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Lecture 8: Introduction.

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# LECTURE 8
# Solvability
\sectionmark{\textit{2019.04.12}}
\subsectionmark{\textit{Lecture 8}}
In this lecture, we will concern ourselves with the question of which groups are solvable. First we recall what it means to be solvable.
**Definition 1.** (Solvable) A finite group $G$ is **solvable** if there exist subgroups of $G$, $S_0, \ldots, S_n$ such that
\begin{align*}
\{\mathrm{id}\} = S_0 \triangleleft S_1 \triangleleft \ldots \triangleleft S_{n-1} \triangleleft S_n = G
\end{align*}
such that every quotient $S_{j+1}/S_j$ is abelian.
*Remark.* Without the requirement of the quotient being abelian, we call such a sequence of normal subgroups a **cascading tower of normal subgroups** or a **subnormal series**.
Sometimes however, it is easier to work with an equivalent characterisation.
**Definition 2.** (Solvable) A finite group $G$ is **solvable** if there exists a cascading tower of normal subgroups in $G$ with $S_0 = \{\mathrm{id}\}$ and $S_n = G$ such that every quotient $S_{j+1}/S_j$ is cyclic.
*Remark.* In both definitions, the quotient is well defined, as the subgroups are normal to their neighbours.
*Remark.* (Normality is not transitive) In a cascading tower of normal subgroups we only require $S_j$ to be normal in $S_{j+1}$ (for all $j$). This however does **not** mean that for any $i<j$ $S_i$ is normal in $S_j$.
The second definition however, only really shows it potential in a third equivalent definition of solvability. First we will quickly define a technical tool to compactly describe this.
**Definition.** (Commutator) Let $G$ be a group. Then we define the **commutator** of $a, b \in G$ to be
\begin{align*}
[a, b] := aba^{-1}b^{-1}
\end{align*}
**Definition.** (Commutator Subgroup) Let $G$ be a finite group. Then we define the **commutator subgroup** of $G$ to be the subgroup generated by commutators
\begin{align*}
[G, G] := \langle \{ [a, b] : a, b \in G \} \rangle.
\end{align*}
We often call the commutator subgroup of a group just simply *the* commutator.
**Lemma.** *The conjugate of a commutator is the commutator of the conjugate.*
*Proof.* Let $G$ a finite group and $C$ its commutator. Now let $c \in G, aba^{-1}b^{-1} \in C$, then we have
\begin{align*}
c [a, b] c^{-1}
& = c(aba^{-1}b^{-1})c^{-1} \\
& = ca(c^{-1}c)b(c^{-1}c)a^{-1}(c^{-1}c)b^{-1}c^{-1} \\
& = (cac^{-1})(cbc^{-1})(ca^{-1}c^{-1})(cb^{-1}c^{-1}) \\
& = [cac^{-1}, cbc^{-1}]
\end{align*}
**Corollary.** The commutator of a finite group is a normal subgroup.
**Definition.** (Derived Series) Let $G$ be a finite group. We define the **derived series** of $G$ to be
\begin{gather*}
G^{(0)} = G \\
G^{(n+1)} = [G^{(n)}, G^{(n)}]
\end{gather*}
*Remark.* It immediately follows from the definition of the derived series, that the derived series form a sequence of normal subgroups
\begin{align*}
G = G^{(0)} \triangleright G^{(1)} \triangleright G^{(2)} \triangleright \ldots
\end{align*}
**Theorem.** (The quotient group by the commutator is abelian) Let $G$ be a finite group, and $C$ its commutator. Then $G/C$ is an abelian group.
*Proof.* As $C$ is normal in $G$, he quotient is surely a group. Now let $\overline{a}, \overline{b} \in G/C$. Then
\begin{align*}
\overline{a} \overline{b}
& = aCbC \\
\end{align*}
@ror
**Definition 3.** (Solvable) A finite group $G$ is **solvable** if there exists an $n \in \mathbb{N}$ such that
\begin{align*}
G^{(n)} = \{\mathrm{id}\}
\end{align*}
**Theorem.** (These definitions are equivalent)
*Proof.* As every cyclic group is abelian, the second definition clearly implies the first. Now to proving the converse. We will do this by adding in seperate
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