Any element $g$ of $GL(2,p)$ of order $p$, $p$ prime, is conjugate to $\begin{bmatrix}1&1\\0&1\end{bmatrix}$

Question

Any element $g$ of $GL(2,p)$ of order $p$, $p$ prime, is conjugate to $\begin{bmatrix}1&1\\0&1\end{bmatrix}.$

I showed that $\langle g\rangle $ acts on the set $X$ of vectors with entries in $ F_p$ and hence that $g$ fixes some non-zero element of $X$ (By Orbit-Stabiliser, since $|X| = p^2$ and $|\langle g\rangle|=p$). In an exercise, I am then asked to deduce from this the statement above, which I am stuck on.

Answer 1

It sounds like you want a mixture of group actions and linear algebra.

---

Let $g\in G=GL_2(\Bbb{F}_p)$ be an element of order $p$. Let's denote $H=\langle g\rangle$, and $X=\Bbb{F}_p^2$ the set of (column) all vectors on which both $H$ and $G$ act by matrix multiplication from the left.

1. By the orbit-stabilizer theorem the orbits of $H$ have sizes $1$ and $p$ only.
2. Because the zero vector forms an $H$-orbit of size $1$ and $|X|=p^2$, there must be other $H$-orbits of size $1$. Let $x\in X$ form such a singleton orbit of $H$.
3. The scalar multiples of $x$ are then all eigenvectors of $g$ belonging to eigenvalue $\lambda=1$. Let $y\in X$ be a vector that is linearly independent from $x$. Then $y$ cannot belong to the eigenvalue $1$ for then we would have $g=1_G$.
4. The vectors $x$ and $y$ form a basis $\mathcal{B}$ of $X$ over $\Bbb{F}_p$, so we have $g\cdot x=x$ and $g\cdot y=ax+by$ for some $a,b\in\Bbb{F}_p$. The matrix of $g$ with respect to $\mathcal{B}$ looks like $$ M_{\mathcal{B}}(g)=\left(\begin{array}{cc} 1&a\\ 0&b \end{array}\right). $$
5. We have $\det(g)=b$, so $1=\det(g^p)=b^p=b$.
6. Because $g\neq 1_G$ we have $a\neq0$.
7. With respect to the basis $\mathcal{B}'=\{ax,y\}$ the matrix of $g$ thus looks like $$ M_{\mathcal{B}'}(g)=\left(\begin{array}{cc} 1&1\\ 0&1 \end{array}\right). $$

Answer 2

Here is a solution using linear algebra over $\mathbb F_p$.

$g$ satisfies $0=g^p-I=(g-I)^p$. Therefore, $1$ is the only eigenvalue of $g$. Since $g$ is a $2\times 2$ matrix, these are the only possible Jordan forms for $g$: $$ \begin{bmatrix}1&0\\0&1\end{bmatrix} \quad\text{and}\quad \begin{bmatrix}1&1\\0&1\end{bmatrix} $$ But $I$ does not have order $p$.

Answer 3

You have shown that all elements of order $p$ are in the same conjugacy class. But $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ has order $p$.

Answer 4

This gives a solution by group action theory, as you wish, but I couldn't avoid using the result (*) hereunder, which is seemingly as much non-elementary as the tools in the other answers/comments (for its proof, see in this site e.g. here).

---

Let's define $X:=\{M\in \operatorname{GL}_2(\mathbb{F}_p)\mid M^p=I\}$

Lemma. $G\in \operatorname{GL}_2(\mathbb{F}_p)\wedge M\in X \Longrightarrow GMG^{-1}\in X$.

Proof. $(GMG^{-1})^p=GM^pG^{-1}=GIG^{-1}=GG^{-1}=I \Longrightarrow GMG^{-1}\in X$.

$\Box$

Therefore, $\operatorname{GL}_2(\mathbb{F}_p)$ acts by conjugation on $X$. Note that $\tilde M:=\begin{bmatrix}1&1\\0&1\end{bmatrix}\in X$, because $\tilde M^p=I$, and thence we can apply the Orbit-Stabilizer Theorem to $\tilde M$:

$$|O(\tilde M)||\operatorname{Stab}(\tilde M)|=|\operatorname{GL}_2(\mathbb{F}_p)|=(p^2-1)(p^2-p)=p(p+1)(p-1)^2 \tag 1$$

Now, $\operatorname{Stab}(\tilde M)=\{G\in \operatorname{GL}_2(\mathbb{F}_p)\mid G\tilde M=\tilde MG\} = \Biggl\{G\in \operatorname{GL}_2(\mathbb{F}_p)\mid G=\begin{bmatrix}a&b\\0&a\end{bmatrix}, a,b\in \mathbb{F}_p\Biggr\}$, whence $|\operatorname{Stab}(\tilde M)|=p(p-1)$ and finally, by $(1)$, $|O(\tilde M)|=(p+1)(p-1)=p^2-1$. Now, $|X|=p^2-1$ (*) and $O(\tilde M) \subseteq X$, whence $O(\tilde M)=X$, and the action is transitive.