residually finite groups with the same finite quotients - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:59:47Z http://mathoverflow.net/feeds/question/90885 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90885/residually-finite-groups-with-the-same-finite-quotients residually finite groups with the same finite quotients ali tavakoli 2012-03-11T07:32:02Z 2012-03-11T23:36:20Z <p>Let $G , H$ be two finitely generated residually finite groups such that $F(G)=F(H)$. Where $F(G)$ denotes the isomorphism classes of finite quotients of $G$. Can we say that $G\cong H$? </p> http://mathoverflow.net/questions/90885/residually-finite-groups-with-the-same-finite-quotients/90895#90895 Answer by HW for residually finite groups with the same finite quotients HW 2012-03-11T11:29:54Z 2012-03-11T11:29:54Z <p>In fact (see Agol's link), $F(G)=F(H)$ if and only if the profinite completions $\widehat{G}$ and $\widehat{H}$ are isomorphic. I believe that there are non-isomorphic finitely generated virtually abelian (in particular, residually finite) groups with isomorphic profinite completions---see Mark Sapir's answer in Agol's link for some references.</p> <p>Furthermore, <a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=AUCN&amp;pg6=PC&amp;pg7=ALLF&amp;pg8=ET&amp;r=1&amp;review_format=html&amp;s4=bridson&amp;s5=grunewald&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq" rel="nofollow">Bridson and Grunewald</a> answered a question of Grothendieck by constructing examples of pairs of finitely presented, residually finite groups $H\subseteq G$ such that $H$ is a proper subgroup of $G$ but the inclusion $H\to G$ induces an isomorphism of profinite completions $\widehat{H}\to\widehat{G}$. </p> <p>Very recently, Bridson and I have used these kinds of constructions to prove that the isomorphism problem for profinite completions of finitely presented, residually finite groups is undecidable.</p> http://mathoverflow.net/questions/90885/residually-finite-groups-with-the-same-finite-quotients/90937#90937 Answer by Igor Rivin for residually finite groups with the same finite quotients Igor Rivin 2012-03-11T21:45:33Z 2012-03-11T21:45:33Z <p>As originally shown by Steve Humphries (J. of Algebra, 1988, I believe) there exist free finitely generated subgroups of $SL(n, \mathbb{Z})$ which surject under every quotient modulo $m.$ In fact it is true (but not yet published) that a random two-generator subgroup of $SL(n, \mathbb{Z})$ has the Humphries properties with probability bounded away from zero, for $n > 2.$ This gives a (large) family of counterexample to the OP's question.</p> http://mathoverflow.net/questions/90885/residually-finite-groups-with-the-same-finite-quotients/90939#90939 Answer by Lior Bary-Soroker for residually finite groups with the same finite quotients Lior Bary-Soroker 2012-03-11T22:09:22Z 2012-03-11T22:09:22Z <p>If $G,H$ are arithmetic groups, then Aka studies when $F(G)=F(H)$, see <a href="http://arxiv.org/abs/1107.4147" rel="nofollow">http://arxiv.org/abs/1107.4147</a>.</p> http://mathoverflow.net/questions/90885/residually-finite-groups-with-the-same-finite-quotients/90949#90949 Answer by Mark Sapir for residually finite groups with the same finite quotients Mark Sapir 2012-03-11T23:36:20Z 2012-03-11T23:36:20Z <p>There are infinitely many metabelian groups with the same finite quotients, see Pickel, P. F. Metabelian groups with the same finite quotients. Bull. Austral. Math. Soc. 11 (1974), 115–120.</p> <p>On the other hand, for many relatively free groups, including free metabelian groups, the genus (i.e. the number of groups with the same finite quotients) is finite, see Gupta; Noskov, G. A. On the genus of certain metabelian groups. Algebra Colloq. 5 (1998), no. 1, 49–66. </p> <p>See also Grunewald, Fritz; Zalesskii, Pavel; Genus for groups. J. Algebra 326 (2011), 130–168. </p>