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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?

edited Mar 17 2012 at 15:14

asked Mar 14 2012 at 19:37

ali tavakoli
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The first condition is redundant - all infinite, finitely generated groups are countable. – HW Mar 14 2012 at 21:43

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ali, your title is very terse. It would be better to change it to something more descriptive, eg 'Is residual finiteness a profinite property?'. – HW Mar 15 2012 at 6:54

HW · Answer 1 · 2012-03-14 21:45:56Z UTC

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.

Steve D · Answer 2 · 2012-03-14 19:50:38Z UTC

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

answered Mar 14 2012 at 19:50

Steve D
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Is residual finiteness a profinite property?

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