From: Fred W. Helenius Subject: Re: about finite generated group Date: Tue, 17 Oct 2000 12:36:33 -0400 Newsgroups: sci.math Summary: [missing] "Shih-Wei Yang" wrote: >can someone given an example that: >Let G be a finite generated group contains a subgroup H >which is not finitely generated. Let f:Q -> Q and g:Q -> Q be functions on the rationals defined by f(x) = 2x and g(x) = x + 1. The group generated by f and g contains as a subgroup all functions h of the form h(x) = x + m/2^n, where m and n are integers. This subgroup (which is isomorphic to the additive group of dyadic rationals) is not finitely generated. -- Fred W. Helenius ============================================================================== From: Fred Galvin Subject: Re: about finite generated group Date: Tue, 17 Oct 2000 16:51:57 -0500 Newsgroups: sci.math On Wed, 18 Oct 2000, Shih-Wei Yang wrote: > can someone given an example that: Let G be a finite generated > group contains a subgroup H which is not finitely generated. Let H be the group consisting of the permutations of Z with finite support. H is not finitely generated (every finitely generated subgroup of H is finite), but H is a subgroup of the group G generated by the two permutations (1,2) and (...,-3,-2,-1,0,1,2,3,...). More generally, if H is *any* countable group, then H can be embedded in a group G which is generated by two of its elements. In fact, H can be embedded in a group F such that *every* countable subgroup of F is contained in a 2-generator subgroup of F. We can take F to be the (full) symmetric group on some set. -- It takes steel balls to play pinball. ============================================================================== From: Fred Galvin Subject: Re: about finite generated group Date: Tue, 17 Oct 2000 17:03:09 -0500 Newsgroups: sci.math On Tue, 17 Oct 2000, Fred Galvin wrote: > On Wed, 18 Oct 2000, Shih-Wei Yang wrote: > > > can someone given an example that: Let G be a finite generated > > group contains a subgroup H which is not finitely generated. > > Let H be the group consisting of the permutations of Z with finite > support. H is not finitely generated (every finitely generated > subgroup of H is finite), but H is a subgroup of the group G > generated by the two permutations (1,2) and > (...,-3,-2,-1,0,1,2,3,...). > > More generally, if H is *any* countable group, then H can be > embedded in a group G which is generated by two of its elements. So far, so good. > In fact, H can be embedded in a group F such that *every* > countable subgroup of F is contained in a 2-generator subgroup of > F. Pointless as stated. What I *meant* to say is that an *arbitrary* (not necessarily countable) group can be embedded in a group F with the stated property. (This is an old observation of B. H. Neumann.) > We can take F to be the (full) symmetric group on some set. -- It takes steel balls to play pinball.