Show that $GL_2(\mathbb{F}_p)$ has only one conjugacy class of order $p$.

Question

Let $p$ be a prime number, let $\mathbb{F}_p$ be the finite field with $p$ elements and let $G = GL_2(\mathbb{F}_p)$. Prove that $G$ has only one conjugacy class of order $p$. That means there can't be two or more conjugacy class with $p$ number of elements in each.

My effort : I was trying with possible Jordan form but can't find any useful thing. Any help/hint in this regards would be highly appreciated. Thanks in advance!

Answer 1

First consider the case $p>2$. (Note that in spite of the variable name, we do not assume in the following that $p$ is a prime, it can also be a prime power).

Assume $g=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in G$ is member of a conjugacy class of order $p$. Observe $$\tag1\begin{pmatrix}1&t\\0&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&-t\\0&1\end{pmatrix}=\begin{pmatrix}a+tc&b+td-ta-t^2c\\c&d-tc\end{pmatrix} $$ and $$\tag2\begin{pmatrix}1&0\\t&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\-t&1\end{pmatrix}=\begin{pmatrix}a-tb&b\\c+ta-td-t^2b&d+tb\end{pmatrix} $$ for $t\in\Bbb F_p$. Also, $$\tag3 \begin{pmatrix}1&0\\0&u\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\0&u^{-1}\end{pmatrix}=\begin{pmatrix}a&u^{-1}b\\uc&d\end{pmatrix}$$ for $u\in \Bbb F_p^\times$.

We have the following cases to consider:

• $c\ne 0$. Then the $p$ conjugates in $(1)$ are distinct, hence the complete list. For the elements of $(3)$ to appear among those of $(1)$, we need $c=uc$ for all $u\ne 0$. From $c\ne0$, we conclude $u=1$ for all $u\ne 0$, contradicting $p>2$.
• $b\ne 0$. Leads to the same contradiction by symmetry.
• $b=c=0$ and $a\ne d$. Again, the $p$ conjugates in $(1)$ are distinct hence are the complete list. But then for $(2)$ to occur among these, we need $c+ta-td=c$ for all $t$, contradiction.

We conclude $p=2$. Partitioning the six elements of $G$ into conjugacy classes is left as an exercise.