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Let $\mathbb{F}_p$ be the field with $p$ element. How many elements do the following subgroups of $\text{GL}_3(\mathbb{F}_p)$ have?

  1. $G_1=\left\{ \begin{pmatrix} x & a & b \\ 0 & d & c \\ 0 & 0 & x \end{pmatrix} \right\}$,
  2. $G_2=\left\{ \begin{pmatrix} x & a & b \\ 0 & x & 0 \\ 0 & d & c \end{pmatrix} \right\}$,
  3. $G_3=\left\{ \begin{pmatrix} a & 0 & b \\ d & x & c \\ 0 & 0 & x \end{pmatrix} \right\}$.

Here GL is the general linear group, the group of invertible n×n matrices.

Could someone provide me with a complete solution for one of the three subgroups so that I know how to redo it with the others?

Thank you in advance.

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Let's do (3).

The only constraint on the variables is that the determinant is nonzero. The determinant in this case is $ax^2$, so we must require $a,x\ne 0$ but $b,c,d$ can then be arbitrary. That means $(p-1)^2$ choices for $a,x$ and $p^3$ choices for $b,c,d$ so there are $p^3(p-1)^2$ elements in this subset.

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  • $\begingroup$ Thank you very much. As you have seen, it seems that the three subgroups have the same order $p^3(p-1)^2$. Am I right? $\endgroup$ – نورالدين سنانو Jan 26 at 12:58

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