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 :(

  • $\begingroup$ 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?) $\endgroup$ Nov 12, 2012 at 5:36
  • $\begingroup$ $\prod(P^n-P^(j-1))$, with j running from 1 to n $\endgroup$ Nov 12, 2012 at 5:39
  • $\begingroup$ 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!) $\endgroup$ Nov 12, 2012 at 5:41
  • $\begingroup$ Oh write, sorry didn't notice that. Editing. $\endgroup$ Nov 12, 2012 at 6:04

1 Answer 1


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

  • $\begingroup$ Literally just figured that out 4 seconds before reading this. Thanks anyway. $\endgroup$ Nov 12, 2012 at 6:26
  • $\begingroup$ And the normalizer of that group is fairly easy to calculate. $\endgroup$ Nov 12, 2012 at 6:34
  • $\begingroup$ @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) $\endgroup$ Nov 12, 2012 at 6:46
  • 1
    $\begingroup$ Edit: I think I got it. Is it the group of upper triangular matrices (with any kind of diagonal instead of all 1's)? $\endgroup$ Nov 12, 2012 at 6:54
  • $\begingroup$ @benjamin thats correct $\endgroup$
    – jim
    Nov 12, 2012 at 7:08

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