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For a group $G$, let $\operatorname{Aut}(G)$ denote the group of all automorphisms of $G$ and $\operatorname{Inn}(G)$ denote the subgroup of all autmorphisms which is of the form $f_h(g)=hgh^{-1}, \forall g\in G$, where $h\in G$ . Now if $G_1$ is a group containing $G$ as a subgroup then every $f_h \in \operatorname{Inn}(G)$ extends to an inner automorphism of $G'$ as $f_h(x)=hxh^{-1},\forall x\in G_1$, so in other words, for every $f\in \operatorname{Inn} (G)$ and every group $G_1$ containing $G$ as a subgroup, $\exists \bar f\in \operatorname{Inn}(G_1) \subseteq \operatorname{Aut} (G_1)$ such that $\bar f|_G =f$.

Now my question is : Let $f \in \operatorname{Aut} (G)$ be such that for every group $G_1$ containing $G$ as a subgroup, $\exists \bar f\in \operatorname{Aut}(G_1)$ such that $\bar f|_G =f$. Then is it necessarily true that $f \in \operatorname{Inn}(G)$ ? If this is not true in general, then does some extra condition on $G$ makes it true (like $G$ being finite, or simple)?

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    $\begingroup$ Take a look at George Bergman’s paper An inner automorphism is only an inner automorphism but an inner endomorphism can be something strange; inner automorphisms can be characterized as precisely the automorphisms $f$ of $G$ that can be extended functiorally to an automorphism of $H$ for any morphism $h\colon G\to H$. This seems related to your condition, so you may be able to use this characterization to show that $f$ is inner. $\endgroup$
    – Arturo Magidin
    Jan 3, 2019 at 2:20
  • $\begingroup$ @ArturoMagidin: thanks ... that looks interesting ... will take a look $\endgroup$
    – user521337
    Jan 3, 2019 at 2:23
  • $\begingroup$ Bergman's paper is very interesting, but I think the condition given there is much stronger than the condition you've given here, so the statement seems likely to either be false or hard to prove. Particularly since he says explicitly that he doesn't know whether the stronger condition that an automorphism be "extensible" (my words not his) along all homomorphisms implies that it is inner, which is why the coherence/functorial condition is necessary. (I'm admittedly quite tired, and the paper is certainly worth reading, but this is intended as a brief summary for future readers) $\endgroup$
    – jgon
    Jan 3, 2019 at 2:51

1 Answer 1

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How about this: A Characterization of Inner Automorphisms Paul E. Schupp Proceedings of the American Mathematical Society Vol. 101, No. 2 (Oct., 1987), pp. 226-228

https://www.jstor.org/stable/2045986?seq=1#page_scan_tab_contents

Abstract: It turns out that one can characterize inner automorphisms without mentioning either conjugation or specific elements. We prove the following

THEOREM Let $G$ be a group and let $\alpha$ an automorphism of $G$. The automorphism $\alpha$ is an inner automorphism of $G$ if and only if $\alpha$ has the property that whenever $G$ is embedded in a group $H$, then $\alpha$ extends to some automorphism of $H$.

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  • $\begingroup$ Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group? $\endgroup$
    – the_fox
    Jan 3, 2019 at 23:42
  • $\begingroup$ Well, the paper in the answer has the proof in the general case. Apparently, this was proved earlier in the finite case in "Miller, Charles F., III; Schupp, Paul E. Embeddings into Hopfian groups. J. Algebra 17 1971 171–176." $\endgroup$
    – verret
    Jan 3, 2019 at 23:55
  • $\begingroup$ I've looked at both, but neither contains a satisfactory answer. $\endgroup$
    – the_fox
    Jan 4, 2019 at 0:09
  • $\begingroup$ Why are the proofs in those papers unsatisfactory? $\endgroup$
    – verret
    Jan 4, 2019 at 0:57
  • $\begingroup$ They are not satisfactory to me because I am only interested in finite groups. The methods and techniques involved are often very different from the infinite counterpart. $\endgroup$
    – the_fox
    Jan 4, 2019 at 1:09

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