The Jordan-Hölder Theorem

# The Jordan-Hölder Theorem

Recall from the Composition Series in a Group page that a subnormal series:

(1)
\begin{align} \quad \{ 1 \} = G_0 \triangleleft G_1 \triangleleft ... \triangleleft G_k = G \end{align}

is said to be a composition series if all of the factors $G_{i+1}/G_i$ are simple groups (that is, they have no nontrivial proper subgroups).

We are about to look at a very important result known as the Jordan-Hölder theorem. We will first need a preliminary result.

 Theorem 1 (The Schreier Refinement Theorem): Let $G$ be a group. Then any two subnormal series have equivalent refinements.

We will not prove the Schreier refinement theorem, but it can be proven using The Third Group Isomorphism Theorem.

 Lemma 2: Let $G$ be a group. If $G$ is a finite group then $G$ has a composition series.
• Proof: We prove the result by induction. Suppose that every group of order less than $|G|$ has a composition series.
• Now if $G$ is a simple group then $1 \triangleleft G$ is a composition series. If $G$ is not a simple group then there exists a nontrivial proper normal subgroup of $G$. Since $G$ is a finite group, a maximal nontrivial proper normal subgroup exists. Denote this subgroup by $H$. Since $H$ is a proper subgroup of $G$ we have that $|H| \leq |G|$. By the induction hypothesis, $H$ has a composition series:
(2)
\begin{align} \quad \{1 \} = H_0 \triangleleft H_1 \triangleleft ... \triangleleft H_k = H \end{align}
• But then $H \triangleleft G$ by the maximality of $H$. So:
(3)
\begin{align} \quad \{ 1 \} = H_0 \triangleleft H_1 \triangleleft ... \triangleleft H_k = H \triangleleft G \end{align}
• The above subnormal series is a composition series of $G$. $\blacksquare$
 Theorem 3 (The Jordan-Hölder Theorem): Let $G$ be a group. Then any two composition series in $G$ are equivalent.
• Proof: Denote two composition series in $G$ by:
(4)
\begin{align} \quad \{ 1 \} = G_0 \triangleleft G_1 \triangleleft ... \triangleleft G_k = G \quad (*) \end{align}
(5)
\begin{align} \quad \{ 1 \} = H_0 \triangleleft H_1 \triangleleft ... \triangleleft H_l = G \quad (**) \end{align}
• By the Schreier refinement theorem, these two series have equivalent refinements. But these two series are composition series, and so their factors $G_{i+1}/G_i$ and $H_{j+1}/H_j$ are simple groups. So if we delete repetitions from the refinement of $(*)$ we get $(**)$. And similarly, if we delete repetitions from the refinement of $(**)$ we get $(*)$. So $k = l$ and $G_{i+1}/G_i \cong H_{j+1}/H_j$ for $i, j \in \{ 0, 1, ..., k - 1 = l - 1 \}$. $\blacksquare$