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