§ DS-12 Simple Group and Hölder Program

**# §12 Simple Groups and the Hölder Program

§12.1 Simple Groups

Some groups cannot be built out of other groups. For example, what if $H$ admits no nontrivial normal subgroups? Then there can be no short exact sequence unless $H\cong K$ or $G\cong H$ . In this sense, groups with no nontrivial normal subgroups are the simplest groups.

Definition 12.1 A group $H$ is called a simple group if it has no nontrivial normal subgroups.

Example 12.1 A cyclic group is simple if and only if it is finite of prime order.

If it has prime order, then it is simple because it has no subgroups other than itself and ${1}$ .

Conversely, a cyclic group of order $n$ has a subgroup for every number $n/k$ dividing $n$ ; for example, given any generator $x$ , consider the set

\{1,x^{k},\ldots\}.

Therefore, a cyclic group is simple precisely when it has prime order.

Example 12.2 $\mathbb{Z}$ is not simple.

It has many subgroups, and every subgroup of an Abelian group is normal.

Example 12.3 $A_1,A_2,A_3$ are simple.

$A_1$ and $A_2$ each have one element. By the First Isomorphism Theorem, $A_3$ is a subgroup of index $2$ in $S_3$ —it is a group of order $3$ , and hence a cyclic group of prime order.

Example 12.4 $A_4$ is not simple.

We need to find a nontrivial normal subgroup. The unique nontrivial normal subgroup of this group consists of elements that are products of two cycles. This normal subgroup is isomorphic to the Klein four group, while the quotient group is cyclic of order $3$ .

Theorem 12.1 For $n\geqslant5$ , $A_n$ is simple.

Proof: To prove that the alternating group $A_n$ is simple for $n\geqslant5$ , we must show that $A_n$ has no normal subgroups other than the trivial subgroup ${e}$ and $A_n$ itself.

Let $N$ be a normal subgroup of $A_n$ with

N\neq\{e\}.

We will prove that

N=A_n,

and hence that $A_n$ is simple.

$N$ contains a $3$ -cycle.

Since $N$ is nontrivial, there exists a nonidentity element

\sigma\in N.

Consider the cycle types appearing in the disjoint-cycle decomposition of $\sigma$ .

Case 1: If $\sigma$ contains a $3$ -cycle, then $N$ contains a $3$ -cycle.

Case 2: If $\sigma$ contains no $3$ -cycle, we will show in the following steps that $N$ nevertheless contains a $3$ -cycle.

Conjugation and normality.

Since $N$ is normal in $A_n$ , for every

\tau\in A_n,

we have

\tau\sigma\tau^{-1}\in N.

This property allows us to generate new elements of $N$ from elements already known to lie in $N$ .

Producing $3$ -cycles.

Express elements as products of $3$ -cycles.

Every even permutation can be expressed as a product of $3$ -cycles. For example, a cycle of length $k\geqslant3$ ,

(a_1\,a_2\,\dots\,a_k),

can be written as

(a_1\,a_2\,a_3) (a_1\,a_3\,a_4) \dots (a_1\,a_{k-1}\,a_k).

Use commutators to obtain a $3$ -cycle.

Consider $\sigma$ and a suitable

\tau\in A_n.

The commutator

\gamma = \sigma\tau\sigma^{-1}\tau^{-1}

is an element of $N$ , because $N$ is normal, and it can be a $3$ -cycle.

$N$ contains all $3$ -cycles.

Conjugacy classes.

In $A_n$ , the $3$ -cycles split into two conjugacy classes, but their union is the set of all $3$ -cycles.

Since $N$ contains at least one $3$ -cycle and is normal, it must contain every $3$ -cycle in the corresponding conjugacy class.

$A_n$ is generated by $3$ -cycles.

For

n\geqslant5,

the group $A_n$ is generated by its $3$ -cycles.

Therefore, the subgroup of $A_n$ generated by all $3$ -cycles is $A_n$ itself.

Conclusion.

Since $N$ contains all $3$ -cycles, $N$ generates $A_n$ . Hence

N=A_n.

Therefore, for

n\geqslant5,

$A_n$ is simple.

~\tag*{$\square$}

§12.2 The Hölder Program

Now that we have seen many examples of groups, we would like to begin classifying them.

Can we come up with a general strategy that would allow us to say:

“I know all groups”?

Question: How do we classify all groups?

In some sense, this question has no satisfactory answer.

We can try to understand all simple groups, and then understand all the ways in which they can be assembled.

