Quasiprimitive Group of Type HS, HC and HA are Primitive

Definition of HS, HC and HA

The following property gives the structure of groups of type HS and HC. If $l=1$, the type is named HS; while when $l>1$ the type is named HC.

Proposition

Let $G$ be a quasiprimitive permutation group acting on $\Omega$. If $G$ has two different minimal normal subgroups $M$ and $N$, then there is

$$M:\mathrm{Inn}(M)⩽G⩽M:\mathrm{Aut}(M).$$

Furthermore, $M\cong N\cong T^l$ is isomorphic to a direct product of isomorphic nonabelian simple groups $T$, and $\mathrm{Aut}(M)=\mathrm{Aut}(T)\wr S_l$.

\begin{proof} Since $G$ is quasiprimitive, groups $M$ and $N$ are transitive. By Minimal Normal Subgroups of Permutation Groups, we have $M=C_G(N)$, $\mathrm{soc}(G)=M\times N$ and $M\cong N$. Note that $M\times N$ can be identified by

$$M\times N=\hat{M}\times \check{M}=\hat{M}:\tilde{M}=M:\mathrm{Inn}(M),$$

where the notations come from Three Permutations on G Induced from G. By NC lemma, there is $N_G(N)/C_G(N)=G/M⩽\mathrm{Aut}(M)$ and so $G⩽M:\mathrm{Aut}(M)$. Moreover, $\mathrm{Aut}(M)=\mathrm{Aut}(T)\wr S_l$ is proved here. \end{proof}

Now we define the group of type HA.

If $G$ has an abelian minimal normal subgroup $N$, then $N$ is the unique minimal normal subgroup of $G$ by ^9gl6q2. It deduces that $N=\mathrm{soc}(G)$. Furthermore, $N$ is elementary abelian. By the definition of quasiprimitive, $N$ is transitive on $\Omega$. Then by Frattini’s argument, there is $G=N{G}_{\alpha }$ with $N={\mathbb{Z}}_p^d$. Note that $N$ is regular on $\Omega$, so $G_{\alpha }$ can be identified as a subgroup of $\mathrm{GL}(d,p)$. If $G_{\alpha }$ is reducible, then $N:G_{\alpha }$ has a normal subgroup $W<N$, contradiction. Therefore, $G$ is defined as below.

Definition

Let $\Omega ={\mathbb{Z}}_p^d$ for a prime $p$ and positive integer $d$, and let $G$ be the semidirect product $G=N:G_0$, a subgroup of the affine group $\mathrm{AGL}(d,p)$ on $\Omega$, where $N$ is the group of translations and $G_0$ is an irreducible subgroup of $\mathrm{GL}(d,p)$. Then $G$ is a quasiprimitive group of type HA.

These Groups are Primitive

Lemma

A quasiprimitive permutation group of type HS, HC or HA is primitive.

\begin{proof} i) Let $G=T:H$ be of type HS with simple $T$. By ^04lqgl, we have $\mathrm{Inn}(T)⩽H⩽\mathrm{Aut}(T)$. Then $H=G_{\omega }$ where $\omega =1\in T$. It remains to show $H$ is a maximal subgroup by Equivalent Condition of Primitivity. Otherwise, there is a $K$ such that $H<K<G$, and there exists an element $g\in K\backslash H$ with the form $g=(t,h)\in G$. Then $(t,1)=(t,h)(1,h^{-1})\in K$.

For any element $x\in H$, we have $(1,x^{-1})\in K$ and $(1,x^{-1})(t,1)=(t^x,x^{-1})\in K$. Hence $(t^x,1)=(t^x,x^{-1})(1,x)\in K$ and so $M$ is a subgroup of $K$, where

$$M=⟨(t^x,1):x\in H⟩⩽T\times \{1\}.$$

Since $H⩾\mathrm{Inn}(T)$, the action of $T$ on $M$ by conjugation fixes $M$. So $M⊲T$ and $M=T$. Thus $K⩾⟨M,H⟩=G$, contradiction. Therefore, $G$ is primitive.

ii) The proof is similar to i). Let $G=M:H$ be of type HC with $M\cong T^l$ a direct product of isomorphic simple groups. By ^04lqgl, we have $\mathrm{Inn}(M)⩽H⩽\mathrm{Aut}(M)$. If $H=G_{\omega }$ with $\omega =1\in T$ is not a maximal subgroup of $G$, then there is a $L$ such that $H<L<G$ and $L$ has a subgroup $N:=m^H$ where $(m,h)\in L\setminus H$. Note that $N⊲M$ and $M$ is a minimal normal subgroup of $G$. Thus $N=M$ and so $L=G$, which is impossible. So $G$ is primitive.

iii) When $G$ is of type HA, the primitivity of $G$ is proved here. \end{proof}