4
$\begingroup$

I have been looking for ways to construct examples of finitely generated residually finite groups that are poly-(locally virtually abelian) but not virtually solvable. If $K$ is a finite non-solvable group, then the wreath product $K \wr \mathbb{Z}$ satisfies all of these properties, except that of residual finiteness. While working on an unrelated problem, my co-author came up with a construction that looks like a residually finite version of the wreath product, which we then generalized. Since neither of us are group theorists, we'd like to know whether it is a well-known construction, and possibly find a reference that we could cite in our paper. Our literature searches haven't turned up anything, but it may be that we are just using the wrong keywords or sources.

The construction

Consider two residually finite countable groups $K$ and $H$ with $|K| \geq 2$. Define the set $$ X = \bigcup_{\substack{A \trianglelefteq_{\text{f}} K \\ B \trianglelefteq_{\text{f}} H}} (K/A)^{H/B} $$ where $A$ ranges over the finite-index normal subgroups of $K$, and similarly for $B$ and $H$. An element $f \in X$ is a function from $H/B$ to $K/A$. Define a left action $\phi_K$ of $K$ on $X$ by $$ \phi_K(k, f)(g B) = \begin{cases} k \cdot f(g B), & \text{if $g \in B$,} \\ f(g B), & \text{otherwise,} \end{cases} $$ and a left action $\phi_H$ of $H$ on $X$ by $$ \phi_H(h, f)(g B) = f(h^{-1} g B). $$ Each set $(K/A)^{H/B}$ is invariant under both actions; the idea is that $K$ acts by multiplication on the element at the "origin", while $H$ acts by translating the coordinates. We define $G = \langle \phi_K(K) \cup \phi_H(H) \rangle \leq \text{Sym}(X)$. The resulting group $G$ is residually finite (since it splits $X$ into finite orbits) and finitely generated if both $K$ and $H$ are, and the maps $\phi_K : K \to G$ and $\phi_H : H \to G$ are injective homomorphisms. Furthermore, $\phi_H$ has a retraction $\psi : G \to H$, and $\ker \psi$ is isomorphic to a subgroup of a direct product of copies of $K$. If we choose $K = A_5$ and $H = \mathbb{Z}$, then $G$ is locally finite-by-$\mathbb{Z}$ and hence poly-(locally virtually abelian), and one can show that it is not virtually solvable (see the link above for a proof).

$\endgroup$
3
$\begingroup$

Edit: I expanded this post after reading more carefully your construction and reminding a relevant reference. So it splits into 3 parts.

  1. B.H. Neumann's examples
  2. Comments on your construction
  3. (my original post: a construction with Mann)


1 B.H. Neumann's construction

Reference: B.H. Neumann. Some remarks on infinite groups. J. London Math. Soc. 1(2) (1937) 120--127.

He considered a strictly increasing sequence $(m_n)$, $m_n\ge 5$ tending to infinity, of odd numbers, and $X_n$ a copy of $\{1,\dots,m_n\}$, and $X=\bigsqcup_n X_n$. Let $c$ act as the 3-cycle $(123)$ on each $X_n$, and $t$ act as the full cycle $(1,\dots,m_n)$ on all $X_n$. Consider $\Gamma=\langle t,c\rangle$.

This is residually finite (since orbits are all finite). Neumann checks that $\Gamma$ has a homomorphism onto $\mathbf{Z}$, mapping $t$ to 1 and $c$ to 0. Moreover, its kernel contains a smaller normal subgroup, the finitely supported even permutations preserving all $X_n$ (that is, $\bigoplus_n\mathrm{Alt}_{m_n}$). The quotient is the semidirect product $\mathrm{Alt}(\mathbf{Z})\rtimes\mathbf{Z}$, where $\mathbf{Z}$ acts by shifing.

This construction is an illustration of the fact that every finitely generated LEF group is quotient of a FC-central group, which can be chosen of the form $\bigoplus_n F_n$ with each $F_n$ finite and normal in the whole group.

