Jordan-Hölder Theorem

\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