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What is the smallest group $\widetilde{G}$ containing $G$ and such that every automorphism of $G$ is induced by an inner automorphism of $\widetilde{G}$?


Here are some of my thoughts. For notation, let $G$ be a group and denote $\operatorname{Inn}(G)$ the group of inner automorphisms (i.e. those which are conjugation in $G$) and define $\operatorname{Out}(G) = \operatorname{Aut}(G)/ \operatorname{Inn}(G)$.

The holomorph $\operatorname{Hol}(G)$ is $G \rtimes_{\operatorname{id}} \operatorname{Aut}(G)$ has the desired property, but doesn't seem to be the smallest such group. For instance, $\operatorname{Aut}(D_4) \cong D_4$ with $\operatorname{Out}(D_4) \cong C_2$. Then $\operatorname{Hol}(D_4)$ has order 64, while $D_8 \cong D_4 \rtimes \operatorname{Out}(D_4)$ is smaller, and every automorphism of $D_4$ is inner when we consider it as a subgroup of $D_8$.

Maybe the general construction should be $\widetilde{G} \cong \operatorname{Hol}(G) / \operatorname{Inn}(G)$. Does it have a name? Is there a reason the holomorph is the canonical choice, even if its bigger?

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    $\begingroup$ "Smallest" in what sense? Cardinality? $\endgroup$
    – freakish
    Commented Nov 9, 2022 at 23:01
  • $\begingroup$ „the“ in what sense? isomorphism? Equivalence in the homomorphism preorder? $\endgroup$ Commented Nov 9, 2022 at 23:03
  • $\begingroup$ By small I guess I mean I expect it to be a subgroup or quotient of $\operatorname{Hol}(G)$. I would also be happy to hear a criterion by which $\operatorname{Hol}(G)$ is already the "best" option. $\endgroup$ Commented Nov 10, 2022 at 1:55
  • $\begingroup$ ${\rm Inn}(G)$ is not in general a normal subgroup of ${\rm Hol}(G)$. $\endgroup$
    – Derek Holt
    Commented Nov 10, 2022 at 8:09
  • $\begingroup$ You are only going to get good answers when $Z(G)=1$, I think. It's going to go badly for a variety of reasons in general, e.g., as Derek Holt says above, and also non-uniqueness even when it does work. $\endgroup$ Commented Nov 10, 2022 at 14:14

1 Answer 1

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Any such group $\widetilde{G}$ would have $G \trianglelefteq \widetilde{G}$ such that the map $\widetilde{G} \rightarrow \operatorname{Aut}(G)$ is surjective, giving an isomorphism $\widetilde{G} / C_{\widetilde{G}}(G) \rightarrow \operatorname{Aut}(G)$.

Take for example $G = D_8$, dihedral of order $8$.

You could take $\widetilde{G} = D_{16}$, dihedral of order $16$. This contains two normal subgroups isomorphic to $G$, and for both the map $\widetilde{G} \rightarrow \operatorname{Aut}(G)$ is surjective.

You could also take $\widetilde{G} = \operatorname{SD}_{16}$ (semidihedral group of order $16$), which contains a normal subgroup isomorphic to $G$, such that $\widetilde{G} \rightarrow \operatorname{Aut}(G)$ is surjective.

So it seems to me the $\widetilde{G}$ you are looking for is not unique.

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