# Smallest group containing $G$ with every automorphism of $G$ becomes inner

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?

• "Smallest" in what sense? Cardinality? Commented Nov 9, 2022 at 23:01
• „the“ in what sense? isomorphism? Equivalence in the homomorphism preorder? Commented Nov 9, 2022 at 23:03
• 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. Commented Nov 10, 2022 at 1:55
• ${\rm Inn}(G)$ is not in general a normal subgroup of ${\rm Hol}(G)$. Commented Nov 10, 2022 at 8:09
• 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. Commented Nov 10, 2022 at 14:14

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.