Normalizer of Unipotent subgroup in General Linear group

Let $\mathrm{GL}(n,\mathbb{F}_p)$ be the general linear group over field of order $p$, and $\mathrm{U}(n,\mathbb{F}_p)$ be the subgroup consisting of upper triangular matrices with each diagonal entry $1$. Then

the normalizer of $\mathrm{U}(n,\mathbb{F}_p)$ in $\mathrm{GL}(n,\mathbb{F}_p)$ is the subgroup consisting of upper triangular matrices.

Is there any elementary way to prove this?

(The way I know is by using Bruhat decomposition. But, this decomposition may not be known to a beginner in group theory, it will be difficult to explain the proof by Bruhar decomposition.)

I don't like your notation $U(n,{\mathbb F}_p)$, because that is also used for the unitary group.

Consider the module $V$ defined by the action of $U(n,p)$ on its underlying vector space. The fixed point submodule $V_1$ clearly has dimension $1$. The induced action of $U(n,p)$ on $V/V_1$ is isomorphic as a module to the natural module for $U(n-1,p)$, so $V/V_1$ has $1$-dimensional fixed point submodule $V_2/V_1$. etc.

(In fact $V$ is uniserial with submodules $0 < V_1 < V_2 < \cdots < V_n=V$.)

Now it is routine to show that the normalizer of $U(n,p)$ in ${\rm GL}(n,p)$ fixes each $V_i$ and the subgroup that fixes each $B_i$ consists preciselt of the upper triangular matrices.

• This notation is from Alperin-Bell. – Groups Sep 2 '15 at 8:35