From: rld@math.ohio-state.edu (Randall Dougherty) Subject: Re: About free groups of finite rank: a contradiction. Date: 15 Sep 1999 20:44:33 GMT Newsgroups: sci.math Keywords: subgroups always have a conjugate with trivial intersection In article <7roduc\$bqa\$1@wisteria.csv.warwick.ac.uk>, Dr D F Holt wrote: >Let H be a subgroup of F (free of finite rank) having finite rank and >infnite index. Does there always exist a g in F with > g H g^-1 intersect H = 1 (trivial subgroup)? Yes. Karrass and Solitar (Proc. Amer. Math. Soc. 22 (1969), 209-213) showed that, given F and H as above, there must be a nontrivial normal subgroup K of F having trivial intersection with H. Let g be an element of K other than the identity. Suppose g H g^-1 intersects H nontrivially; so there is a non-identity h in H such that g h g^-1 is an element h' of H. This gives h^-1 g h g^-1 = h^-1 h'. But (h^-1 g h) g^-1 is in K and h^-1 h' is in H, so we must have h^-1 g h g^-1 = h^-1 h' = 1. Hence, g commutes with h, so (since we are in a free group) g and h have a common power; this common power is a nontrivial common element of K and H, contradiction. Randall Dougherty rld@math.ohio-state.edu Department of Mathematics, Ohio State University, Columbus, OH 43210 USA "I have yet to see any problem, however complicated, that when looked at in the right way didn't become still more complicated." Poul Anderson, "Call Me Joe"