###### Example 13.11

Any series of subgroups of an abelian group is a normal series. Consider the following series of groups:

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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 \(H_{i+1}\text{.}\) If each subgroup \(H_i\) is normal in \(G\text{,}\) then the series is called a normal series. The length of a subnormal or normal series is the number of proper inclusions.

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*}

A subnormal series need not be a normal series. Consider the following subnormal series of the group \(D_4\text{:}\)

\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 \(D_4\text{;}\) 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 \(\{ H_i \} \subset \{ K_j \}\text{.}\) That is, each \(H_i\) is one of the \(K_j\text{.}\)

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 \(G\text{,}\) is actually to study the factor groups \(H_{i+1}/H_i\text{.}\) 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 \(\{ K_{j+1}/ K_j \}\text{.}\)

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.

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.

For \(n \geq 5\text{,}\) 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.

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 \(k \in {\mathbb N}\text{.}\) 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 \({\mathbb Z}_{60}\text{,}\) it turns out that any two composition series are related. The factor groups of the two composition series for \({\mathbb Z}_{60}\) are \({\mathbb Z}_2\text{,}\) \({\mathbb Z}_2\text{,}\) \({\mathbb Z}_3\text{,}\) and \({\mathbb Z}_5\text{;}\) that is, the two composition series are isomorphic. The Jordan-Hölder Theorem says that this is always the case.

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

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 \(k\text{,}\) where \(1 \leq k \lt n\text{.}\) 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 \(G\text{.}\) 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 \(K_{j+1} \cap H_{n-1}\text{:}\)

\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 \(H_{i+1} \cap K_{m-1}\text{,}\) 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 \(H_i (H_{i+1} \cap K_{m-1})\text{.}\) 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 \(H_i/H_i\text{.}\) That is, \(H_i (H_{i+1} \cap K_{m-1})\) must be either \(H_i\) or \(H_{i+1}\text{.}\) 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 \(H_{n-1}\text{.}\) 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 \(H_{n-1} = K_{m-1}\text{,}\) 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 \(H_{n-1}\text{.}\) 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.

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 \(n \geq 5\text{.}\)