Almost Simple Groups

Definition

A group $G$ is almost simple if there is a (non-abelian) simple group $T$ such that $T⊲G⩽\mathrm{Aut}(T)$.

Proposition

$G$ is an almost simple group iff its socle is simple.

\begin{proof} Suppose $G$ is almost simple. Then there is a simple group $T$ such that $T⊲G\le \mathrm{Aut}(T)$. Since $T$ is a minimal normal subgroup, $T\le \mathrm{soc}G$. Assume $T^{\prime }$ is another minimal normal subgroup. Then $[T,T^{\prime }]⩽T\cap T^{\prime }=\{1\}$ yields $T^{\prime }⩽C_G(T)$. For any $t\in T$, define ${\phi }_t\in \mathrm{Aut}(T)$ by ${\phi }_t:t_0\mapsto t^{-1}{t}_{0}t$. Then $g^{-1}{\phi }_{t}g={\phi }_t$ for any $g\in C_G(T)$. And for any $x\in T$, $t^{-1}xt=x^{{\phi }_t}=x^{g^{-1}{\phi }_{t}g}=(t^{-1}{x}^{g^{-1}}t{)}^g=(t^g{)}^{-1}x{t}^g$yields $t(t^g{)}^{-1}\in Z(T)=1$. Hence $t^g=t$, i.e. $g\in \mathrm{Aut}(T)$ acts on $T$ trivially. Therefore, $g=1$ and so $C_G(T)=1$. So $T^{\prime }$ is trivial and $T$ is the unique minimal normal subgroup. In the other word, $\mathrm{soc}G=T$ is simple.

Conversely, suppose $\mathrm{soc}(G):=N$ is simple. If $C_G(N)\ne 1$, then $C_G(N)⊲G$ contains a minimal normal subgroup of $G$, which must be $N$. It is impossible because $N$ is non-abelian. Therefore, $C_G(N)=1$. By NC lemma there is

$$G=G/C_G(N)=N_G(N)/C_G(N)\lesssim \mathrm{Aut}(N)$$

and so $G$ is almost simple. \end{proof}

Proposition

Let $G$ be an almost simple group with $\mathrm{soc}(G)=T$. Then $C_G(T)=\{1\}$.

\begin{proof} For any $g\in C_G(T)=C_G(\mathrm{Inn}(\mathrm{T}))$ and ${\varphi }_t\in \mathrm{Inn}(\mathrm{T})$, ${\varphi }_t^g={\varphi }_t$ and then $t^{-1}xt=x^{{\varphi }_t}=x^{g^{-1}{\varphi }_{t}g}=(t^{-1}{x}^{g^{-1}}t{)}^g=(t^g{)}^{-1}x{t}^g=x^{t^g}$. Thus $t=t^g$ for any $t\in T$, i.e. $g$ is the identity in $\mathrm{Aut}(T)$ and so for $G$. Therefore, $C_G(T)=1$. \end{proof}