Five responses to Mike Cole's group theory question, followed by one
response to Turygin's question about group actions.......DMD
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Subject: Re: three postings
From: Bob OLIVER <bobol@math.univ-paris13.fr>
Date: Tue, 29 Mar 2005 15:12:49 +0200 (CEST)

A quick answer to Mike Cole.  Fix $n$, and look at the pair $S_{n-1}<S_n$ 
(symmetric groups).  Then any group which acts freely and transitively on 
{1,...,n} is a complement; there is thus a complement isomorphic to any 
group of order $n$.

Bob Oliver

>> Subject: group theory question
>> From: "Michael Cole" <Michael.Z.Cole@hofstra.edu>
>> Date: Mon, 28 Mar 2005 23:44:41 -0500
>> 
>> This is really a pure algebra question rather than a topology question.
>> Recall that two subgroups H,K of G are complements if HK=G and they
>> intersect in the trivial subgroup.  In general, neither of H,K need be
>> normal; example: the Sylow 2 and 3 subgroups of the symmetric group S4.
>> Question:  If H has two complements K1 and K2 in the finite group G must
>> they be isomorphic?  That will clearly be the case if H is normal and we
>> have the semidirect product situation, but I don't see why that would
>> have to be true in general.  If it is not true, does anybody know the
>> simplest counter example and/or smallest group G for which a
>> counterexample occurs.
>> 
>> Happy Easter to all,
>> Mike Cole
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Subject: response to Michael Cole's question
From: Tyler Lawson <tlawson@math.mit.edu>
Date: Tue, 29 Mar 2005 10:30:28 -0500 (EST)

The smallest group for which a subgroup H can have two nonisomorphic 
complements is when G has size 8.

Take G = D_8, the dihedral group of symmetries of a square.  Let H be 
the subgroup generated by reflection along some diagonal.  Then H has 
two nonisomorphic complements: the subgroup K1 of rotations (which is cyclic), 
and the subgroup K2 generated by non-diagonal reflections (which is the 
Klein four-element group).
-- 
Tyler.
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Subject: Re: group theory question
From: Robert Bruner <rrb@math.wayne.edu>
Date: Tue, 29 Mar 2005 12:12:57 -0500 (EST)

Here's the answer to Mike's question:

The smallest counterexample G would be the dihedral group of order 8.
Let H be a subgroup generated by a reflection, so H has order 2.
Then the subgroup of rotations is a cyclic complement, and there
is another complement that is noncyclic.  Both complements are
normal.
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Subject: Re: three postings
From: Ian Leary <leary@math.ohio-state.edu>
Date: Tue, 29 Mar 2005 16:08:35 -0500 (EST)

Let G be the dihedral group of order 8, let H be the subgroup of
order two generated by a reflection.  Let K_1 be the cyclic
subgroup of order 4 and let K_2 be the non-cyclic subgroup
consisting of the central element and the two reflections not
conjugate to the one in H.

If you think of G as the symmetries of a square, H is generated
by (say) a reflection through 2 vertices, K_1 consists of
rotations, and K_2 consists of rotation through pi together
with reflections in lines parallel to the sides of the square.

      Best wishes,
       Ian Leary
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Subject: Re: three postings
From: Jeff Strom <jeff.strom@wmich.edu>
Date: Tue, 29 Mar 2005 17:21:17 -0500

I posed the group theory question to David Richter, an
algebraist/geometer at WMU, and he provided this example.

Let  G = Z/2 x Sym(4),  (symmetric group on 4 letters)
and denote elements of the group as
+\sigma  or  -\sigma  (i.e.  Z/2 =  { +1, -1  } ).

Then let   H  be the subgroup generated by  -(12)  and  (123).
Then the subgroups

   K,  generated by   (12)(34)  and  (1234)

and

    L,  generated by  +(1234), -(1234)

are both complements for   H   in   G.  But

   K = D_4  (dihedral group)    and    L =  Z/2 +  Z/4.

Jeff