\begin{definition}
A subnormal series of a group $G$ is a chain of subgroups $G = G_0 \geq G_1 \geq \cdots \geq G_n$ such that $G_{i+1} \trianglelefteq G_i$ for $i \in \{0,1,\cdots,n\}$. The factors of the series are the quotient groups $G_i/G_{i+1}$. The length of the series is the number of strict inclusions. A subnormal series such that $G_i \trianglelefteq G$ for all $i$ is said to be a normal series.
\end{definition}

As an example, the derived series
$G \geq G^{(1)} \geq \cdots \geq G^{(n)}$
is a normal series for any group
$G$. If
$G$ is nilpotent, the
ascending central series
$C_1(G) \leq C_2(G) \leq \cdots \leq C_n(G) = G$
is a normal series for
$G$.

\begin{definition}
Let $G = G_0 \geq G_1 \geq \cdots \geq G_n$ be a subnormal series. A one-step refinement of this series is any series of the form $G = G_0 \geq \cdots \geq G_i \geq N \geq G_{i+1} \geq \cdots \geq G_n$ or $G = G_0 \geq \cdots \geq G_n \geq N$ where $N \trianglelefteq G_i$ and $G_{i+1} \trianglelefteq N$ for $i < n$. A refinement of a subnormal series $S$ is any subnormal series obtained from $S$ by a finite sequence of one-step refinements. A refinement is said to be proper iff its length is larger than the length of $S$.
\end{definition}\begin{definition}
A subnormal series $G = G_0 \geq G_1 \geq \cdots \geq G_n = \langle e \rangle$ is a composition series iff each factor $G_i/G_{i+1}$ is simple. A subnormal series $G = G_0 \geq G_1 \geq \cdots \geq G_n = \langle e \rangle$ is a solvable series iff each factor is abelian.
\end{definition}

A commonly used fact for composition series: When
$G \neq N$,
$G/N$ is simple iff
$N$ is maximal in the set
of all normal subgroups
$M$ of
$G$ with
$M \neq G$.

\begin{theorem}
\begin{enumerate}[label=(\roman*)]
\item Every finite group $G$ has a composition series.
\item Every refinement of a solvable series is a solvable series.
\item A subnormal series is a composition series iff it has no proper refinements.
\end{enumerate}
\end{theorem}

Proof. (i) Let
$G_1$ be a maximal normal
subgroup of
$G$. Then
$G/G_1$ is simple. Let
$G_2$ be a maximal normal
subgroup of
$G_1$ and so on. Since
$G$ is finite, this
process must end with
$G_n = \langle e \rangle$.
Thus
$G > G_1 > G_2 > \cdots > G_n = \langle e \rangle$
is a composition series.

(ii) If
$G_i/G_{i+1}$ is abelian,
$G_{i+1} \trianglelefteq H \trianglelefteq G_i$,
then
$H/G_{i+1}$ is abelian
since it is a subgroup of
$G_i/G_{i+1}$ and
$G_i/H$ is abelian since
it is isomorphic to
$(G_i/G_{i+1})/(H/G_{i+1})$.

(iii) If
$G_{i+1} \triangleleft H \triangleleft G_i$,
$H/G_{i+1}$ is a proper
normal subgroup of
$G_i/G_{i+1}$. All proper
normal subgroups of
$G_i/G_{i+1}$ have the
form
$H/G_{i+1}$ for some
$G_{i+1} \triangleleft H \triangleleft G_i$. ◻

\begin{theorem}
A group $G$ is solvable iff it has a solvable series.
\end{theorem}

Proof. If
$G$ is solvable,
$G \geq G^{(1)} \geq G^{(2)} \geq \cdots \geq G^{(n)} = \langle e \rangle$
is a solvable series. If
$G \geq G_1 \geq G_2 \geq \cdots \geq G_n = \langle e \rangle$
is a solvable series for
$G$, then
$G/G_1$ abelian implies
that
$G_1 \geq G^{(1)}$.
$G_1/G_2$ abelian implies
$G_2 \geq G'_1 \geq G^{(2)}$.
Proceeding by induction conclude
$\forall i, G_i \geq G^{(i)}$.
In particular,
$\langle e \rangle = G_n \geq G^{(n)}$
so $G$ is solvable. ◻

\begin{proposition}
A finite group $G$ is solvable iff $G$ has a composition series whose factors are cyclic of prime order.
\end{proposition}

