Let $G$ be a (possibly infinite) non-centerless group, i.e. such that $Z(G) \ne \lbrace e \rbrace$. Left and right multiplications establish the subgroups $\Theta:=\lbrace \theta_a \mid a \in G \rbrace \le \operatorname{Sym}(G)$ and $\Gamma:=\lbrace \gamma_a \mid a \in G \rbrace \le \operatorname{Sym}(G)$, such that:

  • $G \cong \Theta$ and $G \cong \Gamma$;
  • $\Theta\Gamma=\Gamma\Theta \le \operatorname{Sym}(G)$;
  • $Z(G) \cong \Theta \cap \Gamma$;
  • $\operatorname{Inn}(G)= \Theta\Gamma \cap \operatorname{Aut}(G)$.

Can we state anything about $Z(\operatorname{Aut}(G))$ and, in particular, about its relationships with $\Theta \cap \Gamma$ (inclusion, intersection, etc.)?

  • 2
    $\begingroup$ What do you mean by the second sentence? $G\cong \Theta\cong \Gamma$? $\endgroup$ – tomasz Mar 20 at 15:11
  • $\begingroup$ Yes, what you have written. $\endgroup$ – Luca Mar 20 at 15:14
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    $\begingroup$ These are very strange conditions. What is the motivation? $\endgroup$ – tomasz Mar 20 at 15:14
  • $\begingroup$ It looks like you intend to have $\Theta$ and $\Gamma$ be the $G$ acting on itself by left and right translation. Is that right? $\endgroup$ – tomasz Mar 20 at 15:26
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    $\begingroup$ ${\rm Aut}(G)$ fixes the identity element, so it has trivial intersection with both $\Theta$ and $\Gamma$. $\endgroup$ – Derek Holt Mar 20 at 20:22

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