Finite Simple Groups

This post is about the classification of finite simple groups and its uses. It will be mostly expository and contain relatively few proofs.

The goal of understanding all finite groups is an ambitious one. Given a big problem, one tries to break it up into smaller pieces and solving them individually. For groups, this takes the form of quotienting out by normal subgroups until one can't go any further, then trying to put these pieces back together to understand the big group. When one can't go any further, these pieces are 'simplest' in some sense, i.e. have no non-trivial normal subgroups, hence they are called simple groups. Amazingly, we now understand these completely: 

    Classification: A finite simple group belongs to one of the following categories:

1. finite cyclic groups of prime order
2. alternating groups
3. finite groups of Lie type
4. sporadic groups

We discuss each of these very briefly in turn but will say nothing about the proof of the classification, which has taken decades and 10,000 pages. It is in the process of a major revision, but even then the end product will no doubt be a fearsome beast not to be trifled with. Note that there has been at least one whole book devoted to fleshing out the details of what follows, i.e. stating what the main characters are, so naturally the following discussion barely scratches the surface.

Finite cyclic groups

A finite cyclic group of order $n$ has as normal subgroups cyclic groups of order dividing $n$, so it follows immediately that the simple ones are precisely the ones of prime order.

Alternating groups

The symmetric group $S_n$ is the set of permutations of $[n]:=\{1, \cdots, n\}$ . To a permutation $\sigma$ we can associate the sign, which is the parity of number of transpositions $(a b)$ that multiply together to give $\sigma$. This assignment is in fact a group homomorphism, and the alternating group $A_n$ is then defined to be the kernel. For $n \geq 5$, these are simple. Whether or not there is a good proof of this fact is a matter of taste, but there are certainly ones which I dislike more than others. See this MO thread, which mentions Iwasawa's criterion, discussed below. Out of Keith Conrad's list, my favourite is the second proof, which is reproduced below:

One checks simplicity of $A_5$ and $A_6$ by an extremely simple-minded technique: show that there is no set conjugacy classes whose sizes add up to a divisor of the size of the group.

 We show that if a subgroup $H$ contains a $3$-cycle, then every $3$-cycle is in $H$. Then we are done since $A_n$ is generated by $3$-cycles. For concreteness, suppose we know $(a\; b\; c) \in H$, and we want to show $(1\; 2\; 3) \in H$.

Since they have the same cycle type, so we have $\sigma \in S_n$ such that $(a\; b\; c) = \sigma(1\; 2\; 3) \sigma^{-1}$. If $\sigma$ is even, i.e.\ $\sigma \in A_n$, then we have that $(1\; 2\; 3) \in \sigma^{-1}H\sigma = H$, by the normality of $H$ and we are trivially done.

If $\sigma$ is odd, replace it by $\bar{\sigma} = \sigma \cdot (4\; 5)$. Here is where we use the fact that $n \geq 5$ (we will use it again later). Then we have

\[\bar{\sigma} (1\; 2\; 3) \bar{\sigma}^{-1} = \sigma (4\; 5)(1\; 2\; 3)(4\; 5) \sigma^{-1} = \sigma(1\; 2\; 3) \sigma^{-1} = (a\; b\; c),\]

using the fact that $(1\; 2\; 3)$ and $(4\; 5)$ commute. Now $\bar{\sigma}$ is even. So $(1\; 2\; 3) \in H$ as above.

Now suppose $n \geq 7$ and set $N$ to be a normal subgroup of $A_n$. We show that $N$ contains a $3-$cycle.

Let \( \sigma \) be a non-identity element of \( N \). It moves some number. By relabeling, we may  

suppose \( \sigma(1) = 1 \). Let \( \sigma = (ijk) \), where \( i, j, k \) are not 1 and \( \sigma(1) = i, j, k \). Then  

\[\sigma^{-1}(1) = \sigma^{-1}(\sigma(1)) = 1,\]

so we see that \( \sigma^{-1} \) is a 3-cycle. Since \( N \) is normal, it contains all conjugates of its elements. Now write  

\[\sigma^{-1} = \tau \sigma \tau^{-1}.\]

Let \( \rho = \tau \sigma \tau^{-1} \), so that \( \rho = \sigma(1) \). Writing  

\[\rho = (\alpha \beta \gamma)\]

we see that  

\[\tau^{-1} \sigma \tau = (\tau^{-1} \alpha \ \tau^{-1} \beta \ \tau^{-1} \gamma).\]

Thus, \( \sigma^{-1} \) is also a 3-cycle. Therefore, \( N \) moves at most 6 numbers in \( S_{12} \) and is a product of two 3-cycles.  

