A *subnormal series* of a group \(G\) is a finite sequence of subgroups
\begin{equation*}G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \},\end{equation*}
where \(H_i\) is a normal subgroup of \(H_{i+1}\). If each subgroup \(H_i\) is normal in \(G\), then the series is called a *normal series*. The *length* of a subnormal or normal series is the number of proper inclusions.

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Example13.11

Any series of subgroups of an abelian group is a normal series. Consider the following series of groups:
\begin{gather*}
{\mathbb Z} \supset 9{\mathbb Z} \supset 45{\mathbb Z} \supset 180{\mathbb Z} \supset \{0\},\\
{\mathbb Z}_{24} \supset \langle 2 \rangle \supset \langle 6 \rangle \supset \langle 12 \rangle \supset \{0\}.
\end{gather*}

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Example13.12

A subnormal series need not be a normal series. Consider the following subnormal series of the group \(D_4\):
\begin{equation*}D_4 \supset \{ (1), (12)(34), (13)(24), (14)(23) \} \supset \{ (1), (12)(34) \} \supset \{ (1) \}.\end{equation*}
The subgroup \(\{ (1), (12)(34) \}\) is not normal in \(D_4\); consequently, this series is not a normal series.

A subnormal (normal) series \(\{ K_j \}\) is a *refinement of a subnormal (normal) series* \(\{ H_i \}\) if \(\{ H_i \} \subset \{ K_j \}\). That is, each \(H_i\) is one of the \(K_j\).

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Example13.13

The series
\begin{equation*}{\mathbb Z} \supset 3{\mathbb Z} \supset 9{\mathbb Z} \supset 45{\mathbb Z} \supset 90{\mathbb Z} \supset 180{\mathbb Z} \supset \{0\}\end{equation*}
is a refinement of the series
\begin{equation*}{\mathbb Z} \supset 9{\mathbb Z} \supset 45{\mathbb Z} \supset 180{\mathbb Z} \supset \{0\}.\end{equation*}

The best way to study a subnormal or normal series of subgroups, \(\{ H_i \}\) of \(G\), is actually to study the factor groups \(H_{i+1}/H_i\). We say that two subnormal (normal) series \(\{H_i \}\) and \(\{ K_j \}\) of a group \(G\) are *isomorphic* if there is a one-to-one correspondence between the collections of factor groups \(\{H_{i+1}/H_i \}\) and \(\{ K_{j+1}/ K_j \}\).

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Example13.14

The two normal series
\begin{gather*}
{\mathbb Z}_{60} \supset \langle 3 \rangle \supset \langle 15 \rangle \supset \{ 0 \}\\
{\mathbb Z}_{60} \supset \langle 4 \rangle \supset \langle 20 \rangle \supset \{ 0 \}
\end{gather*}
of the group \({\mathbb Z}_{60}\) are isomorphic since
\begin{gather*}
{\mathbb Z}_{60} / \langle 3 \rangle \cong \langle 20 \rangle / \{ 0 \} \cong {\mathbb Z}_{3}\\
\langle 3 \rangle / \langle 15 \rangle \cong \langle 4 \rangle / \langle 20 \rangle \cong {\mathbb Z}_{5}\\
\langle 15 \rangle / \{ 0 \} \cong {\mathbb Z}_{60} / \langle 4 \rangle \cong {\mathbb Z}_4.
\end{gather*}

A subnormal series \(\{ H_i \}\) of a group \(G\) is a *composition series* if all the factor groups are simple; that is, if none of the factor groups of the series contains a normal subgroup. A normal series \(\{ H_i \}\) of \(G\) is a *principal series* if all the factor groups are simple.

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Example13.15

The group \({\mathbb Z}_{60}\) has a composition series
\begin{equation*}{\mathbb Z}_{60} \supset \langle 3 \rangle \supset \langle 15 \rangle \supset \langle 30 \rangle \supset \{ 0 \}\end{equation*}
with factor groups
\begin{align*}
{\mathbb Z}_{60} / \langle 3 \rangle & \cong {\mathbb Z}_{3}\\
\langle 3 \rangle / \langle 15 \rangle & \cong {\mathbb Z}_{5}\\
\langle 15 \rangle / \langle 30 \rangle & \cong {\mathbb Z}_{2}\\
\langle 30 \rangle / \{ 0 \} & \cong {\mathbb Z}_2.
\end{align*}
Since \({\mathbb Z}_{60}\) is an abelian group, this series is automatically a principal series. Notice that a composition series need not be unique. The series
\begin{equation*}{\mathbb Z}_{60} \supset \langle 2 \rangle \supset \langle 4 \rangle \supset \langle 20 \rangle \supset \{ 0 \}\end{equation*}
is also a composition series.

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Example13.16

For \(n \geq 5\), the series
\begin{equation*}S_n \supset A_n \supset \{ (1) \}\end{equation*}
is a composition series for \(S_n\) since \(S_n / A_n \cong {\mathbb Z}_2\) and \(A_n\) is simple.

