From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Subgroup Construction Date: 26 Mar 1998 19:54:44 GMT In article <6f8r47\$de4\$1@geraldo.cc.utexas.edu>, Bruce W. Colletti wrote: >Given H a proper subgroup of a finite group G. > >Let H intersect H^x = {identity}, for all x not in H and where ^ is >conjugation (a^x = inverse(x)*a*x). > >What name is given to this type of subgroup and is there a way to >construct it? The set K of elements not in any conjugate of H, together with the identity, is a subgroup of G, necessarily normal. Then G is a Frobenius group, K the Frobenius kernel. An example is given by the symmetries of a triangle or pentagon: take H to be a single reflection; K will be the group of rotations. Try computing some more examples; do you notice that the Frobenius kernels seem to be nilpotent? Try to prove that. :-) dave ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Subgroup Construction Date: 29 Mar 1998 07:30:04 GMT _Frobenius groups_ are those with a subgroup H having >>> Let H intersect H^x = {identity}, for all x not in H and where ^ is >>> conjugation (a^x = inverse(x)*a*x). I had written >> The set K of elements not in any conjugate of H, together with the >> identity, is a subgroup of G, necessarily normal. which is a nontrivial result in finite group theory, requiring character theory (at least, I don't think there is an "elementary" proof, which is either frustrating or delightful, depending on your perspective). I further teased >> Try computing some more examples; do you notice that the Frobenius kernels >> seem to be nilpotent? Try to prove that. which was pretty unfair of me. Nilpotence follows from solvability fairly easily, but the proof of solvability required a young John Thompson (later a Fields medalist). "Nick Halloway" wrote: >If |H| is even then it seems K must be abelian; take an element h in H of >order 2, then K is a Frobenius group; for k in K h*k has order 2 so >h*k*h^(-1) = k^(-1), so K is abelian. Right. The usual way of thinking about K is to observe it's a group with a collection of fixed-point-free automorphisms on it, forming the group H. There are oodles of results describing groups which admit f-p-f automorphisms of various sorts. >What about if |G| is infinite? Is K a group then? Given my previous comments you wouldn't expect there to be a proof that K must be a subgroup of G, but I didn't really know anything more. I did find these results in Math Reviews, which suggest that the answer is "no". (Just offhand, I can't think of any group-theory results with these properties 1. The result can be stated identically for finite and infinite groups (i.e., statement does not refer to order or imply finitude) 2. The result is proven for finite groups 3. No counterexample is known for infinite groups, nor a proof of the result valid for infinite groups. That is, if you can't prove something without referring to order, there's probably a good reason for that...) dave ============================================================================== © Copyright American Mathematical Society 1998 91i:20003 20B07 Collins, Michael J.(4-OXUC) Some infinite Frobenius groups. (English) J. Algebra 131 (1990), no. 1, 161--165. _________________________________________________________________ Generalizations are sought of Frobenius' theorem on finite permutation groups to the infinite case. Since the usual proof uses character theory it is evident that either some radically different methods are required, or else sufficiently strong conditions must be imposed to enable the finite argument to be carried over. O. H. Kegel and B. A. F. Wehrfritz [Locally finite groups, North-Holland, Amsterdam, 1973; MR 57 #9848] showed that local finiteness is a sufficient hypothesis and gave an example of a strong failure of the theorem in the general case. Let us say that a Frobenius group is a transitive permutation group \$G\$ in which some nonidentity element fixes a point, but none fixes two or more. A regular normal subgroup of \$G\$ is called a Frobenius kernel. As a weakening of the local finiteness condition it is shown that there is a Frobenius kernel in a Frobenius group \$G\$ acting on an infinite set \$\Omega\$ if either (i) any two elements of \$G\$ generate a finite subgroup, or (ii) any two fixed-point-free elements generate a finite subgroup, there is at least one such, and \$G\$ is doubly transitive on \$\Omega\$. In addition if (ii) holds, the Frobenius kernel is abelian; and if any three elements of \$G\$ generate a finite subgroup then the Frobenius kernel satisfies an Engel condition. Reviewed by John K. Truss 81h:20024 20D15 20D45 Huhro, E. I. A solvable group admitting a regular splitting automorphism of prime order is nilpotent. (Russian) Algebra i Logika 17 (1978), no. 5, 611--618, 623. _________________________________________________________________ Ju. M. Gorchakov (same journal 4 (1965), no. 1, 15 - 29; MR 33 #1369) termed an automorphism phi of order n of the group G splittable if xx {sup} phi x{sup}(phi {sup}2) ...x{sup}(phi {sup}n{sup} - {sup}1)=1 for any element x in G. An automorphism phi is said to be regular if x{sup} phi {neq}x for any nonidentity element x. The author gives an affirmative answer to Gorchakov's question (Kourovka notebook: unsolved problems in the theory of groups (Russian), fifth edition, Section 1.10, Akad. Nauk SSSR Sibirsk. Otdel. Inst. Mat., Novosibirsk, 1976; see MR 56 #5692): Is every solvable group admitting a regular splitting automorphism of prime order nilpotent? At the same time he proves the nilpotency of the normal kernel of certain infinite Frobenius groups. (English translation: Algebra and Logic 17 (1978), no. 5, 402 - 406 (1979).) Reviewed by S. S. Levishchenko