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 <mareg@primrose.csv.warwick.ac.uk> 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"