Subnormal Series in a Group

# Subnormal Series in a Group

 Definition: Let $G$ be a group and let $1$ denote the identity in $G$. A Subnormal Series in $G$ of Length $k$ is a collection of subgroups $G_0, G_1, ..., G_k$ such that $\{ 1 \} = G_0 \subseteq G_1 \subseteq ... \subseteq G_k = G$ and such that $G_i$ is a normal subgroup of $G_{i+1}$ for all $i \in \{ 0, 1, ..., k-1 \}$. We denote such a subnormal series by $\{ 1 \} = G_0 \trianglelefteq G_1 \trianglelefteq ... \trianglelefteq G_k = G$.

For example, consider the group $(\mathbb{Z}_6, +)$. Let $G_0 = \{ 0 \}$, $G_1 = \{ 0, 3 \}$, and $G_2 = \mathbb{Z}_6$. Then a subnormal series in $\mathbb{Z}_6$ is:

(1)
\begin{align} \quad G_0 \trianglelefteq G_1 \trianglelefteq G_2 \end{align}

Clearly $G_0 \trianglelefteq G_1$] and $G_1 \trianglelefteq G_2$ since $G_2$ is an abelian group.

 Definition: Let $G$ be a group and let $1$ denote the identity in $G$. If $G$ has a subnormal series $\{ 1 \} = G_0 \trianglelefteq G_1 \trianglelefteq ... \trianglelefteq G_k = G$ then the Factors of this subnormal series are the quotient groups $G_{i+1} / G_i$ for all $i \in \{ 0, 1, ..., k-1 \}$.

Since $G_i \trianglelefteq G_{i+1}$ for all $i \in \{ 0, 1, ..., k-1 \}$ the factors $G_{i+1}/G_i$ are well-defined.

In the example above we have that the factors of the subnormal series $G_0 \trianglelefteq G_1 \trianglelefteq G_2$ are:

(2)
\begin{align} \quad G_1 / G_0 & \cong \mathbb{Z}_2 \\ \quad G_2 / G_1 & \cong \mathbb{Z}_3 \end{align}

This is because $G_1 / G_0 = G_1 \cong \mathbb{Z}_2$, which we can easily see by defining the isomorphism $\phi : \mathbb{Z}_2 \to G_1$ by $\phi(0) = 0$ and $\phi(1) = 3$.

And because $G_2 / G_1 \cong \mathbb{Z}_3$, which we can easily see by defining the isomorphism $\phi : \mathbb{Z}_3 \to G_2/G_1$ by $\phi(0) =  = $, $\phi(1) =  = $, and $\phi(2) =  = $.

 Definition: Let $G$ be a group and let $1$ denote the identity in $G$. If $\{ 1 \} = G_0 \trianglelefteq G_1 \trianglelefteq ... \trianglelefteq G_k = G$ is a subnormal series in $G$ then a Refinement of such a subnormal series is another subnormal series in $G$, $\{ 1 \} = H_0 \trianglelefteq H_1 \trianglelefteq ... \trianglelefteq H_l = G$ such that there exists a nondecreasing function $f : \{ 0, 1, ..., k \} \to \{ 0, 1, ..., l \}$ for which $G_i = H_{f(i)}$.

In other words, if we have a subnormal series in $G$ then a refinement of such a subnormal series can be obtained by adding more subgroups to the subnormal series.

 Definition: Let $G$ be a group and let $1$ denote the identity in $G$. A subnormal series $\{ 1 \} = G_0 \trianglelefteq G_1 \trianglelefteq ... \trianglelefteq G_k = G \}$ is said to be Irredundant if for all $i, j \in \{ 0, 1, ..., k \}$ with $i \neq j$ we have that $G_i \neq G_j$. A subnormal series is said to be Redundant if it is not irredundant.

Thus, a subnormal series is irredundant if and only if no two subgroups in the series are equal. The example at the top of the page is an irredundant subnormal series. An example of a redundant subnormal series is:

(3)
\begin{align} \quad \{ 0 \} \triangleleft \{ 0, 3 \} \trianglelefteq \{ 0, 3 \} \trianglelefteq \mathbb{Z}_6 \end{align}