From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Frobenius Complement Date: 21 Apr 1998 20:36:19 GMT In article <6h6ukq\$opa\$1@geraldo.cc.utexas.edu>, Bruce W. Colletti wrote: >Does S(n), the symmetric group on n letters, have any Frobenius >Complements? If so, what's an example? Do you mean you would like to find a proper H of G=S(n) with H \intersect H^g = {1} for all g in G \ H? You asked about these before and I mentioned >The set K of elements not in any conjugate of H, together with the >identity, is a subgroup of G, necessarily normal. Note that |K| = |G| - (|H|-1)[G:H] means |G|=|K|*|H|. For n>4 this rules out the existence of such an H, since then only normal subgroups of S(n) are then {1}, A(n), and S(n). The cases K={1}, H=S(n) and K=S(n), H={1} are usually excluded by definition, so you must have K=A(n), |H|=2. But it's easy to check that an involution in S(n) always has a centralizer of order larger than 2 (if n>3). For n=1 and n=2 there are no candidate subgroups H. For n=3, S(n) = Dihedral(6) does have a Frobenius complement, H=<(12)>. For n=4, we can rule out most candidates for K as above, but there is an additional normal subgroup of S(4): K = <(12)(34), (13)(24)> of order 4. But this requires |H|=6, which would make H conjugate to S(3) in S(4). But e.g. (12) lies in both S(3) and its conjugate under g=(34), so S(3) isn't a Frobenius complement either. So it looks like the answer to your question is, "yes, iff n=3". dave