Is it $Z(\operatorname{Aut}(G)) \cap \operatorname{Inn}(G) \cong H/Z(G)$ for some $H \le G$?

Question

Could you check if this proof is correct, please? (I'm not even sure about the result itself, whence the title.)

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Proposition. Let $G$ be a group. Then: $$Z(\operatorname{Aut}(G)) \cap \operatorname{Inn}(G) \cong H/Z(G)$$ where $H=\lbrace a \in G \mid \sigma(a) \in Z(G)a, \forall \sigma \in \operatorname{Aut}(G) \rbrace$.

Proof. Let $\varphi: G \rightarrow \operatorname{Aut}(G)$ be the homomorphism induced by conjugacy, namely $\varphi_a(g):=a^{-1}ga$. We get:

\begin{alignat}{1} \varphi_a \in Z(\operatorname{Aut}(G)) &\Leftrightarrow \varphi_a\sigma=\sigma\varphi_a, \forall \sigma \in \operatorname{Aut}(G) \\ &\Leftrightarrow (\varphi_a\sigma)(b)=(\sigma\varphi_a)(b), \forall b \in G, \forall \sigma \in \operatorname{Aut}(G) \\ &\Leftrightarrow \varphi_a(\sigma(b))=\sigma(\varphi_a(b)), \forall b \in G, \forall \sigma \in \operatorname{Aut}(G) \\ &\Leftrightarrow \varphi_a(\sigma(b))=\sigma(a^{-1}ba), \forall b \in G, \forall \sigma \in \operatorname{Aut}(G) \\ &\Leftrightarrow \varphi_a(\sigma(b))=\sigma(a^{-1})\sigma(b)\sigma(a), \forall b \in G, \forall \sigma \in \operatorname{Aut}(G) \\ &\Leftrightarrow \varphi_a(\sigma(b))=\sigma(a)^{-1}\sigma(b)\sigma(a), \forall b \in G, \forall \sigma \in \operatorname{Aut}(G) \\ &\Leftrightarrow \varphi_a(\sigma(b))=\varphi_{\sigma(a)}(\sigma(b)), \forall b \in G, \forall \sigma \in \operatorname{Aut}(G) \\ &\Leftrightarrow (\varphi_a\sigma)(b)=(\varphi_{\sigma(a)}\sigma)(b), \forall b \in G, \forall \sigma \in \operatorname{Aut}(G) \\ &\Leftrightarrow \varphi_a\sigma=\varphi_{\sigma(a)}\sigma, \forall \sigma \in \operatorname{Aut}(G) \\ &\Leftrightarrow \varphi_a=\varphi_{\sigma(a)}, \forall \sigma \in \operatorname{Aut}(G) \\ &\Leftrightarrow \sigma(a) \in (\operatorname{ker}\varphi)a, \forall \sigma \in \operatorname{Aut}(G) \\ &\Leftrightarrow a \in H \\ \end{alignat}

where $H:= \lbrace a \in G \mid \sigma(a) \in Z(G)a, \forall \sigma \in \operatorname{Aut}(G) \rbrace $. Thence, $H=\varphi^{\leftarrow}\lbrace \operatorname{Inn}(G) \cap Z(\operatorname{Aut}(G)) \rbrace$ and, by the Correspondence Theorem: $H \le G$, $H \supseteq Z(G)$, $H/Z(G) \cong \operatorname{Inn}(G) \cap Z(\operatorname{Aut}(G))$. $\Box$

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EDIT:

Corollary

1. $Z(\operatorname{Aut}(G)) \cap \operatorname{Inn}(G) = \lbrace \iota \rbrace \Leftrightarrow H=Z(G)$: this holds if $G$ is abelian (trivially, being then $\operatorname{Inn}(G)=\lbrace \iota \rbrace$). Are there nonabelian $G$s such that $H=Z(G)$?
2. If $G$ is centerless $(Z(G)=\lbrace e \rbrace)$, then: $$Z(\operatorname{Aut}(G)) \cap \operatorname{Inn}(G) \cong \lbrace a \in G \mid \sigma(a)=a, \forall \sigma \in \operatorname{Aut}(G) \rbrace = \bigcap_{\sigma \in \operatorname{Aut}(G)}\operatorname{Fix}(\sigma)$$ where $\operatorname{Fix}(\sigma):=\lbrace g \in G \mid \sigma(g)=g \rbrace$.

Answer 1

This all looks correct.

For more on your corollaries:

1. I doubt we can easily classify all $G$ with $H=Z(G)$ but an obvious collection of groups for which this is true is those $G$ for which $G/Z(G)$ is centerless - this includes simple and, more generally, quasi-simple groups.
2. This is a special case of $G/Z(G)$ centerless. In particular, in this case, $Z({\rm Aut}(G))\cap {\rm Inn}(G)$ is trivial so $\bigcap\limits_{\sigma\in{\rm Aut}(G)}{\rm Fix}(\sigma)$ is trivial.

Answer 2

It seems okay to me.

Having $\varphi:G\to\operatorname{Aut}(G)$ be induced by conjugacy isn't all that distinct from $\varphi\in\operatorname{Inn}(G)$ for me for some reason. Perhaps you could be a little more detailed there. EDIT: But I see now that it is the result of being in the intersection, right?

The iff section is flawless.

However, your use of the Correspondence Theorem could be clearer, although I suppose only for those inexperienced with it like me.

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Disclaimer: Please take this answer with a pinch of salt. This is advanced stuff.