1.2 Factorization of Permutation Groups

Problem: find all factorization of $S_n$.

Thm. Let $G=S_n$ and $G=HK$ such that $H,K\ngtr A_n$. Then

• either $A_{n-m}⊲H\le S_{n-m}\times S_m$ and $K$ is $m$-homogeneous where $m\le 5$,
• or $n=6,8,10$ and there are a few exception.

m-homogeneous/m-transitive

Def. A group $G\le Sym(\Omega )$ is $m$-homogeneous if $G$ is transitive on ${\Omega }^{\{m\}}=\{m$-subset of $\Omega \}$ and is $m$-transitive if $G$ is transitive on ${\Omega }^{(m)}=\{m$-tuples of $\Omega \}$.

We have known these group well:

Thm. (Burnside) If finite group $G\le Sym(\Omega )$ is $6$-transitive, then $G=A_n$ or $S_n$.

Thm. All $m$-transitive finite groups with $m\ge 2$ are explicitly known.

Thm. (Kantor) Finite $m$-homogenuous permutation groups with $m\ge 2$ but not $m$-transitive are explicitly known.

One factor of $G\le S_n$ transitive

Suppose $G=S_n=HK$.

If $K$ is intransitive, then $K\le Sym(\Delta )\times Sym(\Omega -\Delta )=S_m\times S_n$, where $|\Delta |=m$. Then $H$ is $m$-homogeneous and so it is explicitly known.

However, if $K$ is transitive, it’s hard to describe $H$. So we have a lemma:

Lemma. Suppose $G=H\times K<Sym(\Omega )$. If $K\curvearrowright \Omega$ is transitive, then:

• $H$ is semi-regular;
• if $H$ transitive, then $H$ and $K$ are both regular; moreover, $H\cong K$;
• if $K$ is regular, then $H\lesssim K$;
• for any $\omega \in \Omega$, let $B={\omega }^H$. Then $K_B\cong K_{\omega }:H$.

Proof. i) Suppose there is a non-identity $h\in H$ satisfying ${\omega }^h=\omega$ for some $\omega \in \Omega$. Then for any ${\omega }^{\prime }\in \Omega$, there is a $k\in K$ such that ${\omega }^{\prime }={\omega }^k$. Then ${\omega }^{\prime k^{-1}hk}={\omega }^{\prime }$, i.e. $k^{-1}hk\in H_{{\omega }^{\prime }}$. Since $k^{-1}hk=h$ by $G=H\times K$, $h$ fixes every element of $\Omega$. Thus $h=id$, contradiction.

ii) If $H$ is transitive, by i) $H$ is regular and $K$ is semi-regular. Then $K$ is also regular.

For any $\omega \in \Omega$, $G_{\omega }=\{(h,k):{\omega }^{hk}=\omega \}$. Define $\varphi :H\to K$, $h\mapsto k$ such that $(h,k)\in G_{\omega }$. It’s easy to verify $\varphi$ is well-defined and is homomorphic. By $H,K$ regular, $|H|=|\Omega |=|K|$ and so $\varphi$ is a bijective. Therefore, $H\cong K$.

iii) Use the map $\varphi$ defined in ii), $\varphi$ is still well-defined and homomorphic, but now it is injective. So $H\lesssim K$ by $\varphi$.

iv) If $H$ is transitive, then $H,K$ are regular and $H\cong K=K_{\Omega }/K_{\omega }$.

Now assume $H$ is intransitive. Consider $\mathcal{B}=\{H$-orbits on $\Omega \}=\{B_1\cdots ,B_n\}$, where $B_i=\{{\omega }_i^h:h\in H\}$. Then $\mathcal{B}$ is a block system for $G$ on $\Omega$, because $B_i^g=\{{\omega }_i^{hg}:h\in H\}=\{{\omega }_i^{g(g^{-1}hg)}:h\in H\}=B_j$, where ${\omega }_i^g={\omega }_j$. Thus $K$ acting on each $B_i$ is transitive.

For any $\omega \in \Omega$ and $B={\omega }^H\in \mathcal{B}$, define $K_B$ as the stabilizer of $K$ acting on $\mathcal{B}$ . Note that for any ${\omega }^{\prime }\in B$, there is $h\in H$ such that ${\omega }^h={\omega }^{\prime }$. Then $K_{{\omega }^{\prime }}=K_{\omega }^h=K_{\omega }$, i.e. $K_{\omega }$ fixes all points in $B$. By $K_B$ actiong on $B$ transitive and $H$ acting on $B$ regular, we have $K_B/K_{\omega }\cong H$, i.e. $K_B=K_{\omega }:H$. $■$