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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 .)

Charles Rezk
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  • 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 01 '20 at 01:13
  • 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 01 '20 at 01:40
  • Those are inner automorphisms. For any nonabelian finite simple group $G$, $Aut(G)$ has no outer automorphisms, I guess by the classification. But for some simple groups, the outer automorphisms do act on the Schur multiplier, eg, $PSL_n(\mathbb F_p)$ or $A_6$. – Ben Wieland Oct 01 '20 at 02:13
  • The proof that $Aut(G)$ is complete for a nonabelian finite simple group does not need the classification: it's contained in this comment for example. – Steve D Oct 01 '20 at 07:50
  • @BenWieland I don't understand. I'm thinking of the non-trivial central extension $G$ of $S_n$ with center of order 2. This $G$ has an non-identity automorphism covering the identity of $S_n$, which is not inner. – Charles Rezk Oct 01 '20 at 16:06
  • @SteveD Why does this need finiteness? – Charles Rezk Oct 01 '20 at 16:19
  • @BenWieland Why do outer automorphisms of $PU(n,p^2)$ act trivially on the Schur multiplier? – Charles Rezk Oct 01 '20 at 16:37
  • I misread the outer automorphism group from wikipedia. There's a part of the same size as the Schur multiplier and a Galois part. I have no idea what the first is, but it is unlikely to act. I thought that for $\mathbb F_{q^2}/\mathbb F_q$ it was the Galois group of $\mathbb F_q/\mathbb F_p$, but actually it's for $\mathbb F_{q^2}/\mathbb F_p$. So that would act on the Schur multiplier, leaving only 2-torsion. – Ben Wieland Oct 01 '20 at 20:08

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