Let \( H \) be the copy of \( A_6 \) inside \( A_n \) corresponding to the even permutations of those 6 numbers  

(possibly augmented to 6 arbitrarily if in fact \( N \) moves fewer numbers). Then \( N \cap H \) is nontrivial  

(it contains \( \sigma \)) and is a normal subgroup of \( H \). Since \( H = A_6 \), which is simple,  

we conclude that \( N \cap H = H \). Thus, \( H \subseteq N \), so \( N \) contains a 3-cycle. $\blacksquare$

Finite groups of Lie type

It is an unfortunate fact of life that basic terminology in mathematics is often ambiguous due to the inevitable messiness of how the theory develops. It is unclear what finite groups of Lie type means, so we will just have to pick a definition and hope not to offend too many people. See also this MO thread.

The classification of Dynkin diagrams

This is inevitably linked to the theory of Lie algebras, and will involve some concepts that we won't be able to discuss properly for reasons of length. Chevalley's construction goes as follows: semisimple Lie algebras over $\mathbb{C}$ are classified by Dynkin diagrams. Associate to a Lie algebra a $\mathbb{Z}-$form $G_{\mathbb{Z}}$. Over a field $k$ of characteristic $p$ build the tensor product $G_{\mathbb{Z}} \otimes _{\mathbb{F_p} k}$, and exponentiate the 'nilpotent' parts this analogously to how you can exponentiate nilpotent elements in Lie algebras of Lie groups. This gives the unipotent subgroups $U^{\pm}$ that generate the untwisted Chevalley group over the field $k$ $G(k)$ of a given type. 

The Dynkin diagrams admit graph automorphisms $\tau$, to which one can associate subgroups $U_{\tau}^{\pm}$ of the unipotent subgroups. These subgroups generate the twisted Chevalley groups.

An alternative viewpoint is the following. A variety is the set of zeros of a polynomial. An algebraic group is a variety with a group structure such that all the basic operations are polynomials. The classic example is $SL_n(k)$, which is the root of the polynomial $det=1$. The classification of semisimple algebraic groups (i.e. it has only trivial soluble connected closed normal subgroups) directly produces abstract root subgroups (without exponentiating from some Lie algebra). The relations between the abstract root subgroups can then be shown to satisfy the restrictions corresponding to a root system, so that the Cartan-Killing classification of root systems could be invoked, to give essentially the same set of answers for algebraic groups as for Lie algebras. This is one of the miracles of mathematics, that Dynkin diagrams classify both things. For other things with an ADE classification, see the wikipedia page or this article. (Pure speculation: given Scholze's program of showing that most things which seem analytic, like the properties of complex manifolds, are in fact consequences of the underlying algebraicity, might there be some deeper explanation of this coincidence, in the condensed framework or otherwise?)
Setting $q=p^a$ and defining the Frobenius automorphism $\phi: x \mapsto x^p$, the fixed points under $\phi^a$ of the algebraic group $G(\overline{\mathbb{F}_p})$ are $G({\mathbb{F}_q})$. The twisted groups arise as the fixed points of a product between $\phi^a$ with a graph automorphism $\tau'$, which are determined by the automorphisms of the Dynkin diagram.

It is easier to see the simplicity from this viewpoint: in the relevant cases one can use Tits' theory of BN-pairs to show that $G({\mathbb{F}_q})/ Z(G({\mathbb{F}_q}))$ is simple, which is why we always have to pass to the projective matrix group.

Below we discuss Iwasawa's criterion and indicate how it gives an elementary proof of the simplicity of $PSL_n(k)$ for any field $k$ when $n>2$ and for any field of size at least 4 when $n=4$, following Keith Conrad's article.