The strategy that began in the nineteenth century is called the Hölder program.

It looks very natural.

The problem is that we do not know how to carry it out.

We cannot even classify all simple groups.

Can we at least understand all finite simple groups and their extensions?

That would classify all finite groups.

We still do not know how to do this.

By around 1985, we were able to classify all finite simple groups, but we still do not know how to solve the problem of classifying their extensions.

To give you some sense of how difficult the classification problem is, consider the following theorem, which helped earn Thompson a Fields Medal:

Theorem 12.2 (Feit-Thompson Theorem, or Odd Order Theorem)

Every finite non-Abelian simple group has even order.

Thus, for example, if you give me a non-Abelian group of odd order, I immediately know that it is not simple.

Definition 12.2 The Hölder program for classifying groups is:

(1) Classify all simple groups.

(2) Classify all ways of constructing extensions of simple groups.

We do not know how to complete this program.

For example, we do not know how to carry out Step (1) in general.

We only know how to classify finite simple groups, and this classification was not completed until around 1985.

Even for finite groups, we have not completed Step (2).

§12.3 Solvable Groups

Following the Hölder program's emphasis on understanding and decomposing complicated group structures, we now study solvable groups, which can be systematically decomposed into simpler Abelian components.

Definition 12.3 A group $G$ is called solvable if it has a finite sequence of subgroups

G=G_{0}\triangleright G_{1}\triangleright G_{2}\triangleright\cdots\triangleright G_{n}=\{e\}

such that each $G_i$ is normal in $G_{i-1}$ and the corresponding quotient group

G_{i-1}/G

is Abelian.

Remark A group is called “solvable” because it can be decomposed into smaller and simpler pieces, eventually reaching the trivial group.

Example 12.5 Every cyclic group is Abelian, and therefore it is trivially solvable.

Example 12.6

$S_3$ is solvable.

A normal series is

S_3\triangleright A_3\triangleright\{e\}.

The quotient groups are

S_3/A_3 \cong \mathbb{Z}/2\mathbb{Z},

and

A_3/\{e\} \cong \mathbb{Z}/3\mathbb{Z}.

Both are Abelian, so $S_3$ is solvable.

Example 12.7 $S_5$ is not solvable.

The absence of a suitable normal series highlights the difference between solvable groups and more complicated groups.

Proposition 12.3 Every subgroup of a solvable group is solvable.

Proof: A solvable group $G$ has a chain of subgroups

G=G_{0} \triangleright G_{1} \triangleright \dots \triangleright G_{n} = \{e\},

where each quotient group

G_{i-1}/G_i

is Abelian.

Let

H\subseteq G

be a subgroup.

We may intersect $H$ with each group in the chain:

H\cap G_{0} \triangleright H\cap G_{1} \triangleright \dots \triangleright H\cap G_{n} = \{e\}.

Each quotient group

(H\cap G_{i-1})/(H\cap G_i)

is a subgroup of the Abelian quotient group

G_{i-1}/G_i,

and is therefore itself Abelian.

Thus we obtain a normal series for $H$ whose quotient groups are Abelian.

Therefore,

H

is solvable.

~\tag*{$\square$}

Proposition 12.4 A quotient group of a solvable group is solvable.

Proof: Suppose $G$ is solvable.

Then it has a chain of normal subgroups

G=G_{0} \triangleright G_{1} \triangleright \dots \triangleright G_{n} = \{e\},

where each quotient

G_{i-1}/G_i

is Abelian.

Let

N\triangleleft G

be a normal subgroup. We want to prove that

G/N

is solvable.

Use the series in $G$ to form a new series in the quotient:

G_{0}/N \triangleright G_{1}/N \triangleright \dots \triangleright G_{n}/N = \{e\}.

Each new quotient

(G_{i-1}/N)/(G_i/N) \cong G_{i-1}/G_i

is Abelian because the original quotient

G_{i-1}/G_i

is Abelian.

Therefore,

G/N

is solvable.

~\tag*{$\square$}

Our discussion of solvable groups illustrates how groups can be decomposed into simpler components.

Although solvable groups have a clear and manageable structure, the study of nonsolvable groups, such as simple groups, remains essential for understanding the broader landscape of group theory.

In the next chapter, we will continue to explore more specialized results in group theory, such as the Sylow theorems, which provide deeper insight into the structure of finite groups. **

Contents

§12.1 Simple Groups

§12.2 The Hölder Program

§12.3 Solvable Groups