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

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

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: $$H \le G$$, $$H \supseteq Z(G)$$, $$H/Z(G) \cong \operatorname{Inn}(G) \cap Z(\operatorname{Aut}(G))$$. $$\Box$$

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 $$\operatorname{Fix}(\sigma):=\lbrace g \in G \mid \sigma(g)=g \rbrace$$.

This all looks correct.

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.

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.

• Your warning on the Correspondence Theorem is appropriate, as I'm not very familiar with it. About the other point, $\operatorname{Inn}(G)$ is just the image of $\varphi$, that is normal in $\operatorname{Aut}(G)$, as $Z(\operatorname{Aut}(G))$ is: here I wish to study their intersection. Please, keep it.