Embedding $G$ in a $Z(G)$ extension of $\operatorname{Aut}G$ This question follows up a question I asked on math.SE. This is a refinement and a reference request.

For what groups $G$ does there exist a $Z(G)$-extension of $\operatorname{Aut}G$ (call it $\tilde G$) that contains $G$ normally, in such a way that $\tilde G\rightarrow \operatorname{Aut}G$ describes $\tilde G$'s conjugation action on $G$?
Do you know if this question has been studied? If so, can you point me to references?

Comments

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*It is trivial that the class $\mathscr{C}$ of groups admitting such an embedding includes centerless groups (take $\tilde G = \operatorname{Aut} G$).


*Also trivially, $G\in\mathscr{C}$ if $\operatorname{Out}G = 1$. Take $\tilde G = G$.


*Also easily, $G\in\mathscr{C}$ if $G$ is abelian. Take $\tilde G$ to be $G$'s holomorph.


*It seems to me that also $G\in\mathscr{C}$ if $G$ has any faithful irreducible representation of a unique dimension and $\operatorname{Aut}G$ has trivial Schur multiplier, for the following reason. Let $\zeta:G\rightarrow GL(V)$ be the representation in question. $\operatorname{Aut}G$ acts on $G$'s representations but must fix $\zeta$ since it is the unique irreducible representation of its dimension; so any automorphism of $G$ is induced by conjugation by an element of $GL(V)$, which is determined up to a scalar factor; this gives us a faithful projective representation of $\operatorname{Aut}G$ on $V$, which lifts to an ordinary representation $\xi$ because the Schur multiplier of $\operatorname{Aut}G$ is trivial. Then it seems to me that the subgroup of $GL(V)$ generated by the images of $\zeta$ and $\xi$ can be taken to be $\tilde G$.


*Actually in this construction (when $G$ has a faithful irreducible representation $\zeta$ of a unique dimension), the assumption of trivial Schur multiplier for $\operatorname{Aut} G$ is overly restrictive. All we need is that the particular projective representation of $\operatorname{Aut} G$ arising from considering its action on the image of $\zeta$ (as above) represents the trivial class in the Schur multiplier. I am not sure how to tell when this happens.


*Even this latter condition seems too much to ask for this construction to work, because it is actually asking for $\tilde G \rightarrow\operatorname{Aut}G$ to split. It seems to me that the construction will work as long as the class of $H^2(\operatorname{Aut}G,\mathbb{C}^\times)$ corresponding to the projective representation of $\operatorname{Aut}G$ lies in the subgroup $H^2(\operatorname{Aut}G, \zeta(Z(G)))$. Again, I am not sure how to tell when this happens.
 A: As Derek Holt points out, the existence of such an extension is equivalent to the universal cohomology class in $H^3(\text{Out}(G);Z(G))$ being zero. The Eilenberg-MacLane paper is the original reference, a modern reference is K. Brown: Cohomology of Groups.
Some time ago I wrote some Magma programs to check whether a cohomology class in H^3 vanishes or not. To sum up, while the desired extension exists for many small groups, it does not exist for all finite groups. The smallest counterexamples are the dihedral group $D_{16}$ and the quaternion group $Q_{16}$.
There is also an old theorem of de Siebenthal which says that if $G$ is a compact connected Lie group, then the epimorphism $\text{Aut}(G)\rightarrow \text{Out(G)}$ splits, so in this case the desired extension always exists.
A: Yes, this situation has been studied. I think the first relevant reference is 
S. Eilenberg and S. MacLane, Cohomology theory in abstract groups II, Ann. of Math. 48 (1947), 326-341.
The main result IIRC is that a group $H$ with $I={\rm Inn} G \le H \le A={\rm Aut} G$ determines an element $\sigma \in H^3(H/I,Z(G))$ with the property that a group $\tilde H$ exists with $Z \le \tilde I \le \tilde H$, $Z \cong Z(G)$, $G \cong \tilde I$, and $\tilde H/Z \cong H$, if and only if $\sigma = 0$. So a nonzero $\sigma$ can be thought of as an obstruction to the existence of the extension.
Of course, this is just expressing the problem in a different language rather than solving the problem, But if you know that $H^3(A/I,Z(G))=0$ (for example if $A/I$ and $Z(G)$ have coprime orders), then you know that the extension exists.
