Sylow P-Subgroups of Symmetric Groups

Date: 05/13/2009 at 17:58:55
From: Karen
Subject: Sylow p-subgroups of Symmetric Groups

Let p be an odd prime.  First, find a set of generators for a p-Sylow
subgroup K of S_p^2 (the symmetric group with degree p^2).  Then find
the order of K and determine whether it is normal in S_p^2 and if it
is Abelian.

The hardest part for me is finding the generators.  I think if I knew
that I might be able to get the rest of the question.  What I have
gotten so far is that the order of the whole group is p^2! so p^2 is
the highest power of p that divides the order of the group.  So the
Sylow p-subgroup has order p^2 and hence must be Abelian.

Date: 05/14/2009 at 06:58:15
From: Doctor Jacques
Subject: Re: Sylow p-subgroups of Symmetric Groups

Hi Karen,

I believe it is easier to compute the order of K first, and to find
generators after that.

The order of K is the highest power of p that divides (p^2)!, as you
write.  However, this is larger than p^2.  If we look at the expansion
of (p^2)!:

  (p^2)! = 1 * 2 *... * p * ... * 2p * ... * p^2

we see that there are p multiples of p (p, 2p, ..., p^2), and there is
an additional factor p in p^2.  This means that the product contains
(p+1) factors p, and the order of K is p^(p+1).  This is a particular
case of the technique described in:

  Finding Number of Trailing Zeros in Factorials
    http://mathforum.org/library/drmath/view/66749.html 

I know that the problem states that p is odd, but there is no harm
looking at the case p = 2.  In that case, K is isomorphic to D4 (the
dihedral group of 8 elements).  This group is not Abelian, and is not
normal in S4 (it has three conjugates).  As we may expect the case 
p = 2 to be the simplest, we may also expect that K will never be
Abelian or normal for p > 2.

Let us now try to find a generating set for K.  As Sylow subgroups are
conjugate (hence isomorphic) to each other, we only need to find a
subgroup of order p^(p+1) in S_(p^2).

Let us arrange the numbers 1 through p^2 in a square:

  1       2    .....    p
  p+1     p+2  .....    2p
  ........................
  p(p-1)................p^2

Let a_i be the permutation that permutes row 'i' cyclically:

  a_1 = (1, 2, ..., p)
  a_2 = (p, p+1, ..., 2p)

and so on.  Each of those permutations generates a cyclic subgroup of
order p.  These cyclic subgroups intersect pairwise trivially, and the
generators commute with each other.  This means that the subgroup L
generated by the a_i is isomorphic to the direct product of p copies
of C_p:

  L = <a_1, ..., a_p> = C_p X ... X C_p

and L is an Abelian subgroup of order p^p.

Now, consider the element h that permutes cyclically and 
simultaneously all the columns of the square:

  h = (1, p+1, ..., p(p-1)) (2, p+2, ...) ...

The subgroup H = <h> is cyclic of order p.  Can you try to prove that
the subgroup T generated by L and H has order p^(p+1)?  Try to prove
first that L is normal in T, by looking at the action of h by
conjugation (incidentally, this will also show that T is not Abelian).

This is a particular case of a construction called "wreath product".

To prove that T is not normal use the known fact that A_n is simple
for n > 4.

Please feel free to write back if you require further assistance.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/