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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.