtained by writing it backwards. For
example $(123)(321) = (1)$. The inverse has the same cycle
structure, and Insol can choose it as part of her permutation.
Second, the copy tactic can be applied to any odd-length cycle, so
Insol can preserve an odd-length cycle. For instance, if
$(1234567)$ comes up, then Insol copies with $(1234567)$ which
gives $(1234567) \cdot (1234567) = (1357246)$. Third, a long
even-length ($2k \geq 4$) cycle can be reduced to a
$(2k-1)$-cycle, since, for instance, $(12345678)\cdot(21345678) =
(2468357)$. Finally, a 2-cycle can be turned into a 3-cycle, as we
saw previously. Combining all these tactics gives
\begin{thm}
Insol has a winning strategy in $n$-Galois-lite for any $n \geq
3$.
\end{thm}
So much for the lite version of the game, which appears to be a
bit easy for Insol! Let's toughen it up a bit.
\begin{defn}[restricted $n$-Galois]
Again we have $n$ objects and Sol selects a non-identity
permutation $x_0$ to start. Again Insol selects a permutation
$y_0$ with the same cycle structure as
$x_0$. Then we set $x_0 y_0 x_0^{-1} y_0^{-1} = x_1$.
Insol chooses $y_1$ with the same cycle structure as $x_1$,
and we form $x_1 y_1 x_1^{-1} y_1^{-1} = x_2$.
Play continues; if Insol can keep away from
$(1)$ indefinitely, she wins; otherwise Sol wins.
\end{defn}
Actually we can notice that $axa^{-1}$ always has the same cycle
structure as $x$, which is in turn the same as $x^{-1}$. The three
permutations $y_n x_n^{-1} y_n^{-1}$, $x_n^{-1}$ and $x_n$ have
the same cycle structure. In group theoretic language they are
conjugates. If we allowed Insol to choose \emph{any} permutation
for $y_n$, the game would be identical to Galois-lite; hence the
epithet ``restricted".
\noindent {\bf Game 10.} $n=2$. Sol can only choose $(12)$ to
start and Insol can only choose $(12)$ to obtain $(1)$. Sol wins.
\noindent {\bf Game 11.} $n=3$. Sol can start with, say, $(123)$
or $(12)$. Suppose she chooses $(123)$. Then Insol must choose
$(123)$ or $(132)$, both of which yield $(1)$, and Sol wins. On
the other hand, if Sol starts with $(12)$, then $y_0 x_0^{-1}
y_0^{-1}$ will be a 2-cycle; the product of two 2-cycles is always
of the form $(abc)$ or $(1)$. In any case, Sol wins.
$$
\begin{array}{ccccc}
(12) & & \rightarrow & & (123) \\
&\searrow& &\swarrow & \\
& & (1) & & \\
\end{array}
$$
\noindent {\bf Game 12.} $n=4$. Sol has 4 choices, namely
$(1234)$, $(123)$, $(12)(34)$, and $(12)$. We can consider the
following facts, gleaned from 4-Galois-lite.
The product of two 4-cycles is either of the form $(abc)$,
$(ab)(cd)$ or $(1)$. This was illustrated for 4-Galois-lite above;
in restricted Galois even fewer moves are available. Similarly the
product of two 2-cycles is either of the form $(abc)$, $(ab)(cd)$
or $(1)$. For similar reasons, once arriving at a cycle of the
type $(123)$, $(12)(34)$ or $(1)$, Insol cannot get back to a
$(1234)$ or $(12)$.
Thus we only need consider permutations like $(123)$, $(12)(34)$
or $(1)$. From $(123)$, can Insol maintain a 3-cycle? The only way
to do this is if $y_0 (321) y_0^{-1} = (123)$, where $y_0$ is a
3-cycle. The 3-cycle cannot involve $4$, and so must be $(123)$ or
$(132)$. But the conjugate of $(321)$ by both possibilities is
again $(321)$. So Insol cannot maintain a 3-cycle, and must go to
a permutation of the form $(12)(34)$ or $(1)$. Similarly, Insol
cannot maintain $(12)(34)$: from $(12)(34)$ all roads lead to
$(1)$.
Therefore, having considered all cases, we conclude that Insol
fails, and Sol has a winning strategy for $n=4$. There is a
diagram of possible moves, which is the same as figure 1, with the
loops removed.
\section{The big game.}
We now come to the final duel between Sol and Insol: the case
$n=5$. As we will see, it is on the outcome of this game that the
fate of the solubility of the quintic polynomial turns.
