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^{\varphi_t}=x^{g^{-1}\varphi_tg}=(t^{-1}x^{g^{-1}}t)^g=(t^g)^{-1}xt

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 $T'$ is trivial and $T$ is the unique minimal normal subgroup. In the other word, $\mathrm{soc}\space G=T$ is simple. Conversely, suppose $\mathrm{soc}(G):=N$ is simple. If $C_G(N)\neq 1$, then $C_G(N)\lhd G$ 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 $C_G(T)=\{1\}$. ^kbvpvv `\begin{proof}` For any $g\in C_G(T)=C_G(\mathrm{Inn(T)})$ and $\phi_t\in\mathrm{Inn(T)}$, $\phi_t^g=\phi_t$ and then $t^{-1}xt=x^{\phi_t}=x^{g^{-1}\phi_tg}=(t^{-1}x^{g^{-1}}t)^g=(t^g)^{-1}xt^g=x^{t^g}$. 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}`