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The Borel subgroup $B$ of GL(2,p) is the subgroup group of upper triangular matrices. It is easy to see that it is (internal) semi-direct product of two subgroups:

$B=U\rtimes T$

where $U$ is the normal subgroup of $B$ consisting of unitriangular matrices, and $T$ is the subgroup of $B$ consisting of diagonal matrices.

In other words, $B\cong C_p\rtimes (C_{p-1}\times C_{p-1})$

How to get presentation of $B$ from this?

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up vote 7 down vote accepted

The group is generated by the three elements $$ \sigma=\begin{pmatrix}1 & 1\\\\0&1\end{pmatrix}, x=\begin{pmatrix}\alpha & 0\\\\0&1\end{pmatrix}, \text{ and }y=\begin{pmatrix}1 & 0\\\\0&\alpha\end{pmatrix}, $$ where $\alpha$ is a generator of $(\mathbb{Z}/p\mathbb{Z})^\times$. To get a presentation, you need to determine how these three elements conjugate with each other. Of course, as you have already said, $x$ and $y$ commute, so you just need to compute $x\sigma x^{-1}$ and $y\sigma y^{-1}$. E.g. $$ x\sigma x^{-1} = \begin{pmatrix}\alpha & 0\\\\0&1\end{pmatrix}\begin{pmatrix}1 & 1\\\\0&1\end{pmatrix}\begin{pmatrix}\alpha^{-1} & 0\\\\0&1\end{pmatrix}= \begin{pmatrix}1 & \alpha\\\\0&1\end{pmatrix}=\sigma^{a}, $$ for any lift $a$ of $\alpha\in\mathbb{Z}/p\mathbb{Z}$ to $\mathbb{Z}$. Similarly, you will find that $y\sigma y^{-1}=\sigma^{b}$ where $b$ is a lift of $\alpha^{-1}$ to $\mathbb{Z}$. This gives you the presentation $$ B=\langle x,y,\sigma|x^{p-1}=y^{p-1}=\sigma^p = 1,xy=yx,x\sigma x^{-1}=\sigma^a,y\sigma y^{-1}=\sigma^{b}\rangle $$ The last relation can also be replaced by $y^{-1}\sigma y = \sigma^a$, if you don't want to introduce $b$.

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