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:

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

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