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Here's the question:

Find a Sylow p-subgroup of $GL_n(F_p)$, and determine the number of Sylow p-subgroups.

So far here's what I've got:

  • Order of $GL_n(F_p)$, which is $\prod_{j=1}^n p^n-p^{j-1}$ with j running from 1 to n,and thus the order of Sylow p-subgroup of it, and also its index.
  • From the index, $\prod_{j=1}^n (p^{n-j+1}-1)$, we have the clue that the number of Sylow p-subgroups, s, both is congruent to 1 mod p and divides $\prod_{j=1}^n (p^{n-j+1}-1)$.

But this doesn't seem to carry me any further... Please help :(

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Try to do this for $n=2$ first. Then for $n=3$. Then you'll see the pattern. (What is the order of the group, by the way?) –  Mariano Suárez-Alvarez Nov 12 '12 at 5:36
    
$\prod(P^n-P^(j-1))$, with j running from 1 to n –  Benjamin Lu Nov 12 '12 at 5:39
    
Edit the question and add that information there. (Also, notice that the numebr that you are claiming is the index is actually larger than the number you are claiming to be the order of the group!) –  Mariano Suárez-Alvarez Nov 12 '12 at 5:41
    
Oh write, sorry didn't notice that. Editing. –  Benjamin Lu Nov 12 '12 at 6:04

1 Answer 1

up vote 0 down vote accepted

the order of sylow psubgroup is $p^\frac{n(n-1)}{2}$ a sylow p subgroup would be the subgroup of upper triangular matrices with diagonal entries 1

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Literally just figured that out 4 seconds before reading this. Thanks anyway. –  Benjamin Lu Nov 12 '12 at 6:26
    
And the normalizer of that group is fairly easy to calculate. –  Geoff Robinson Nov 12 '12 at 6:34
    
@GeoffRobinson Could you elaborate? I was just thinking about that too, it's pretty much the only way I can think of to calculate the number of Sylow p-subgroups (which are conjugates) –  Benjamin Lu Nov 12 '12 at 6:46
    
Edit: I think I got it. Is it the group of upper triangular matrices (with any kind of diagonal instead of all 1's)? –  Benjamin Lu Nov 12 '12 at 6:54
    
@benjamin thats correct –  jim Nov 12 '12 at 7:08

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