Do there exist acyclic simple groups of arbitrarily large cardinality?

Question

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

Questions:

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.

Notes:

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.

Answer 1

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.

Answer 2

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.

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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).