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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 .)

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  • $\begingroup$ 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. $\endgroup$
    – YCor
    Sep 30, 2020 at 18:27
  • $\begingroup$ OK. I see. It's related at least. $\endgroup$
    – Charles Rezk
    Sep 30, 2020 at 18:28
  • $\begingroup$ I've been infected by homotopy type theory, so sometimes I think of isomorphism as a kind of equality :D $\endgroup$
    – Charles Rezk
    Sep 30, 2020 at 18:36
  • $\begingroup$ 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. $\endgroup$
    – Ben Wieland
    Oct 1, 2020 at 1:13
  • $\begingroup$ 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. $\endgroup$
    – Charles Rezk
    Oct 1, 2020 at 1:40

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