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 <bcolletti@mail.utexas.edu> 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