This question doesn't seem that hard but it has stumped me for some time.

First, note that the Fundemental Theorem of Permutation Groups (or
Fundemental Theorem of Group Actions, if you prefer) states:

Let the group G act on the set S, then for all s in S, |Orb(s)| =
[G:Stab(s)] and if G is finite then by Lagrange,
|Orb(s)|=|G|/|Stab(s)|

Notation: Orb(s) is the G-orbit of the element s, Stab(s) is the
stablizer of s in G, |X| is the cardinality of the set X, [G:H] is the
index of the group H within G, <= means "less than or equal to"

Now the question becomes, given a finite group G of order n and d_1,
d_2, ..., d_m a list of all the distinct divisors of n, can we find a
collection of sets {S_i} for 1 <= i <= m such that for all s in S_i
where |Stab(s_i)| = d_i the equation |Orb(s)| = |G|/|Stab(s)| is
satisfied?

In other words, can we associate with each divisor of n a set S such
that for all s in S, |Orb(s)| = [G : Stab(s)]?

Now, I suspect this to be true but I haven't been able to prove it so
far and it finally started to annoy me.  It "may" be that it is false
but I highly doubt it.  In any case all I am asking for is a proof or
a disproof.

Please note: I need 100% complete detail with the answer, please don't
bandy about with the words clearly, trivial, obviously, etc. I don't
mean that I need it to be answered from first principals, certainly
not.  I just need sufficient detail with the answer.  And if further
clarification is needed I'll be happy to provide it.

---

Clarification of Question by sartoris-ga on 11 Oct 2005 05:49 PDT

Ah, let me try to clear this up a little.  You are certainly correct,
G does act upon itself by conjugation and satisfies
|Orb(s)|=|G|/|Stab(s)| for all s in G.  However, consider the
collection of all elements in G denoted by {g_i} for 1 <= i <= n. 
Now, we have the associated set T := {|Stab(g_1)|, |Stab(g_2)|, ...,
|Stab(g_n)|}.  What we would ask then is, can I find for every d_i, a
t in T such that t = d_i?

The problem is not simply finding some S such that for all s in S
|Orb(s)|=|G|/|Stab(s)|, the problem is finding an S (and a G-action)
such that the cardinalities of all the Stablizers of the elements in S
runs over all divisors of |G|.

Now, I do believe that by letting G act on itself by conjugation you
can show that this is true, however that is what I'm having trouble
doing.

I hope this helps.

Thanks

I'm having some trouble understanding what your problem is.

You ask: "Given a finite group G of order n and d_1,
d_2, ..., d_m a list of all the distinct divisors of n, can we find a
collection of sets {S_i} for 1 <= i <= m such that for all s in S_i
where |Stab(s_i)| = d_i the equation |Orb(s)| = |G|/|Stab(s)| is
satisfied?"

But you've already quoted a theorem that says that the part after
"such that" is always true for any group action. So all you need to
find is a set for G to act on, and it can always act on itself by
conjugation, so what's the problem?

I'm not sure, but I believe the problem can be restated in this way:

Given a finite group G, does there exist a group action of G on set S
such that for every divisor d of |G|, there exists s in S such that
|Stab(s)| = d ?

Regardless of what set S is chosen for G to act upon, we would be
requiring that G have a subgroup, Stab(s), of any order dividing the
order of G, i.e. a converse to Lagrange.  This is generally false,
though it can be proven for finite abelian groups and some other
classes of interest.

[Lagrange's Theorem -- Wikipedia]
http://en.wikipedia.org/wiki/Lagrange's_theorem

"The smallest example is the alternating group G = A_4 which has 12
elements but no subgroup of order 6."

So in particular for G = A_4 we cannot find a group action of G on any
set S such that the stabilizer of some element of S has order 6.

Hope this sheds some light...

regards, mathtalk-ga

Let's try that link again, this time with a bit of HTML magic:

[Lagrange's Theorem -- Wikipedia]
http://en.wikipedia.org/wiki/Lagrange's_theorem

<crossing fingers>

Nope, sorry.  Can't think of a workaround... I'll ask the Editors to take a look.

[Lagrange's Theorem -- Wikipedia]
en.wikipedia.org/wiki/Lagrange's_theorem