Lemma 1: If \( G \) acts doubly transitively on \( X \), then the stabilizer subgroup of each point in \( X \) is a maximal subgroup of \( G \).

A maximal subgroup is a proper subgroup contained in no other proper subgroup.

Proof: Pick \( x \in X \) and let \( H_x = \text{Stab}_x \).

Step 1: For each \( g \notin H_x \), we have \( G = H_x \cup H_x g H_x \).

For \( g \in G \) such that \( g \notin H_x \), we will show \( g \in H_x g H_x \). Both \( gx \) and \( g^{-1} x \) are not \( x \), so by double transitivity with the pairs \( (x, gx) \) and \( (x, g^{-1}x) \), there is some \( g' \in G \) such that

\[g' x = x \quad \text{and} \quad g' (gx) = g^{-1} x.\]

The first equation implies \( g' \in H_x \), so let us write \( g' = h \). Then

\[h (gx) = g^{-1} x,\]

so \( g = h g^{-1} h^{-1} H_x \), which shows \( g \in H_x g H_x \).

Step 2: \( H_x \) is a maximal subgroup of \( G \).

The group \( H_x \) is not all of \( G \), since \( H_x \) fixes \( x \) while \( G \) moves \( x \) to each element of \( X \), and \( |X| \geq 2 \). Let \( K \) be a subgroup of \( G \) strictly containing \( H_x \) and pick \( g \in K \setminus H_x \). By step 1, we have \( G = H_x \cup H_x g H_x \). Since both \( H_x \) and \( H_x g H_x \) are in \( K \), we conclude \( G = K \), proving the maximality of \( H_x \). $\blacksquare$

Lemma 2: Suppose \( G \) acts doubly transitively on a set \( X \). Any normal subgroup \( N \triangleleft G \) acts on \( X \) either trivially or transitively.

Proof: Suppose \( N \) does not act trivially, meaning there exists \( n \in N \setminus \{1\} \) such that \( nx \neq x \) for some \( x \in X \). Pick arbitrary \( y, y' \in X \) with \( y \neq y' \). By double transitivity, there is \( g \in G \) such that

\[g x = y \quad \text{and} \quad g(nx) = y'.\]

Then

\[y' = (g n g^{-1})(g x) = (g n g^{-1})(y).\]

Since \( g n g^{-1} \in N \), this shows that \( N \) acts transitively on \( X \).

Iwasawa's criterion: Let \( G \) act doubly transitively on a set \( X \). Assume the following:

1. For some \( x \in X \), the group \( \text{Stab}_x \) has an abelian normal subgroup whose conjugate subgroups generate \( G \).
2. \( [G, G] = G \).

Then \( G/K \) is a simple group, where \( K \) is the kernel of the action of \( G \) on \( X \).

Proof: To show \( G/K \) is simple, we will show that the only normal subgroups of \( G \) lying between \( K \) and \( G \) are \( K \) and \( G \).

Let \( K \subseteq N \triangleleft G \). Let \( H = \text{Stab}_x \), so \( H \) is a maximal subgroup of \( G \) (by Lemma 1). Since \( N \) is normal, the subgroup $N H = \{ nh \mid n \in N, h \in H \}$

is a subgroup of \( G \) and contains \( H \). By maximality, either \( N H = H \) or \( N H = G \).

By Lemma 2, \( N \) acts either trivially or transitively on \( X \). If \( N H = H \), then \( N \subseteq H \), so \( N \) fixes \( x \). Therefore, \( N \) does not act transitively on \( X \), so \( N \) must act trivially, implying \( N \subset K \). Since \( K \subseteq N \) by assumption, we obtain \( N = K \).

