The present question follows up this one, in which I accidentally asked for less than I actually wanted. Given a group $G$, I would like to find an extension $\tilde G$ of its automorphism group $\operatorname{Aut}G$ by its center $Z(G)$, into which $G$ embeds in such a way that the restriction of the surjection $\tilde G\rightarrow\operatorname{Aut}G$ to $G$ maps $g\mapsto \rho_g$, where $\rho_g$ is conjugation by $g$.

With a little more precision: starting from a group $G$, I want an exact sequence

$$1\rightarrow Z(G) \rightarrow\tilde G\xrightarrow{\varphi} \operatorname{Aut}G\rightarrow 1$$

and an embedding $i:G\hookrightarrow \tilde G$, such that $\varphi|_{i(G)}$ is the map $i(g)\mapsto \rho_g$.

My question is this:

Does the desired $\tilde G$ always exist, and for what $G$'s is it unique? Is it always unique?

I am happy to assume that $G$ is finite if that is helpful.

Thanks in advance.


(1) I think that finding my desired $\tilde G$ is equivalent to finding an element of $H^2(\operatorname{Aut}G,Z(G))$ whose restriction to $H^2(\operatorname{Inn}G,Z(G))$ is the class of $G$.

(2) the difference between this question and the previous one I asked (which was correctly answered by the holomorph) is that here I am requiring that the kernel of the map $\tilde G\rightarrow \operatorname{Aut}G$ be exactly $Z(G)$.


No, such an extension does not exist for $G = {\rm SL}(2,9)$, for example (but I don't know whether that is the smallest such example). In the notation of the ATLAS of Finite Groups, $G = 2.A_6$ is a perfect central extension of $A_6 \cong {\rm PSL}(2,9)$, and ${\rm Aut}(G) = {\rm P \Gamma L}(2,9) = A_6.2^2$.

There are three extensions of $A_6$ by automorphisms of order $2$, namely $S_6$, ${\rm PGL}(2,9)$ and $M_{10}$. The combined extensions $2.S_6$ and $2.{\rm PGL}(2,9)$ both exist, but $2.M_{10}$ does not and hence neither does $2.A_6.2^2$. There is a lot of information about this question for simple groups in the ATLAS.

More generally, for any odd prime $p$, the extension $2.{\rm PSL}(2,p^2).2^2$ does not exist.

  • $\begingroup$ Thanks so much. Unsure of a piece of vocabulary you're using - what do you mean an "extension of $A_6$ by automorphisms of order $2$"? (I see you do not mean a $\mathbb{Z}_2$ extension of $A_6$ because $S_6$ is not one, but you also do not mean an $A_6$ extension of $\mathbb{Z}_2$ because $PGL(2,9)$ is not one...) $\endgroup$ – Ben Blum-Smith Nov 15 '14 at 1:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.