1.7 Normal and Subnormal Series

subnormal/normal series

A finite subnormal series of $G$ is

$$G=G_0>G_1>\cdots >G_n$$

where $G_i⊳G_{i+1}$ for all $i$.

• The factors are $G_i/G_{i+1},i⩾0$.
• The length of the series equals the number of $i$ such that $G_i\ne G_{i+1}$.

Furthermore, if $G_i⊲G_j$ for all $i>j$, then it is called a normal series.

Example. Derived series and upper central series are normal series.

refinement

Let $G=G_0>G_1>\cdots >G_n$ be a subnormal series. An one-step-refinement is a series $G=G_0>\cdots >G_i>N>G_{i+1}>\cdots >G_n$ or $G=G_0>\cdots >G_n⊳N$. A refinement is the series after finite times of one-step-refinement. It is called a proper refinement if the length is strictly bigger.

composition/solvable series

A subnormal series $G=G_0>\cdots >G_n=\{e\}$ is called a composition series if $G_i/G_{i+1}$ is simple for all $i$, and is called solvable series if $G_i/G_{i+1}$ is abelian for all $i$.

Theorem

The followings hold.

• Every finite group has a composition series.
• Every refinement of a solvable series is solvable.
• A subnormal series is a composition series iff it has no proper refinement.

\begin{proof} Easy. \end{proof}

Theorem

$G$ is solvable iff $G$ has a solvable series.

\begin{proof} Assume that $G$ is solvable, then its derived series is a solvable series.

Suppose $G=G_0>\cdots >G_n$ is a solvable series. Note that $G/G_1$ is abelian yields that $G_1⩾G^{(1)}$ and $G_1/G_2$ is abelian yields that $G_2⩾G_1^{(1)}⩾G^{(2)}$. Repeat this procedure, we have that $G^{(n)}⊲G_n=\{e\}$ and so $G$ is solvable. \end{proof}

Definition

Two subnormal series $S$ and $T$ are called equivalent if there exists $1$-$1$ correspondence between the non-trivial factors of $S$ and $T$, which are isomorphic pairwise.

Zassenhaus

Let $A^{\ast }⊲A<G$ and $B^{\ast }⊲B<G$. Then:

• $A^{\ast }(A\cap B^{\ast })⊲A^{\ast }(A\cap B)$;
• $B^{\ast }(A^{\ast }\cap B)⊲B^{\ast }(A\cap B)$;
• $A^{\ast }(A\cap B)/A^{\ast }(A\cap B^{\ast })\simeq B^{\ast }(A\cap B)/B^{\ast }(A^{\ast }\cap B)$.

\begin{proof} Let $D=(A\cap B^{\ast })(A^{\ast }\cap B)$. We now define an epimorphism $f:A^{\ast }(A\cap B)\twoheadrightarrow (A\cap B)/D$ with $\ker f=A^{\ast }(A\cap B^{\ast })$, then $A^{\ast }(A\cap B)/\ker f\simeq (A\cap B)/D$. Similarly we can prove $B^{\ast }(A\cap B)/B^{\ast }(A^{\ast }\cap B)\simeq (A\cap B)/D$.

Define $f:a\cdot c\mapsto Dc$.

• We check $f$ is well defined. If $ac=a{c}_1$, then $c{c}_1^{-1}\in A^{\ast }$ and $c{c}_1^{-1}\in A\cap B$, which yields that $c{c}_1^{-1}\in A^{\ast }\cap B\subseteq D$. Hence, $f(c)=f(c_1)$.
• Note that $f(a_1{c}_1{a}_2{c}_2)=f(a_1{a}_2^{c_1^{-1}}{c}_1{c}_2)=D{c}_1{c}_2=f(a_1{c}_1)f(a_2{c}_2)$, so $f$ is a homomorphism.
• $ac\in \ker f$ iff $c\in D$ iff $c=a_1{c}_1$ with $a_1\in A\cap B^{\ast }$ and $c_1\in A^{\ast }\cap B$ iff $ac\in A^{\ast }(A\cap B^{\ast })$.

Now we finish the proof. \end{proof}

Schreier

Any two subnormal (rep. normal) series of $G$ has two subnormal (rep. normal) refinements that are equivalent to each other.

\begin{proof} Let $G=G_0>\cdots >G_n$ and $G=H_0>\cdots >H_m$ be two subnormal (rep. normal) series, and let $G_{n+1}=H_{m+1}=\{e\}$. For $0⩽i⩽n$, consider

$$G_i=G_{i+1}(G_i\cap H_0)>G_{i+1}(G_i\cap H_1)>\cdots >G_{i+1}(G_i\cap H_{m+1})=G_{i+1}.$$

By ^f5c5fc, this is still a subnormal (rep. normal) series. Denote $G(i,j)=G_{i+1}(G_i\cap H_j)$, so we get a refinement of $G=G_0>\cdots >G_n$:

$$G=G(0,0)>G(0,1)>\cdots >G(0,m+1)=G(1,0)>\cdots >G(n,m).$$

Similarly, set $H(i,j)=H_{j+1}(G_i\cap H_j)$ and get

$$G=H(0,0)>\cdots >H(n,0)>H(0,1)>\cdots >H(n,m).$$

Notice that $G(i,j)/G(i,j+1)\cong H(i,j)/H(i+1,j)$, then these two series are equivalent. \end{proof}

Jordan-Holder

Any two composition series are equivalent.

\begin{proof} By ^fdb53f. \end{proof}