Incidentally, the group $S_5$ of permutations on $5$ elements is
isomorphic to the group of isometries of a regular icosahedron. So
this game can also be interpreted in terms of moving around
(possibly reflecting at times) an icosahedron. Taking 3 pairs of
opposite edges of an icosahedron appropriately, it is possible to
form three mutually perpendicular golden rectangles. This can be
done 5 different ways. The objects being permuted as the
icosahedron is transformed are these 5 triads of rectangles. The
relationship between the icosahedron and the quintic was explored
by Felix Klein in \cite{Klein}.
\noindent {\bf Game 13.} $n=5$. Sol has a six-shooter, which can
shoot $(123)(45)$, or $(12)$, or $(1234)$, or $(12345)$, or
$(12)(34)$, or $(123)$.
\begin{verse}
First shot: Sol fires $(123)(45)$. But Insol chooses $(124)(35)$
to obtain $(134)$ --- a 3-cycle.
Second shot: Sol fires $(12)$. But Insol deflects the shot with a
$(23)$, which results in $(123)$.
Third shot: Sol fires $(1234)$. Insol retaliates with $(1243)$ to
obtain $(143)$.
Fourth shot: Sol fires $(12345)$. But a quick $(12354)$ ends with
$(254)$ and evades the blast.
Fifth shot: Sol fires $(12)(34)$. Another conjugation from the
hip, this time choosing $(12)(35)$, gives $(354)$.
Final shot: We may assume now that Insol is faced with an incoming
3-cycle, without loss of generality $(123)$. But Insol now
delivers the coup de gras, and chooses $(145)$ to obtain $(153)$.
Insol has stabilised at a 3-cycle and is now invincible.
\end{verse}
\begin{thm}
Insol wins the duel. That is, Insol has a winning strategy for
restricted $5$-Galois.
\end{thm}
The actual game involved in determining whether or not a general
quintic polynomial is soluble is the following more complicated
game. Play is identical to restricted $n$-Galois and
$n$-Galois-lite, but the choices for Insol at each stage lie
somewhere between the many choices of the lite version, and the
fewer choices of the restricted version.
\begin{defn}[$n$-Galois]
Sol chooses $x_0 \neq (1)$. Insol may select
any permutation for $y_0$, then
$x_1 = x_0 y_0 x_0^{-1} y_0^{-1}$. Call all the possible permutations for $x_1$,
in any possible game, \emph{level-1} permutations. For $y_1$,
Insol selects a level-1 permutation and we form $x_2 = x_1 y_1
x_1^{-1} y_1^{-1}$. The set of all possibilities for $x_2$ in
any possible game are called \emph{level-2} permutations. Then
$y_2$ must be a level-2 permutation, and so play continues. If
Insol can avoid the identity, she wins; otherwise Sol wins.
\end{defn}
Those who know some group theory will see that the level-1
positions form $S_n'$, the level-2 positions are $S''_n$, and so
on; so Insol has a winning strategy if and only if $S_n$ is
insoluble.
However, because the $n$ objects are indistinguishable, clearly
all the permutations with the same cycle structure will be in the
same levels. Therefore Insol has more moves than in restricted
$n$-Galois, but less than in Galois-lite. So if Insol can win
restricted $n$-Galois, she can win $n$-Galois.
\begin{cor}
Insol has a winning strategy for $5$-Galois.
\end{cor}
\begin{cor}
The general quintic polynomial is insoluble by radicals.
\end{cor}
And, at last, the score is settled. The forces of insolubility win
the day, at least for $n=5$. I'll leave the other cases for you.
\begin{thebibliography}{Blah}
\bibitem{Bell} E. T. Bell, Men of Mathematics (1937), 406--423
\bibitem{Conway} John Conway, On Numbers and Games (1976)
\bibitem{Dalmas} Andr\'{e} Dalmas, \'{E}variste Galois
\bibitem{Galois} \'{E}variste Galois, Oeuvres Math\'{e}matiques
(1846)
\bibitem{Klein} Felix Klein, Lectures on the Icosahedron and the
Solution of Equations of the Fifth Degree (1884) (English
translation 1913 by G. G. Morrice)
\bibitem{Stewart} Ian Stewart, Galois Theory: Second Edition (1989)
\begin{flushright}
--- Daniel Mathews
\end{flushright}
Department of Mathematics, Stanford University, 94305 CA, USA
Email: mathews@math.stanford.edu
Web: http://math.stanford.edu/\~{}mathews
\end{thebibliography}
\end{document}