The question, which is somewhat open-ended, is this: under which conditions can we guarantee that for a finite group $G$, $\operatorname{Aut}(G/Z(G))$ is isomorphic to a subgroup of $\operatorname{Aut}(G)$?

This is sometimes possible, but not always. It is clearly possible if $G$ has trivial centre or if $G$ is abelian. But it is also possible for $G \cong Q_8$, since $Q_8/Z(Q_8) \cong C_2 \times C_2$ so $\operatorname{Aut}(Q_8/Z(Q_8)) \cong S_3$ whereas $\operatorname{Aut}(Q_8) \cong S_4$. On the other hand, $D_8/Z(D_8) \cong C_2 \times C_2$ again, but $\operatorname{Aut}(D_8) \cong D_8$.

Another case where the embedding I am asking about is possible is when $G/Z(G)$ is a complete group, i.e. it has trivial centre and no outer automorphisms. In that case, $$\operatorname{Aut}(G/Z(G)) \cong G/Z(G) \cong \operatorname{Inn}(G)$$ is a normal subgroup of $\operatorname{Aut}(G)$, but that's not very interesting.

I should, perhaps, clarify that I am mainly interested in what (if anything) we can say from knowledge of $G$ and its subgroup structure alone, without assuming knowledge of $\operatorname{Aut}(G)$. If nothing interesting can be said, however, ignore this restriction.

  • $\begingroup$ Given a group $G$, a characteristic subgroup $N$ and an automorphism $\phi: G \to G$ then you can define $\bar{\phi} : G/N \to G/N$ on the quotient. Now take $N=Z(G)$, you have a map $\Phi: \operatorname{Aut}(G) \to \operatorname{Aut}(G/Z(G)$... so in the opposite direction! You can study specific cases in which this map is injective or surjective. $\endgroup$ – AnalysisStudent0414 Jul 9 '19 at 12:51
  • $\begingroup$ What does the existence of $\Phi$ imply in the context of my question? $\endgroup$ – the_fox Jul 9 '19 at 13:00
  • $\begingroup$ You want $A \lhd B$, and I'm telling you there is a map $B \to A$... if this map is surjective (is it?) then $A$ is a homomorphic image of $B$., so instead of subgroups you should be looking at subquotients! $\endgroup$ – AnalysisStudent0414 Jul 9 '19 at 13:08
  • $\begingroup$ Well, I am interested in subgroups. Even if the map is surjective, whether the target group is isomorphic to a subgroup of the domain is not clear. (It will, of course, be isomorphic to a quotient of the domain.) In other words, if $N$ is a normal subgroup of $X$, then $X$ may or may not have a subgroup isomorphic to $X/N$. $\endgroup$ – the_fox Jul 9 '19 at 13:19
  • $\begingroup$ Well, yes, that is my point. $\endgroup$ – AnalysisStudent0414 Jul 9 '19 at 13:20

If $H$ is a perfect group with trivial centre, and $G$ is the Schur covering group of $H$, then $G/Z(G) \cong H$ and ${\rm Aut}(G) \cong {\rm Aut}(H)$.

For example, if $H$ is the simple group ${\rm PSL}(n,q)$ with $n>1$, then with a small number of exceptions (such as ${\rm PSL}(2,9)$ and ${\rm PSL}(3,4)$), we have $G = {\rm SL}(n,q)$.

To explain the above, let $G$ be any group and write $G = F/R$ with $F$ free. Then an automorphism $\tau$ of $G$ lifts to a homomorphism (not necessarily an automorphism) $\rho:F \to F$ with $\rho(R) \le R$, and so $\rho$ induces a homomorphism $\bar{\rho}:F/[F,R] \to F/[F,R]$.

Since $R/[F,R] \le Z(F/[F,R])$, the restriction, $\sigma$ say, of $\bar{\rho}$ to $[F,F]/[F,R]$ is uniquely determined by $\tau$ - i.e. it is does not depend on the chosen lift to $\rho$.

Also, it is easy to see that applying the same process to $\tau^{-1}$ results in the inverse of $\sigma$ on $[F,F]/[F,R]$, so $\sigma$ is an automorphism. Note that $\sigma$ induces an automorphism of the Schur multiplier $M(G) = ([F,F] \cap R)/[F,R]$ of $G$, and in fact we get an induced homomorphism ${\rm Aut}(G) \to {\rm Aut}(M(G))$.

The above is true for any group $G$. But if $G$ is perfect, then $[F,F]/([F,F] \cap R) \cong G$, and $[F,F]/[F,R]$ is the unique Schur cover of $G$.

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  • $\begingroup$ You probably mean: "...Schur covering group of $H$,..." . Why is $\operatorname{Aut}(G) \cong \operatorname{Aut}(H)$? $\endgroup$ – the_fox Jul 9 '19 at 14:44
  • $\begingroup$ Yes, thanks. Roughly speaking the fact that $G$ is uniquely defined when $H$ is perfect implies that all automorphisms of $H$ lift to $G$. I can try and write down a more formal proof some time if you like, but it won't be for a day or two. Conversely, it is not hard to show that the only automorphism of $G$ that induces the identity on $G/Z(G)$ is the identity. That is because such an automorphism induces the identity on $[G,G] = G$. $\endgroup$ – Derek Holt Jul 9 '19 at 19:00
  • $\begingroup$ Thanks. That's ok, no hurry. If you have the time at some point and want to write down a few more details, I'd appreciate it. (Fine too would be a reference.) I suppose not much more than that can be said in general, right? $\endgroup$ – the_fox Jul 10 '19 at 3:29
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    $\begingroup$ I have added a few more details. $\endgroup$ – Derek Holt Jul 10 '19 at 7:49

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