# Order of normalizer of upper triangular matrices subgroup in $GL_n(\mathbb{F}_p)$

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.

• All the diagonal non-singular matrices are in this normaliser. – Lord Shark the Unknown Sep 22 '18 at 18:36
• 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? – creepyrodent Sep 22 '18 at 18:58
• 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. – Derek Holt Sep 22 '18 at 19:46

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.