# Sylow Subgroups of Symmetric Groups

## Homework Statement

Find a set of generators for a p-Sylow subgroup K of Sp2
.
Find the order of K and determine whether it is normal in Sp2 and if it is abelian.

## The Attempt at a Solution

So far I have that the order of Sp2 is p2!. So p2 is the highest power of p that divides the order of the group. Thus the Sylow p-subgroup has order p2 and because of that, has to be abelian. The other parts I'm not so sure on.

matt grime
Homework Helper
Are you sure you're right so far? Take p=2, S_4 has order 24, so its Sylow 2 subgroup has order 8. You seem to have assumed that p^2!=p!^2.

Sorry, it's supposed to be restricted to just odd primes, in which case I think that still holds, correct?

matt grime
Homework Helper
If you believe your conjecture, then it is the matter of no time at all to check if it's true for p=3 (and it obviously isn't). If you can't prove something in general try an example or 2.

Okay, that's true. Sorry.
I guess I really just have no idea how to find generators for this.
I was trying with p=3 for my example and found that 3^4 is the highest power of 3 that divides 9! since there are 4 factors of 3 in 9*8*7*6*5*4*3*2*1. So the Sylow p-subgroup would have 81 elements. So in general for odd prime p the order of the Sylow subgroup would be p^(p+1)?

I'm working on generators right now...

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Here is my idea although it may be way off the mark.
Still looking at p=3, the subgroup could contain powers of 10 different 9-cycles that way you'd have 80 different elements plus the identity. I think that works.

Although I guess we can't be sure that it's closed.

Well, I'm basically out of ideas now...