# Can every non-inner automorphism of a group with residually finite outer automorphisms be realised as an non-inner automorphism of a finite quotient?

First some motivation: most proofs that show that the group of outer automorphisms is residually finite do not only show that the subgroup of inner automorphisms is closed in the profinite topology, they actually show something stronger: that the group of inner automorphisms is closed in the congruence topology, i.e. that a non-inner automorphism can be realised as an non-inner automorphims of some finite quotient. I would like to know whether there is a way to go back.

If $$N \unlhd G$$ is characteristic, the natural projection $$\pi_N \colon G \to G/N$$ induces a homorphism $$\tilde\pi_N \colon \mathop{Aut}(G) \to \mathop{Aut}(G/N)$$ given by $$\tilde\pi_N(\phi)(gN) = \phi(g)N$$ where $$g \in G$$ and $$\phi \in \mathop{Aut}(G)$$.

If $$G$$ is finitely generated, one can easily check that for every $$K \unlhd_{f.i.} G$$ there is $$L \unlhd_{f.i.} G$$ characteristic such that $$L \leq K$$, i.e. the profinite topology on $$G$$ is generated'' by characteristic subgroups of finite index.

The congruence topology on $$\mathop{Aut}(G)$$ is then the topology whose base around $$\mathop{id}_G$$ is given by $$\mathop{Cong}(\mathop{Aut}(G)) = \{\ker{\tilde\pi_N} \mid N \unlhd_{f.i.} G \mbox{ characteristic} \}.$$

One can easily check that the congruence topology is weaker than the profinite topology on $$\mathop{Aut}(G)$$. That is, if a set $$\Xi \subseteq \mathop{Aut}(G)$$ is closed in the congruence topology, then $$\Xi$$ is closed in the profinite topology.

My question is:

if $$G$$ is a finitely generated residually finite group, does $$\mathop{Inn}(G)$$ being closed in the profinite topology on $$\mathop{Aut}(G)$$ imply $$\mathop{Inn}(G)$$ being closed in the congruence topology? Or equivalently, can every non-inner automorphism of $$G$$ be realised as an non-inner automorphism of some finite quotient of $$G$$ if $$\mathop{Out}(G)$$ is residually finite?

• The answer should be negative, due to the following fact: for every countable group $H$ there is a f.g. infinite simple group $G$ such that $H \cong Out(G)$ (arxiv.org/abs/0704.0091). So the residual finiteness of Out(G) may be independent of the residual finiteness of $G$. – Ashot Minasyan Nov 1 '18 at 9:44