Groups with trivial outer automorphism group and prescribed center?

Given an arbitrary abelian group $$A$$, can we find a group $$G$$ such that

• $$\mathrm{Out}(G)=\mathrm{Aut}(G)/\mathrm{Inn}(G)=1$$, and
• $$Z(G)\simeq A$$?

Why is this interesting? Given a group $$G$$, we have its classifying space $$BG$$, which in turn has a topological monoid $$h\mathrm{Aut}(BG)$$ of self-homotopy equivalences. It is straightforward to show that

• $$\pi_0(h\mathrm{Aut}(BG)) = \mathrm{Out}(G)$$,
• $$\pi_1(h\mathrm{Aut}(BG))= Z(G)$$, and
• $$\pi_n(h\mathrm{Aut}(BG)) =0$$ for $$n\geq 2$$.

Thus, I am asking: for which $$A$$ is $$BA$$ equivalent to the space of self-homotopy equivalences of some $$BG$$?

Here are the examples I think I know:

1. $$A=1$$, from $$G=1$$. (Or more generally, from any "complete" group.)
2. $$A=Z/2$$, from $$G=Z/2$$.

[I had more, but they weren't correct, as pointed out by Ben Wieland in comments.]

I think you could construct more by the following procedure: every such $$G$$ is a central extension of some $$K=G/A$$ with $$Z(K)=1$$. So given $$A$$, we can (I think) produce such a $$G$$ if we can find: a group $$K$$ with $$Z(K)=1$$ and $$H^1(K,A)=0$$, and an element $$\kappa\in H^2(K,A)$$ which is not fixed by any non-identity element of $$\mathrm{Out}(K)\times \mathrm{Aut}(A)$$. (Is that right?)

Note that I'm notrequiring $$G$$ to be finite, or even finitely presented, and there are apparently many $$G$$ with trivial $$\mathrm{Out}(G)$$. I have no idea what centers you can get this way. (I don't know much group theory.)

(This question is a variant of Group with finite outer automorphism group and large center .)

• I'm not sure it's a harder version of my question you're linking at, because in my question the constraint that $G$ is finitely generated was the main difficulty. – YCor Sep 30 '20 at 18:27
• OK. I see. It's related at least. – Charles Rezk Sep 30 '20 at 18:28
• I've been infected by homotopy type theory, so sometimes I think of isomorphism as a kind of equality :D – Charles Rezk Sep 30 '20 at 18:36
• The alternating groups do have outer automorphisms, but you can substitute the symmetric groups. Similarly, I believe that although the $PU_n(\mathbb F_{p^2}/\mathbb F_p)$ has outer automorphisms, they act trivially on its Schur multiplier, and thus $Aut(PU_n(\mathbb F_{p^2}/\mathbb F_p))$ is a family of groups with no outer automorphisms and a rich choice of cyclic central extensions. – Ben Wieland Oct 1 '20 at 1:13
• Damn, you're right. But the central extensions of symmetric groups will have extra automorphisms in most cases, since $S_n/[S_n,S_n]$ is order 2. – Charles Rezk Oct 1 '20 at 1:40