Note that it also, by the way, provides examples of residually finite 2-generator groups that are (locally finite)-by-$\mathbf{Z}$ (hence poly-(virtually abelian)) and not virtually solvable (as your example and mine in (3) below), but unlike those, satisfy no group identity.


2 Comments on your construction

First let me rewrite it when $K$ is a simple group and $H=\mathbf{Z}$. Since the singletons $K/K$ play no role, let me remove it. So $$X=\bigsqcup_{n\ge 1}K^n,$$ where $\mathbf{Z}$ acts on $K^n$ by shifting, and $K$ acts on $K^n$ by left multiplication at 0 (that is, $k\cdot (k_0,k_1,\dots,k_{n-1})=(kk_0,k_1,\dots,k_{n-1})$. These two operations on $K^n$ simply provide a simply transitive action of the wreath product $K\rtimes\mathbf{Z}/n\mathbf{Z}$.

When putting things together, one gets a group which, modulo its subgroup of finitely supported permutations, is the wreath product $K\wr\mathbf{Z}$, another typical example of a finitely generated LEF group that is not residually finite.

Actually, this is exactly the kind of construction one gets when one tries to make explicit the proof of the LEF fact mentioned above. A variant would consist in choosing, instead of the simply transitive action of $K\wr\mathbf{Z}/n\mathbf{Z}$, rather its "natural" action, which is on the disjoint union of $n$ copies of $K$, or even, for $K=\mathrm{Alt}_5$ on $n$ copies of $\{1,2,3,4,5\}$.

Another consequence is that these groups are not isomorphic to mine described in (3). Indeed, by construction, your group has a nontrivial FC-center, since it contains nontrivial finitely supported permutations (for an explicit example, let $t$ be the generator of $\mathbf{Z}$, and choose in $K$ two non-commuting elements $b,c$; then $[tbt^{-1},c]$ is supported by the first copy). While my example in (3) has a trivial FC-center.


3 In a note with Avinoam Mann (Some residually finite groups satisfying laws. Geometric group theory, 45–50, Trends Math., Birkhäuser, Basel, 2007.), we constructed a few residually finite groups including the following one, which answers your initial question:

For $A$ an abelian group, consider the free group $W_A$ on the generators $(x_n)_{n\in A}$ in the variety generated by the alternating group $\mathrm{Alt}_5$ (or any non-solvable finite group). (If you're not familiar with varieties of groups, this means the quotient of the free group $F$ on $(x_n)_{n\in A}$ by the intersection of all kernels of homomorphisms $F\to\mathrm{Alt}_5$. For instance, $W_A$ has exponent 30. It embeds into some power of $\mathrm{Alt}_5$ and hence is locally finite. By shifting the generators, we obtain an automorphism of $W_{\mathbf{Z}}$ and hence a semidirect product $G=W_{\mathbf{Z}}\rtimes\mathbf{Z}$, generated by 2 elements ($x_0$ and the generator of $\mathbf{Z}$).

Since $G$ is (locally finite)-by-abelian, it is obviously poly-(locally virtually abelian) (in two steps). It admits the wreath product $\mathrm{Alt}_5\wr\mathbf{Z}$ as a quotient and hence is not virtually solvable.

That $G$ is residually finite is proved as in the proof of Proposition 6 of the linked paper (cf Remark 7). Namely, consider an element $wt^n\neq 1$ and let us construct a finite index subgroup in which it does not belong. This is trivial if $n\neq 0$, so consider $w\neq 1$ in $W_{\mathbf{Z}}$. It involves finitely many letters, say in $\{x_i:-m<i<m\}$. Hence $w$ survives in the quotient $W_{\mathbf{Z}/m\mathbf{Z}}$ of $W_{\mathbf{Z}}$; this quotient is equivariant for the $\mathbf{Z}$-action, and hence $w$ survives in the quotient $W_{\mathbf{Z}/m\mathbf{Z}}\rtimes\langle t\rangle$ of $G$. Consider the finite index subgroup $G_m=W_{\mathbf{Z}}\rtimes\langle t^{2m}\rangle$. Then in restriction to $G_m$, $w$ survives in the quotient $W_{\mathbf{Z}/m\mathbf{Z}}\rtimes\langle t^m\rangle$, which is actually a the direct product $W_{\mathbf{Z}/m\mathbf{Z}}\times\langle t^m\rangle$. Hence it survives in its finite quotient $W_{\mathbf{Z}/m\mathbf{Z}}$.

