The strictly upper triangular matrices form a Sylow subgroup in the general linear group over ZZ/(p)

Show that the subgroup \overline{UT}_n(F_p) of strictly upper triangular matrices (i.e. with only 1 in diagonal entries) in GL_n(F_p) is a Sylow p-subgroup.

We know that the order of GL_n(F_p) is \prod_{i=0}^{n-1} (p^n - p^i) = \prod_{i=0}^{n-1} p^i(p^{n-i} - 1) = (\prod_{i=0}^{n-1} p^i)(\prod_{i=0}^{n-1}(p^{n-i}-1)). Note that p does not divide p^{n-i}-1 for any i. Moreover, we have \prod_{i=0}^{n-1} p^i = p^{\sum_{i=0}^{n-1} i} = p^{n(n-1)/2}. Thus (by definition) a Sylow p-subgroup of GL_n(F_p) has p^{n(n-1)/2} elements.

Recall that a matrix is strictly upper triangular if all diagonal entries are 1, all lower entries are 0, and all upper entries can be any field element. Then the number of elements in \overline{UT}_n(F_p) is p^M, where M is the number of entries above the main diagonal. We can see that this number is M = \sum_{i=1}^n n-i = \sum_{i=0}^{n-1} i = n(n-1)/2. Thus \overline{UT}_n(F_p) is a Sylow p-subgroup in GL_n(F_p).

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  • eric  On October 1, 2011 at 6:20 pm

    On the second line should the p^{n-1} be p^{n-i} or have i missed something?

    • nbloomf  On October 1, 2011 at 6:59 pm

      That was a typo. Thanks!


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