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Let $G$ be a finite primitive group of degree $n$, and let $H$ be the socle of $G$. Then if $H$ is isomorphic to a direct power $T^m$ of a nonabelian simple group $T$ then the following holds when $m=1$: $G$ is isomorphic to a subgroup of ${\rm Aut}(T)$. [Reference: Link, the statement of O'Nan-Scott Theorem.]

Does the converse also hold?

Questions: Let $G$ is a primitive subgroup of $S_n$ such that $T \leq G \leq {\rm Aut}(T)$ (and thus $G$ is an almost simple permutation group) then socle of $G$ is $T$ (a simple group)?

Shaun
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Jins
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    I you define the socle as the product of the minimal normal subgroups, then yes, $T$ is the only minimal normal subgroup of $G$. – Brauer Suzuki May 05 '22 at 11:03
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    Note that the socle is independent of the primitive action. – Brauer Suzuki May 05 '22 at 11:12
  • The Socle of $G$ is the subgroup generated by all minimal normal subgroup of $G$. As you said in your comment, since $T$ is the only one (minimal normal subgroup of $G$), the socle is $T$ itself. Am I correct ? @BrauerSuzuki – Jins May 05 '22 at 11:14

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The answer to your question is yes. In fact for any finite group $G$ such that $T\le G\le$ Aut$(T)$, where $T$ is a non-abelian simple group, the socle of $G$ is $T$. Why? Since $T$ is a minimal normal subgroup, $T\le$ soc$(G)$. Now, show that we have $C_{G}(T)=C_G($Inn$(T))=1$. By the definition,$\quad$ soc$(G)$ is a direct product of minimal normal subgroups and only the identity centralizes $T$, so there is nothing more in soc$(G)$.

pansh1n
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