1.1 Factorization of Groups

Definition

Let $G$ be a finite group. $H,K<G$ is called a factorization if $G=HK$, i.e. any $g\in G$ has the form $g=hk$, where $h\in H$, $k\in K$. Groups $H,K$ are called factors, and $K$ is called a supplement of $H$ in $G$. Furthermore, if $H\cap K=\{1\}$, then $G=HK$ is an exact factorization.

There are two questions to answer:

• Given a group $G$ and $H,K<G$, under what condition $G=HK$?
• Given $G$ and $H<G$, under what condition there exists $K<G$ such that $G=HK$?

Two lemmas are used to answer those questions:

By Order of Group

Lemma

$G=HK$ iff $|G|=|H||K|/|H\cap K|$.

\begin{proof} $HK=\{hk:h\in H,k\in K\}$ is a set whose cardinal number is $|H||K|/|H\cap K|$. By $HK\subseteq G$, $G=HK$ iff $|G|=|HK|/|H\cap K|$. \end{proof}

Frattini’s argument

It is mentioned here.

Lemma

Suppose $G$ is a transitive on $\Omega$. Then $H<G$ is transitive on $\Omega$ iff $G=H{G}_{\omega }$ for $\omega \in \Omega$.

\begin{proof} Suppose that $H$ is transitive on $\Omega$. For any $\omega \in \Omega$ and any $g\in G$, define ${\omega }^{\prime }:={\omega }^g$. Since $H$ is transitive, there is $h\in H$ such that ${\omega }^h={\omega }^{\prime }={\omega }^g$. So we have $h^{-1}g\in G_{\omega }$ and $g=h(h^{-1}g)\in H{G}_{\omega }$. Therefore, $G=H{G}_{\omega }$.

Conversely, assume that $G=H{G}_{\omega }$. For any $\omega ,{\omega }^{\prime }\in \Omega$, there is $g\in G$ such that ${\omega }^g={\omega }^{\prime }$. As $g$ can be written as $g=xh$ with $x\in G_{\omega }$ and $h\in H$, there is ${\omega }^g={\omega }^{xh}={\omega }^h={\omega }^{\prime }$, i.e., $H$ acts on $\Omega$ transitively. \end{proof}