I've been trying to solve the following problem

Let $n\geq 2$ and $T\leq GL_n(\mathbb{F}_P)$ be the subgroup of $GL_n(\mathbb{F}_P)$ having all upper triangular matrices with all the entries in the diagonal as 1, i.e.

$$ T = \left\{ \begin{bmatrix} 1 &a_{12} &a_{13} &\dots &a_{1n}\\ 0 &1 &a_{23} &\dots &a_{2n}\\ 0 &0 &1 &\dots &a_{3n}\\ \vdots &\vdots &\vdots &\ddots &\vdots\\ 0 &0 &0 &\dots &1\\ \end{bmatrix} : a_{ij}\in\mathbb{F}_p \right\} $$

Let $N(T)$ be the normalizer of $T$, then we have

$$ \left| N(T) \right| = (p-1)^np^{n(n-1)/2} $$

I got to the result

$$ \left| N(T) \right| = (p-1)^{(n-1)}p^{n(n-1)/2} $$

But I think maybe this is really the case, and the problem is wrong, am I right?

  • 3
    $\begingroup$ Are you counting in SL(n) or GL(n) ? $\endgroup$ – orangeskid Oct 16 '17 at 19:40
  • $\begingroup$ $GL(n)$, I already understood it, thanks! $\endgroup$ – The Giraffe Guy Oct 17 '17 at 0:52

As Orangeskid asks in the comments: where are you doing your counting?

The normalizer $N(T)$ is the subgroup of all upper triangular matrices: $$ \begin{pmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\0&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&a_{nn}\end{pmatrix}. $$ Since the diagonal entries of a matrix in $N(T)$ must be nonzero, you have $(p-1)^n$ choices for these entries, giving the result stated in the problem.

If you are counting in $SL(n)$, you have complete freedom to choose the first $n-1$ diagonal entries, but the determinant 1 condition forces the last entry to be $a_{nn}=\prod a_{ii}^{-1}$. This gives your count.

  • $\begingroup$ I see what I did wrong, thanks! I was counting in $SL(n)$. $\endgroup$ – The Giraffe Guy Oct 17 '17 at 0:53

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