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Let $G$ be a finitely generated residually finite group and let $M$ be a finitely generated $\mathbb{Z}[G]$-module.

Question: Must $M$ be residually finite in the sense that for all nonzero $x \in M$, there exists some submodule $N$ of $M$ such that $x \notin N$ and $M/N$ is finite?

If this is not true in general, is it true if $G$ is also assumed to be nilpotent?

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2 Answers 2

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It's true, and due to Ph. Hall, when $G$ is virtually nilpotent, and more generally (Roseblade) when $G$ is virtually polycyclic.

When $G=\mathbf{Z}\wr\mathbf{Z}$ there exists an infinite simple $\mathbf{Z}G$-module, so it's not residually finite.

Added: The counterexample is due to Ph. Hall. Since I already mentioned the original one for other purposes on this site, let me provide an immediate variant, which entails the result.

Notation: $L_n=(\mathbf{Z}/n\mathbf{Z})\wr\mathbf{Z}$, $\mathbf{F}_p=\mathbf{Z}/p\mathbf{Z}$ (viewed as field).

Proposition Let $p$, $q$ be primes such that $p$ divides $q-1$. Then there exists an infinite simple $\mathbf{F}_q L_p$-module $V$ (which is therefore a simple $\mathbf{Z} L_p$-module, thus not residually finite).

Proof: Fix an element $x$ of order $p$ in the multiplicative group $\mathbf{F}_q^*$. Let $(w_n)_{n\in\mathbf{Z}}$ be valued in $\{1,x\}$, with the property that for every $n$ there exists $m$ such that $w_m=x$ and $w_i=1$ for all $i$ such that $0<|i|\le n$.

Let $V_q=\mathbf{F}_q^{(\mathbf{Z})}$ be the abelian group of finitely supported sequences $\mathbf{Z}\to\mathbf{F}_q$, with basis $(e_m)_{m\in\mathbf{Z}}$. Let $d$ be the diagonal automorphism of $V$: $(x_n)\mapsto (w_nx_n)$. Let $\tau$ be the shift $(x_n)\mapsto (x_{n+1})$. Hence $\tau^nf\tau^{-n}$ is diagonal for all $n$ and $f^p=\mathrm{id}$, so that $\tau$, $d$ define a representation $L_p\to\operatorname{Aut}(V)$. Thus $V_q$ is a $\mathbf{F}_qL_p$-module.

I claim it is a simple $\mathbf{F}_qL_p$-module. Indeed, start from a nonzero $v\in V_q$, and let $W$ be the $L_p$-submodule generated. Let $S$ be the (finite) support of $v$ and fix $m\in S$. Then there exists a translate of $w$ that equals $1$ on $S\smallsetminus\{m\}$ and equals $x$ on $n$. This corresponds to $\tau^n d\tau^{-n}$ for some $n$. Hence $\tau^nd\tau^{-n}v-v$ is a nonzero scalar multiple of $e_m$. So $e_m\in W$, and using $\tau$ it follows that $W$ contains all basis elements. Hence $W=V$, proving simplicity.

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  • $\begingroup$ Can you give a reference for the $\mathbb{Z}\wr\mathbb{Z}$ example? $\endgroup$
    – HJRW
    Commented May 20, 2021 at 8:46
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    $\begingroup$ @HJRW It's due to Ph. Hall, in his 1954 or 1958/1959 paper. I should check. I think I already described the construction on MO (it's actually torsion-free divisible). Actually for this very question I think one can even do it with $(\mathbf{Z}/p\mathbf{Z})\wr\mathbf{Z}$ (with module being a torsion group, i.e. killed by some — other— prime $p'$). $\endgroup$
    – YCor
    Commented May 20, 2021 at 9:21
  • $\begingroup$ Thanks! I'm trying to find the results you reference in the first paragraph. Roseblade's paper seems to be Group rings of polycyclic groups. J. Pure Appl. Algebra 3 (1973), 307–328., which contains a clear statement of the result in question. I'd be interested in understanding Hall's proof in the special case of nilpotent groups (which must be easier than polycyclic ones). My guess is that Hall's paper is On the finiteness of certain soluble groups. Proc. London Math. Soc. (3) 9 (1959), 595–622, but I can't find a clear statement of anything equivalent to my question. Can you help? $\endgroup$
    – Alice
    Commented May 20, 2021 at 18:09
  • $\begingroup$ I actually think this is done in a Springer book by Robinson, I should double check. This is actually essentially equivalent to the statement that finitely generated abelian-by-nilpotent groups are residually finite. $\endgroup$
    – YCor
    Commented May 20, 2021 at 18:09
  • $\begingroup$ @YCor: Thanks! Just to make sure I'm understanding right, is this the argument you have in mind? Let $G$ be a f.g. nilpotent group and let $M$ be a f.g. $\mathbb{Z}[G]$-module. Consider $x \in M$. Let $\Gamma$ be the semidirect product of $G$ by $M$. Then $\Gamma$ is f.g. abelian-by-nilpotent, so we can find a normal subgroup $N \lhd \Gamma$ with $x \notin N$ and $\Gamma/N$ finite. It follows that $N'=N \cap M$ is a submodule of $M$ with $x \notin N'$ and $M/N'$ finite. $\endgroup$
    – Alice
    Commented May 20, 2021 at 18:30
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Here is another example with different groups. Formanek showed that the group ring over any field of a free product of non-trivial groups (and not both order 2) is primitive, has a faithful simple module. That means that $\mathbb F_pG$ is primitive whenever $G=A\ast B$. And of course $G$ is residually finite if both $A$ and $B$ are. Obviously a faithful simple $\mathbb F_pG$-module is cyclic as a $\mathbb ZG$-module and since $G$ is infinite, it cannot be finite. So this gives lots of examples including the free group on two generators.

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  • $\begingroup$ What makes you believe that a free product is easier than a wreath product? In any case the module over the lamplighter group is mechanically a (non-faithful) module over the free group on 2 generators. $\endgroup$
    – YCor
    Commented May 20, 2021 at 19:50
  • $\begingroup$ Note that in Formanek's result you certainly have to exclude the free product of two groups of order 2. $\endgroup$
    – YCor
    Commented May 20, 2021 at 19:51
  • $\begingroup$ @YCor, of course you are right and he puts that in parentheses right at the beginning of the paper. $\endgroup$
    – Benjamin Steinberg
    Commented May 20, 2021 at 19:55
  • $\begingroup$ @YCor, I guess you make a good point. I think of free groups as easy because most students learn about it in a first algebra course while few general algebra books mention wreath products. In any event a faithful simple module is much stronger than just having an infinite simple one. I guess I should delete this answer though. $\endgroup$
    – Benjamin Steinberg
    Commented May 20, 2021 at 19:59
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    $\begingroup$ A datapoint: I definitely learned what a free product was as a student, but I only learned what a wreath product was far later in my career. I still find free products easier to think about. $\endgroup$
    – Alice
    Commented May 20, 2021 at 20:57

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