$G$ non-centerless group. How is $Z(\operatorname{Aut}(G))$ made?

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.)?

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