#####
Example13.17

Not every group has a composition series or a principal series. Suppose that
\begin{equation*}\{ 0 \} = H_0 \subset H_1 \subset \cdots \subset H_{n-1} \subset H_n = {\mathbb Z}\end{equation*}
is a subnormal series for the integers under addition. Then \(H_1\) must be of the form \(k {\mathbb Z}\) for some \(k \in {\mathbb N}\). In this case \(H_1 / H_0 \cong k {\mathbb Z}\) is an infinite cyclic group with many nontrivial proper normal subgroups.

Although composition series need not be unique as in the case of \({\mathbb Z}_{60}\), it turns out that any two composition series are related. The factor groups of the two composition series for \({\mathbb Z}_{60}\) are \({\mathbb Z}_2\), \({\mathbb Z}_2\), \({\mathbb Z}_3\), and \({\mathbb Z}_5\); that is, the two composition series are isomorphic. The Jordan-Hölder Theorem says that this is always the case.

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Theorem13.18Jordan-Hölder

Any two composition series of \(G\) are isomorphic.

##### Proof

We shall employ mathematical induction on the length of the composition series. If the length of a composition series is 1, then \(G\) must be a simple group. In this case any two composition series are isomorphic.

Suppose now that the theorem is true for all groups having a composition series of length \(k\), where \(1 \leq k \lt n\). Let
\begin{gather*}
G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}\\
G = K_m \supset K_{m-1} \supset \cdots \supset K_1 \supset K_0 = \{ e \}
\end{gather*}
be two composition series for \(G\). We can form two new subnormal series for \(G\) since \(H_i \cap K_{m-1}\) is normal in \(H_{i+1} \cap K_{m-1}\) and \(K_j \cap H_{n-1}\) is normal in \(K_{j+1} \cap H_{n-1}\):
\begin{gather*}
G = H_n \supset H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \}\\
G = K_m \supset K_{m-1} \supset K_{m-1} \cap H_{n-1} \supset \cdots \supset K_0 \cap H_{n-1} = \{ e \}.
\end{gather*}
Since \(H_i \cap K_{m-1}\) is normal in \(H_{i+1} \cap K_{m-1}\), the Second Isomorphism Theorem (Theorem 11.12) implies that
\begin{align*}
(H_{i+1} \cap K_{m-1}) / (H_i \cap K_{m-1}) & = (H_{i+1} \cap K_{m-1}) / (H_i \cap ( H_{i+1} \cap K_{m-1} ))\\
& \cong H_i (H_{i+1} \cap K_{m-1})/ H_i,
\end{align*}
where \(H_i\) is normal in \(H_i (H_{i+1} \cap K_{m-1})\). Since \(\{ H_i \}\) is a composition series, \(H_{i+1} / H_i\) must be simple; consequently, \(H_i (H_{i+1} \cap K_{m-1})/ H_i\) is either \(H_{i+1}/H_i\) or \(H_i/H_i\). That is, \(H_i (H_{i+1} \cap K_{m-1})\) must be either \(H_i\) or \(H_{i+1}\). Removing any nonproper inclusions from the series
\begin{equation*}H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \}, \end{equation*}
we have a composition series for \(H_{n-1}\). Our induction hypothesis says that this series must be equivalent to the composition series
\begin{equation*}H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}.\end{equation*}
Hence, the composition series
\begin{equation*}G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}\end{equation*}
and
\begin{equation*}G = H_n \supset H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \}\end{equation*}
are equivalent. If \(H_{n-1} = K_{m-1}\), then the composition series \(\{H_i \}\) and \(\{ K_j \}\) are equivalent and we are done; otherwise, \(H_{n-1} K_{m-1}\) is a normal subgroup of \(G\) properly containing \(H_{n-1}\). In this case \(H_{n-1} K_{m-1} = G\) and we can apply the Second Isomorphism Theorem once again; that is,
\begin{equation*}K_{m-1} / (K_{m-1} \cap H_{n-1}) \cong (H_{n-1} K_{m-1}) / H_{n-1} = G/H_{n-1}.\end{equation*}
Therefore,
\begin{equation*}G = H_n \supset H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \}\end{equation*}
and
\begin{equation*}G = K_m \supset K_{m-1} \supset K_{m-1} \cap H_{n-1} \supset \cdots \supset K_0 \cap H_{n-1} = \{ e \}\end{equation*}
are equivalent and the proof of the theorem is complete.

A group \(G\) is *solvable* if it has a subnormal series \(\{ H_i \}\) such that all of the factor groups \(H_{i+1} / H_i\) are abelian. Solvable groups will play a fundamental role when we study Galois theory and the solution of polynomial equations.

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Example13.19

The group \(S_4\) is solvable since
\begin{equation*}S_4 \supset A_4 \supset \{ (1), (12)(34), (13)(24), (14)(23) \} \supset \{ (1) \}\end{equation*}
has abelian factor groups; however, for \(n \geq 5\) the series
\begin{equation*}S_n \supset A_n \supset \{ (1) \}\end{equation*}
is a composition series for \(S_n\) with a nonabelian factor group. Therefore, \(S_n\) is not a solvable group for \(n \geq 5\).