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Let $p$ a prime and $\mathbb{F}_p$ the finite field of $p$ elements. For $n \geq 2$ and $G=GL_n(\mathbb{F}_p)$, let $T\subseteq_{sg}G$ the subgroup of upper triangular matrices with all entries in principal diagonal equal to $1$. Show that $|N_{G}(T)|=(p-1)^{n}p^{n(n-1)/2}$.

Using a combinatorial argument I already showed that $|T|=p^{n(n-1)/2}$, but I do not know how to find the order of the normalizer and why it is the order of $T$ multiplied by $(p-1)^n$...

Any help would be appreciated.

Thanks.

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    $\begingroup$ All the diagonal non-singular matrices are in this normaliser. $\endgroup$ – Lord Shark the Unknown Sep 22 '18 at 18:36
  • $\begingroup$ And every matrix in the normaliser is a diagonal non-singular? I can see this is true checking for some exemples, but how can I show that? $\endgroup$ – creepyrodent Sep 22 '18 at 18:58
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    $\begingroup$ No it's not true that every matrix in the normalizer is diagonal non-singular. The elements of $T$ are not diagonal. The elements in the normalizer consist of the non-singular upper triangular matrices. $\endgroup$ – Derek Holt Sep 22 '18 at 19:46
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The key to understanding triangular matrices conceptually is relating them to flags. Specifically, let $V_k\subseteq \mathbb{F}_p^n$ be the subspace spanned by the first $k$ standard basis vectors. Note that an invertible matrix $A$ is upper triangular iff $A(V_k)=V_k$ for all $k$. More specifically, a matrix $A$ is in $T$ iff $A-I$ is nilpotent and $A(V_k)=V_k$ for all $k$. Conversely, we can define the $V_k$ from $T$: $V_k$ is the set of vectors that are annihilated by $(A-I)^k$ for all $A\in T$.

This makes it easy to understand conjugates of $T$. In particular, since we can define the $V_k$ in terms of $T$ and vice versa, for any $B\in G$ we have $BTB^{-1}=T$ iff $B(V_k)=V_k$ for all $k$. Explicitly, $BTB^{-1}$ is exactly the set of $A$ such that $A-I$ is nilpotent and $A(B(V_k))= B(V_k)$ for all $k$, and conversely $B(V_k)$ is the set of vectors annihilated by $(A-I)^k$ for all $A\in BTB^{-1}$. More generally, the conjugates of $T$ are in bijection with sequences of subspaces $W_0\subset W_1\subset\dots\subset W_n$ with $\dim W_k=k$ for each $k$, with such a sequence corresponding to $BTB^{-1}$ for any $B$ such that $B(V_k)=W_k$ for each $k$. (Such sequences are called complete flags in $\mathbb{F}_p^n$.)

So, $N_G(T)$ is exactly the set of $B\in G$ such that $B(V_k)=V_k$ for all $k$. That is, it is exactly the set of invertible upper triangular matrices. I will leave it to you to finish the problem by counting how many such matrices there are.

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