Generators of Sylow p-Subgroup of Symmetric Groups (4.5.46)

Dummit and Foote Abstract Algebra, section 4.5, exercise 46:

Find generators for the Sylow p-subgroup of S_p², where p is a prime. Show that this is a non-abelian group of order p^p+1.

Proof: First notice that the order of this symmetric group contains the following factors in its order: p, 2p, ..., p², which are all the factors not coprime to p. Thus, the Sylow p-subgroup is of order p^p+1.

Consider the following series of p-cycles: ( 1 ... p ), ( p+1 ... 2p ), ..., ( p(p-1)+1 ... p² ). These all have order p and commute with each other, so we have that the group H they generate is of order p^p. Proceed by imagining a normalizer K of order p possessing a trivial intersection, so that HK = < H, K > is a group and of order p^p+1. As one can see, the group generated by ( 1 p+1 ... p(p-1)+1 )( 2 p+2 ... p(p-1)+2 )...( p 2p ... p² ) fits these prerequisites, as conjugating H with such a permutation will map the individual p-cycles onto each other, it is clearly a group of order p, and every power of it (i.e. every element in the generated group) moves indices in orbits impossible for any element in H so that H∩K = 1. Therefore, < ( 1 ... p ), ( p+1 ... 2p ), ..., ( p(p-1)+1 ... p² ), ( 1 p+1 ... p(p-1)+1 )( 2 p+2 ... p(p-1)+2 )...( p 2p ... p² ) > ∈ Syl_p(S_p²), and this is clearly a non-abelian group as the last generator doesn't centralize any of the others.

∎

One might compare this method to a slightly more abstracted approach outlined at this wonderful blog. In this case, the largest permutation generator he worked with is actually the product of ( 1 ... p ) and my own, so one easily reconciles both proofs' results.