# Embedding $\operatorname{Aut}(G/Z(G))$ in $\operatorname{Aut}(G)$

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.

• 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. – AnalysisStudent0414 Jul 9 '19 at 12:51
• What does the existence of $\Phi$ imply in the context of my question? – the_fox Jul 9 '19 at 13:00
• 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! – AnalysisStudent0414 Jul 9 '19 at 13:08
• 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 $X/N$. – the_fox Jul 9 '19 at 13:19
• Well, yes, that is my point. – 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 $$G$$.

• You probably mean: "...Schur covering group of $H$,..." . Why is $\operatorname{Aut}(G) \cong \operatorname{Aut}(H)$? – the_fox Jul 9 '19 at 14:44
• 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 $[G,G] = G$. – Derek Holt Jul 9 '19 at 19:00
• 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? – the_fox Jul 10 '19 at 3:29
• I have added a few more details. – Derek Holt Jul 10 '19 at 7:49