6
$\begingroup$

Let $G,H$ be infinite finitely generated groups such that $F(G)=F(H)$. Where $F(G)$ denotes the isomorphism classes of finite quotients of $G$. Let $G$ be residually finite. can we say that $H$ is residually finite too?

$\endgroup$
  • 2
    $\begingroup$ The first condition is redundant - all infinite, finitely generated groups are countable. $\endgroup$ – HJRW Mar 14 '12 at 21:43
  • 1
    $\begingroup$ ali, your title is very terse. It would be better to change it to something more descriptive, eg 'Is residual finiteness a profinite property?'. $\endgroup$ – HJRW Mar 15 '12 at 6:54
10
$\begingroup$

As Steve D observes, the answer is 'no'. This is a very general phenomenon. Let $G$ be any finitely generated residually finite group, and let $S$ be any finitely generated group with no finite quotients (Higman gave an example, or use an infinite simple group). Then $F(G)=F(G\times S)$, but $G\times S$ is certainly not residually finite.

$\endgroup$
8
$\begingroup$

$\mathbb{Z}\times\mathbb{Z}$ and Thompson's group $F$ have the same finite quotients, because $F'$ is the minimal normal subgroup of $F$.

$\endgroup$

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.