Recall that a group $G$ is acyclic if its group homology vanishes: $H_\ast(G; \mathbb Z) = 0$. Equivalently, $G$ is acyclic iff the space $BG$ is acyclic, i.e. $\tilde H_\ast(BG;\mathbb Z) = 0$.

In order to tie up a loose end over at this question, I wonder


  1. Do there exists arbitrarily large simple acyclic groups?

  2. More generally, do there exist arbitrarily large simple groups $G$ such that there exists an acyclic space $X(G)$ with $\pi_1(X(G)) = G$?

  3. Do there exist arbitrarily large simple groups $G$ with $H_2(G; \mathbb Z) = 0$ -- or equivalently (I think) for which there are no nontrivial central extensions?

  4. Heck, what is one example of a simple nonabelian group $G$ with $H_2(G;\mathbb Z) = 0$?

(2) is all I really need, for which (3) will suffice (see below); (1) is just a natural strengthening.


  1. There is a proper class of simple groups; e.g. the alternating group on any set is simple (though not acyclic).

  2. There are also acyclic spaces with arbitrarily large fundamental group, cf. Kan-Thurston, but the constructions I've seen don't produce spaces with simple fundamental group.

  3. In the comments at the above-linked question, Tom Goodwillie points out that a positive answer to (3) implies a positive answer to (2) by taking $X(G)$ to be the fiber of $BG \to BG^+$.

I've included the "model theory" and "logic" tags mostly because I suspect maybe the people who know the most about very large simple groups might just be logicians. But if these tags seem inappropriate, I wouldn't object too strongly to removing them.

  • 1
    $\begingroup$ This is probably unimportant, but your formulation of the question is sensitive to class-theoretic issues. I guess you mean a proper class of pairwise non-isomorphic such groups? Depending on your set-theoretic framework and issue of global choice, this is not necessarily the same as saying that there are arbitrarily large such groups (since you have to pick them, which is not AC but global AC). But with global choice, this issue evaporates. But perhaps simpler just to ask, as in your title: are there arbitrarily large such groups? $\endgroup$ – Joel David Hamkins Dec 18 '19 at 15:51
  • $\begingroup$ @JoelDavidHamkins Ah, good point. I should definitely specify "non-isomorphic", and indeed, I really do just mean "arbitrarily large". I suppose I tend to forget about this subtlety -- I think my usual implicit mathematical metatheory is ZFC + universes, but with "class" meaning "bigger than some fixed universe". $\endgroup$ – Tim Campion Dec 18 '19 at 15:53

I just realized this is indeed, as Neil Strickland and Tom Goodwillie predicted, not hard, thanks to the fact that a directed union of simple groups is simple. Since homology commutes with direct limits, acyclic groups are also closed under directed unions.

So start with a group $G = G_0$ of sufficiently large cardinality. Embed it in a simple group $G_1$. Then use the Kan-Thurston result to embed $G_1$ in an acyclic group $G_2$. Repeat, obtaining a chain $G_0 \subseteq G_1 \subseteq G_2 \subseteq \dots$. The union $G_\infty$ is simple, since it's the union of the $G_{2i+1}$'s, and acyclic, since it's the union of the $G_{2i}$'s.

Thus every group $G$ embeds in a group $G_\infty$ which is simple and acyclic. In particular, there are simple acyclic groups of arbitrarily large cardinality, and the answers to all the above questions are affirmative.


Here are some explicit examples.

Let $\alpha$ be a cardinal $\ge\aleph_1$ and $X$ a set of cardinal $\alpha$ (we can choose $X=\alpha$). Let $G_\alpha=S_\alpha/D_\alpha$, where $S_\alpha$ (resp.\ $D_\alpha$) is the group of permutations of $\alpha$ whose support has cardinal $\le\aleph_1$ (resp. $<\aleph_1$). This is a simple group (particular case of a result of Baer).

Claim: $G_\alpha$ is acyclic.

Indeed, in a paper of P. de la Harpe and D. McDuff (CMH 1983), one has the definition (given below) of a "flabby" group, with the lemma, attributed to Wagoner "every flabby group is acyclic".

I claim:

$G_\alpha$ is flabby for every $\alpha\ge\aleph_2$. Hence this is a simple acyclic group (of cardinal $\ge\alpha$, namely the same as the set of subsets of $\alpha$ of cardinal $\le\aleph_1$).

I start with the definition: $G$ is flabby if there exist homomorphisms: $\sqcup:G\times G\to G$ ("concatenation") and $\tau:G\to G$ ("countable repetition") satisfying:

for every finite subset $F\subset G$, there exist $u,v,w\in G$ such that $g\sqcup 1=ugu^{-1}$ and $1\sqcup g=vgv^{-1}$, and $g\sqcup \tau(g)=w\tau(g)w^{-1}$ for every $g\in F$.

Indeed, let $s$ be a bijection $X\to X\times\omega$; think of $X\times\{n\}$ as the $n$-th copy of $\alpha$. Define $g\sqcup h$ as "$g$ on the $0$-th copy, $h$ on the $1$-st copy, and identity on other copies, and $\tau(g)$ as "$g$ on each copy". Note that $\tau$ is well-defined (if we were modding out the finitely supported subgroup, this would fail).

Now fix $F$ finite ($F$ of cardinal $<\alpha$ would also work); the union $X_F$ of supports of all $g\in F$ has cardinal $\alpha$. Extend the inclusion $X_F\to X_F\times\{0\}$ to a bijection $U:X\to X\times\omega$ and define $u=s^{-1}\circ U$. Then it satisfies the required equality. The other two conjugacy are obtained similarly.

Notes: let $S(\alpha,\beta)$ be the group of permutations of $\alpha$ with support of cardinal $<\beta$ (it is understood that $\beta$ is infinite or $1$). Noyte that $G_\alpha=S(\alpha,\aleph_2)/S(\alpha,\aleph_1)$. The argument works without change to prove that for all cardinals $\alpha,\beta,\gamma$, the group $S(\alpha,\beta)/S(\alpha,\gamma)$ is flabby, acyclic if $\beta\le\alpha$ and $\gamma$ has uncountable cofinality. Probably the conclusion that it is acyclic holds for $\beta=\alpha^+$ (for $\gamma=1$ this is done in Harpe-McDuff).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.