# 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?

• 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.

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.

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.

• +1 Thank you! I assume by $Q_16$ you mean $\langle x,y\mid x^8=y^4=1, x^4=y^2, yxy^{-1}=x^{-1}\rangle$, right? – benblumsmith Jan 18 '15 at 20:16
• Yes, I should perhaps have written "generalized quaternion group" instead of "quaternion group". The dihedral and quaternion groups of orders $20$ and $32$ also give examples where the extension doesnt exist. – Kasper Andersen Jan 20 '15 at 20:09