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I have shown that an almost simple group has a non-abelian simple socle, and that a finite group with a non-abelian simple socle is almost simple, but is this true for infinite groups?

To prove that a finite group with a non-abelian simple socle is almost simple, I used the argument in this question - however, this relies on showing $\mathrm{C}_{G}(\mathrm{Soc}(G))$ is trivial by contradiction, saying that it is normal and so must contain the unique minimal normal subgroup $\mathrm{Soc}(G)$. In an infinite group, this may not necessarily hold, since there might not be a minimal normal subgroup inside $\mathrm{C}_{G}(\mathrm{Soc}(G))$. Is it true that an infinite group with a non-abelian simple socle is almost simple; if so then how would I go about proving this, and if not is there a counterexample?

Here I am taking $G$ being almost simple to mean that $\mathrm{Inn}(S) \subseteq G \subseteq \mathrm{Aut}(S)$ for a non-abelian simple group $S$.

Michael
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    Isn't $A_5 \times ({\mathbb Z},+)$ a counterexample? – Derek Holt Feb 17 '25 at 14:35
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    @DerekHolt: I think so, and more generally $G = H \times A$ with $H$ non-abelian finite simple and $A$ infinite abelian. – testaccount Feb 17 '25 at 14:41
  • @DerekHolt Thank you for your help. I guess that $A_{5}$ is the socle in your example. I can believe that is true, since $A_{5}$ is its own unique minimal normal subgroup and $\mathbb{Z}$ has no minimal normal subgroups, but I am having some difficulty proving it since it doesn't suffice to consider the two parts of the direct product separately. Is there a straightforward way to show this? – Michael Feb 17 '25 at 15:12

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$G=A_5 \times ({\mathbb Z},+)$ is a counterexample.

I claim that ${\rm Soc}(G) = A_5$. To see this, note that $A_5$ is a minimal normal subgroup of $G$. If $N$ is any other minimal normal subgroup, then $N \cap A_5 = 1$, and so $N$ is isomorphic to a nontrivial subgroup of $G/A_5 \cong ({\mathbb Z},+)$. But then $N \cong ({\mathbb Z},+)$ and since all subgroups of $N$ are characteristic in $N$ they are all normal in $G$, but then $N$ is not minimal normal in $G$, contradiction. So $A_5$ is the only minimal normal subgroup of $G$, and ${\rm Soc}(G) = A_5$.

Now $C_G(N) \cong ({\mathbb Z},+)$ is not almost simple.

Derek Holt
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