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By definition an almost simple group is a group $G$ with $$T \cong \mathrm{Inn}(T)\trianglelefteq G \leq \mathrm{Aut}(T),$$ where $T$ is a non-abelian simple group. How would one show that in this case the socle of $G$ is equal to $\mathrm{Inn}(T)$?

I have seen one argument in a book which says that from the above we get $C_{G}(\mathrm{Inn}(T))=1$ (the centraliser in $G$ of $\mathrm{Inn}(T)$) and thus $\mathrm{Inn}(T)$ is the unique minimal normal subgroup of $G$ from which the result follows. The only part of this argument I fail to understand is how we obtain $C_{G}(\mathrm{Inn}(T))=1$ from the definition.

the_fox
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Ishika
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1 Answers1

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Suppose $g\in C_G(\mathrm{Inn}(T))$. Then $g^{-1}\varphi_tg = \varphi_t$ for any $t\in T$. For any $x\in T$, we have \begin{equation*} t^{-1}xt = x^{\varphi_t} = x^{g^{-1}}\varphi_tg = (t^{-1}(x^{g^{-1}})t)^g = (t^g)^{-1}xt^g \end{equation*} and so $x = t(t^g)^{-1}xt^gt^{-1}$, which implies $t(t^g)^{-1}\in Z(T) = \{1\}$. Therefore, $t = t^g$ for any $t\in T$. So $g$ is trivial in $\mathrm{Aut}(T)$ and in $G$.

Groups
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