Minimal Normal Subgroups of Permutation Groups

Lemma

Every finite permutation group $G$ on $\Omega$ has at most two distinct transitive minimal normal subgroups.

If $K_1$ and $K_2$ are distinct transitive minimal normal subgroups of $G$, then $C_G(K_1)=K_2$, $C_G(K_2)=K_1$.

Moreover, there is an involution of $N_{\mathrm{Sym}(\Omega )}(G)$ that interchanges $K_1$ and $K_2$, and which also centralizes a point stabilizer of $G$.

\begin{proof} #todo Suppose $G$ has three distinct transitive minimal normal subgroups $K_1$, $K_2$ and $K_3$. Then each pair of the $K_i$ intersect trivially and hence $K_2,K_3⩽C_G(K_1)$ by Normal Subgroups that Intersect Trivially. However, $C_G(K_1)$ is semiregular by Equivalent Condition of Semiregular. So $K_2$, $K_3$ and $C_G(K_1)$ are regular as $K_i$ are transitive. Consequently, both $K_2$ and $K_3$ equal $C_G(K_1)$, which is impossible.

Suppose now that $K_1$ and $K_2$ are distinct transitive minimal normal subgroups of $G$. As was noted above, $K_1=C_G(K_2)$ and $K_2=C_G(K_1)$, and they are both regular on $\Omega$.

(to be continued) \end{proof}

ref: Finite Permutation Groups with a Transitive Minimal Normal Subgroup, Lemma 5.1

Theorem

Suppose $G$ is a solvable group and suppose $G$ acting on $\Omega$ is primitive and transitive. Then the minimal normal subgroup of $G$ is unique.

\begin{proof}Suppose $M,N$ are two different minimal normal subgroups of primitive $G$. Then $M,N$ are transitive by here. By Abelian Transitive Action is Regular, we know that $|M|=|N|=|\Omega |$. Note that $M\cap N=\{1\}$. Since $[M,N]\subseteq M\cap N$ by normal, $[M,N]=1$ and so $M$ is in the centralizer of $N$.

Let $Z$ be the centralizer of $N$. Since $N$ is abelian, $Z\ge N$ and so $Z$ is transitive. For any $\alpha ,\beta \in \Omega$, there is $n\in N$ such that ${\alpha }^n=\beta$. Then $Z_{\alpha }=Z_{\alpha }^n=Z_{\beta }$. By the arbitrary of $\beta$, $Z_{\alpha }=1$. Thus $Z$ is also regular and so $Z=N$. It contradicts to $M\le Z$. Hence the minimal normal subgroup is unique. \end{proof}

Corollary

Let $X$ be a quasiprimitive permutation group on a finite set $\Omega$. Then $\mathrm{soc}(B)\cong T^k$ with $k\ge 1$ where $T$ is a simple group.

\begin{proof} By Lemma 1, $X$ has at most two minimal normal subgroup $K_1,K_2$ and they are interchanges by an involution. Thus $K_1\cong K_2$ and so $B=K_1\times K_2\cong T^k$ with $k\ge 1$ where $T$ is a simple group. \end{proof}