${\rm Soc}(G) $ simple implies $ G $ almost simple

Question

Recall that a group $ G $ is called almost simple if there exists a (non-abelian) simple group $ S $ such that $ S \leq G \leq {\rm Aut}(S) $.

If $ G $ is almost simple then $ {\rm Soc}(G) $, the socle of $ G $, is (non-abelian) simple.

See The socle of an almost simple group or Is the Socle of an almost simple group a simple group?

What about the converse? In other words, if ${\rm Soc}(G) $ is (non-abelian) simple then must $ G $ be almost simple?

Answer 1

Yes.

Say $N \trianglelefteq G$ with $N$ non-abelian simple and $\operatorname{soc}(G) = N$. If $C_G(N) \neq 1$, then $C_G(N)$ contains a minimal normal subgroup of $G$, which must be $N$. But this is impossible since $N$ is non-abelian, so $C_G(N) = 1$.

Thus the conjugation action of $G$ on $N$ defines an injective homomorphism $G \rightarrow \operatorname{Aut}(N)$. So $G$ embeds into $\operatorname{Aut}(N)$ with $N \leq G \leq \operatorname{Aut}(N)$.