Proof. A composition series with cyclic factors is a
solvable series. Conversely, assume
$G \geq G_1 \geq G_2 \geq \cdots \geq G_n = \langle e \rangle$
is a solvable series for
$G$. If
$G_0 \neq G_1$, let
$H_1$ be a maximal normal
subgroup of
$G$ containing
$G_1$. If
$H_1 \neq G_1$, let
$H_2$ be a maximal normal
subgroup of
$G$ containing
$G_1$. Continue until we
obtain a series
$G > H_1 > H_2 > \cdots > H_k > G_1$
with each subgroup a maximal normal subgroup of the preceding, whence
each factor is simple. This series terminates as in, eventually
$H_{k+1} = G_1$ since
$G$ is finite. Doing this
for each pair
$(G_i, G_{i+1})$ gives a
solvable refinement
$G = N_0 > N_1 > \cdots > N_r = \langle e \rangle$
of the original series. Each factor of this series is abelian and simple
hence cyclic of prime order. ◻

\begin{definition}
Two subnormal series $S$ and $T$ of a group $G$ are equivalent iff there is a one-to-one correspondence between the nontrivial factors of $S$ and the nontrivial factors of $T$ such that corresponding factors are isomorphic groups.
\end{definition}\begin{lemma}
If $S$ is a composition series of a group $G$, then any refinement of $S$ is equivalent to $S$.
\end{lemma}

Proof. Since
$S$ is a composition
series,
$S$ has no proper
refinements. Thus any refinement of
$S$ is obtained by
inserting additional copies of
$G_i$. Any refinement of
$S$ has the exact same
nontrivial factors as
$S$ and is thus
equivalent to
$S$. ◻

\begin{lemma}[Zassenhaus]Let $A^\ast, A, B^\ast, B$ be subgroups of a group $G$ such that $A^\ast \trianglelefteq A, B^\ast \trianglelefteq B$.
\begin{enumerate}[label=(\roman*)]
\item $A^\ast (A \cap B^\ast) \trianglelefteq A^\ast (A \cap B)$
\item $B^\ast (A^\ast \cap B) \trianglelefteq B^\ast (A \cap B)$
\item $A^\ast (A \cap B)/A^\ast (A \cap B^\ast) \cong B^\ast (A \cap B) / B^\ast (A^\ast \cap B)$
\end{enumerate}
\end{lemma}

Proof. Note
$A \cap B^\ast = (A \cap B) \cap B^\ast \trianglelefteq A \cap B$.
Similarly,
$A^\ast \cap B \trianglelefteq A \cap B$.
Thus
$D = (A^\ast \cap B) (A \cap B^\ast) \trianglelefteq A \cap B$.
Also,
$A^\ast (A \cap B) \leq A, B^\ast (A \cap B) \leq B$.
We will define an epimorphism
$f: A^\ast (A \cap B) \to (A \cap B)/D$
with kernel
$A^\ast (A \cap B^\ast)$.
This would imply that
$A^\ast (A \cap B^\ast) \trianglelefteq A^\ast (A \cap B)$
and that
$A^\ast (A \cap B)/A^\ast (A \cap B^\ast) \cong (A \cap B)/D$.
Define
$f : A^\ast (A \cap B) \to (A \cap B)/D$
as follows: If
$a \in A^\ast, c \in A \cap B$,
let
$f(ac) = Dc$.
$ac = a_1 c_1$ implies
$c_1 c^{-1} = a_1^{-1} a \in (A \cap B) \cap A^\ast = A^\ast \cap B \leq D$.
Thus
$f$ is well defined.
$f$ is clearly
surjective.
$$f((a_1 c_1)(a_2 c_2)) = f(a_1 a_3 c_1 c_2) = D c_1 c_2 = D c_1 D c_2 = f(a_1 c_1)f(a_2 c_2)$$
Finally,
$ac \in \operatorname{ker} f \iff c \in D \iff c = a_1 c_1, a_1 \in A^\ast \cap B, c_1 \in A \cap B^\ast$.
Hence
$ac \in \operatorname{ker} f \iff ac = (a a_1) c_1 \in A^\ast (A \cap B^\ast)$.
$\operatorname{ker} f = A^\ast (A \cap B^\ast)$.
A symmetric argument shows (ii) and
$B^\ast (A \cap B)/B^\ast (A^\ast \cap B) \cong (A \cap B)/D$
whence (iii) follows. ◻

\begin{theorem}[Schreier Refinement]Any two subnormal (resp. normal) series of a group $G$ have subnormal (resp. normal) refinements that are equivalent.
\end{theorem}

Proof. Let
$G = G_0 \geq G_1 \geq \cdots \geq G_n$
and
$G = H_0 \geq H_1 \geq \cdots \geq H_m$
be subnormal [resp. normal] series. Let
$G_{n+1} = H_{m+1} = \langle e \rangle$
and for each
$0 \leq i \leq n$,
consider the groups
$$G_i = G_{i+1} (G_i \cap H_0) \geq G_{i+1} (G_i \cap H_1) \geq \cdots \geq G_{i+1} (G_i \cap H_m) \geq G_{i+1} (G_i \cap H_{m+1}) = G_{i+1}$$
For each
$0 \leq j \leq m$, the
Zassenhaus lemma applied to
$G_{i+1}, G_i, H_{j+1}, H_j$
shows that
$G_{i+1} (G_i \cap H_{j+1}) \trianglelefteq G_{i+1} (G_i \cap H_j)$
(if the original series is normal, each
$G_{i+1} (G_i \cap H_{j+1}) \trianglelefteq G$)
Inserting these groups between each
$G_i$ and
$G_{i+1}$, denoting