The Jordan-Hölder Theorem

In order to state the main theorem of this section, we first need two definitions.

Definition 11.3.1   Let $G$ be a group with $N \lhd G$. Then $N$ is called maximal in $G$ if $N\subset G$ properly (i.e., $N\not = G$) and there does not exist any normal subgroup where the inclusions are all meant to be proper.

On the basis of Corollary 8.3.3 another way of characterizing a maximal normal subgroup is as follows: $N$ is a maximal normal subgroup of $G$ if and only if $G/N$ is simple. (See the exercises below.)

We now state our last definition.

Definition 11.3.2   A composition series for a group $G$ is a normal series as in (11.7), where all the inclusions are proper and such that $G_{i+1}$ is maximal in $G_i$ (in other words, each factor is simple).

For example, in the case of the previously given normal series for $S_4$ in (11.8), only the first $\{e\} \subset C_2\subset V_4\subset A_4\subset S_4$ is a composition series for $S_4$. A composition series for $A_4$ would be: $\{e\} \subset C_2\subset V_4\subset A_4$. Note that $A_4 \supset V_4 \supset \{e\}$ would not be a composition series for $A_4$ (Why?). Unlike the case of normal series, it is possible that an arbitrary group does not have a composition series (see exercise 1 for this section) or even if it does have one a subgroup of it may not have one. Of course, a finite group does have a composition series.

We now consider the case in which a group, $G$, does have a composition series, and we prove the following important theorem.

Theorem 11.3.3 (Jordan-Hölder)   : If a group $G$ has a composition series, then any two composition series are equivalent (i.e., the composition factors are unique).

Proof: Suppose we are given two composition series. Applying Schreier's refinement theorem (Theorem 11.2.2), we get that the two composition series have equivalent refinements. But the only refinement of a composition series is one obtained by introducing repetitions. If in the 1-1 correspondnece between the factors of these refinements, the paired factors equal to $\{e\}$ are disregarded (i.e., if we drop the repetitions), we get clearly that the original composition series are equivalent. $\Box$

It was mentioned in the introduction to Chapter 6 that the simple groups are important because ``they play a role in finite group theory somewhat analogous to that of the primes in number theory.'' In particular, an arbitrary finite group, $G$, can be broken down into simple components. These uniquely determined simple components are, according to the Jordan-Hölder, the factors of a composition series for $G$.

We close by giving an application of this theorem. In particular, we use the Jordan-Hölder Theorem to prove the uniqueness part of the Fundamental Theorem of Arithmetic. The Fundamental Theorem of Arithmetic states that every positive integer not equal to a prime can be factored uniquely (up to order) into a product of primes.

First, we claim that such a factorization exists. Indeed, suppose $n$ is composite (i.e., $n > 1$ and $n$ is not a prime). Then an easy induction shows that $n$ has a prime divisor $p$ and we can write $n = pn_1$, where $n_1$ is an integer satisfying $n_1 < n$. If $n_1$ is prime, the claim holds. Otherwise, $n_1$ has a prime factor $p_1$, and $n_1 = p_1n_2$ where $n_2 < n_1$ is an integer. Continuing in this fashion, we must come to an equation $n_{j- 1} = p_{j- 1}n_j$, where $n_j$ is a prime $p_j$, since the sequence of decreasing positive integers

\begin{displaymath}
n > n_1 > n_2 > n_3 > ...
\end{displaymath}

cannot continue indefinitely. We now have that $n = pp_1p_2...p_j$ is a product of primes. This proves the existence claim.

On the basis of the Jordan-Hölder Theorem, we can easily show the other part of the Fundamental Theorem of Arithmetic, i.e., apart from order of the factors, the representation of $n$ as product of primes is unique. To do this suppose that

\begin{displaymath}
n = p_1p_2...p_s,
\end{displaymath}

and

\begin{displaymath}
n = q_1q_2...q_t
\end{displaymath}

where the $p_i$ and $q_j$ are primes. Then denoting, as usual, by $C_k$ the cyclic group of order $k$, we have

\begin{displaymath}
C_n\supset C_{p_2...p_s} \supset C_{p_3...p_s}
\supset ...\supset C_{p_s}\supset \{e\},
\end{displaymath}

and

\begin{displaymath}C_n\supset C_{q_2...q_t} \supset C_{q_3...q_t}
\supset ...\supset C_{q_t}\supset \{e\},
\end{displaymath}

as two composition series for $C_n$. But the Jordan-Hölder Theorem implies these must be equivalent; hence we must have $s = t$ and by suitably arranging $p_i = q_i$, $1 \leq i \leq s$. Thus we have established the unique factorization theorem for positive integers as an application of the Jordan-Hölder Theroem.



Subsections

David Joyner 2007-08-06