Sylow P-Subgroups of Symmetric GroupsDate: 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/ |
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