Now suppose \( N H = G \). Let \( U \) be the abelian normal subgroup of \( H \) from the hypothesis; its conjugate subgroups generate \( G \). Since \( U \triangleleft H \), we get

\[N U \triangleleft N H = G.\]

For \( g \in G \), we see that

\[g U g^{-1} \subseteq g (N U) g^{-1} = N U.\]

This shows \( N U \) contains all conjugates of \( U \). By hypothesis, this implies \( N U = G \). Hence $G/N \equiv U/(N \cap U)$. $U$ is abelian, so the quotient is too, hence $[G,G] \subset N$. Since \( G = [G, G] \) by assumption, we conclude \( N = G \), proving that \( G/K \) is simple. $\blacksquare$

By direct matrix computation one shows that the action of \( SL_2(F) \) on the linear subspaces of \( F^2 \) is doubly transitive and satisfies the criterion. Similarly by direct computation, the action of \( SL_n(F) \) on \( \mathbb{P}^{n-1}(F) \) is doubly transitive, with kernel equal to the center of the group, and the stabilizer of some point has an abelian normal subgroup, hence \( PSL_n(F) \) is simple.

Sporadic groups

In most classifications there are 26 sporadic groups, so named because they don't fit into any infinite family. I know of no particularly good way to describe them apart from by explicit construction, and that has been done in many books so I won't do it here. Whether or not these should be considered 26 exceptional cases is up for debate, but there are some relations between them so it is not entirely unfounded to lump them together. 20 of these form the so-called 'happy family', and the rest are called 'pariahs'. The largest member of the happy family is called the monster group, and the second largest the baby monster. Letting mathematicians name things was clearly a great idea. The monster group is the source of what's known as monstrous moonshine, linking string theory, Lie algebras, and modular forms to finite simple groups. Subsequent research shows that there exist similar connections, or similar moonshines, coming from some of the other simple groups, showing that the pariahs aren't actually pariahs after all. Those who are interested in algebraic coding theory will appreciate the uses of these sporadic groups to create codes, exposed for example in Conway and Sloane's book.

Consequences of the classification

There is an entire book on how to use the classification to prove results in group theory, from which some of the material is drawn. The others are collected from here. We highlight some of the numerous consequences, some not obviously related to finite group theory. Not all purported consequences of CFSG require its full strength: often one can get away with appealing to just a chunk of it: a common chunk to appeal to is the Feit-Thompson theorem that all groups of odd order are soluble. Most of the proofs of the results below are intricate, technical, difficult, rely heavily on case by case analysis of the different categories of finite simple groups, and are far beyond this blogger's patience and capability, so naturally it is left to the interested reader to explore. It seems incredible that it is possible to prove results of this nature at all. 

• (Schreier conjecture) The outer automorphism group of a finite simple group $S$ is soluble
• (Ore conjecture) In a finite simple group, every element is a commutator.
• (Shalev) Every word has width 3 in every sufficiently large finite simple group
• (Frobenius conjecture) If exactly $n$ elements of a group $G$ satisfy $x^n = 1$ (for $n$ which divides $G$), then these elements form a subgroup of $G$. (It is also surprising that no proof that doesn't go via the classification is known.)
• For global fields $L > K$, the relative Brauer group $B(L/K)$ is infinite.
• (Fried) A polynomial \( f(x) \in \mathbb{C}[x] \) is indecomposable if whenever \( f(x) = g(h(x)) \) for polynomials \( g \), \( h \), one of \( g \) or \( h \) is linear. Let \( f, g \) be nonconstant indecomposable polynomials over \( \mathbb{C} \). Suppose that \( f(x) - g(y) \) factors in \( \mathbb{C}[x,y] \). Then either \[ g(x) = f(ax+b) \] for some \( a,b \in \mathbb{C} \), or \[ \deg f = \deg g = 7, 11, 13, 15, 21, \text{ or } 31,\] and each of these possibilities does occur.
• (Guralnick-Thompson conjecture) Fixe a genus \( g \). As \( X \) varies over all branched covers of the Riemann sphere of that genus, only finitely many non-alternating non-abelian simple groups can arise as composition factors of a monodromy group.
• (Nikolov-Segal) Finite-index subgroups of finitely generated profinite groups are open. This implies that the topology on such groups is uniquely determined by the group structure. 

Not all of those words will be meaningful for everybody. Nonetheless, I hope that by now the reader will have some vague idea of what the classification is and how far reaching its consequences are both for the world of finite groups and for mathematics more generally.