Certainly your construction is closely related. Indeed, every group satisfying ($\star$): generated by two elements $t,x$ such that the normal subgroup generated by $x$ embeds into a power of $\mathrm{Alt}_5$ is a quotient of the above $G$, almost by definition. So your group is quite likely to me a quotient of mine. I haven't checked if ($\star$) holds for your construction but it seems true or close to be true.

[added note 1: it's very close to be true. Indeed suppose that one erases the first two copies, namely $K$ and $K\wr\mathbf{Z}/2\mathbf{Z}$ (this amounts to taking the quotient by a subgroup of order $2|K|^3$, when $K$ is simple), and one finds $b,c$ generating $K$ such $\{b,[b,c]\}$ also generates $K$. Then the group of your construction (at least after modding out these first two copies) is generated by $t$ and $btct^{-1}$, and thus is a quotient of $W_{\mathbf{Z}}\rtimes\mathbf{Z}$.]

[added note 2: the FC-center if this group is trivial, i.e., this group is ICC: indeed, if we have $wt^k$ with $w\neq 1$, then the $t^n(wt^k)t^{-n}=(t^nwt^{-n})t^k$ form infinitely elements, since they eventually have disjoint supports. Similarly, $(t^nwt^{-n})t^k(t^nwt^{-n})^{-1}$ form infinitely many conjugates of $t^k$ when $k\neq 0$. Since your groups have an infinite FC-center, this shows they are very different, anyway.]

(By the way, this work with Avinoam was initiated on sci.math.research, an kind of ancester of this site).

$\endgroup$
  • $\begingroup$ Thank you, this looks very comprehensive. We were aware of your and Mann's note, but I hadn't realized the similarity between the constructions. $\endgroup$ – Ilkka Törmä Jan 27 '18 at 10:22
  • $\begingroup$ I assume that LEF means "Locally embeddable into the class of finite groups" in the sense of A. Vershik and E. Gordon (ux1.eiu.edu/~yigordon/papers/Vershik-Gordon.pdf), right? What is the definition of an FC-central group? In your description of the general "LEF fact", I don't see clearly what is the extension group, the normal subgroup and the quotient group. Is it something you proved in one of your papers? $\endgroup$ – Luc Guyot Jan 27 '18 at 22:13
  • $\begingroup$ @LucGuyot yes for LEF. "FC-central" means "contained in the FC-center", the union of finite conjugacy classes. It's folklore. The idea is that if finite $G_n$ tends to $G$ in the space of marked quotients of $F$, the resulting homomorphism $F\to\prod_nF_n$ factors to an injective homomorphism $G\to\prod_nF_n/\bigoplus_nF_n$. Lifting in the product yields a residually finite group whose quotient by some subgroup of the FC-central subgroup $\bigoplus F_n$ is $G$. $\endgroup$ – YCor Jan 27 '18 at 22:32
  • $\begingroup$ @YCor Thanks, I think I get it now. Your sentence "every finitely generated LEF group is quotient of a FC-central group, which ..." still confuses me. So let me try to fill the gap. Statement: "Every finitely generated LEF group is the quotient of a residually finite group by a normal subgroup lying in its FC-center". Proof: Let $G$ be LEF. By simple arguments, we can build an injective homomorphism $\phi$ from $G$ into $\prod_n F_n/\bigoplus_n F_n$. Let $\tilde{G}$ be a subgroup of $\prod_n F_n$ generated by the lift of a finite generating set of $\phi(G)$ ... $\endgroup$ – Luc Guyot Jan 28 '18 at 17:04
  • $\begingroup$ ... then $G \simeq \tilde{G}/N$ where $N = \tilde{G} \cap (\bigoplus_n F_n))$ is FC-central and $\tilde{G}$ is residually finite. Is it what you meant? $\endgroup$ – Luc Guyot Jan 28 '18 at 17:09

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.