Sylow Subgroups of Symmetric Groups

  • Thread starter Obraz35
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  • #1
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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.

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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.
 

Answers and Replies

  • #2
matt grime
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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.
 
  • #3
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Sorry, it's supposed to be restricted to just odd primes, in which case I think that still holds, correct?
 
  • #4
matt grime
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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.
 
  • #5
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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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  • #6
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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...
 

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