I am working on this problem and not sure about my solution so I would like some help:
Find all matrices in the group $\operatorname{GL}_2(\Bbb{F}_5)$ that are in the stabilizer of $\begin{pmatrix}
0\\
1
\end{pmatrix}$ under left multiplication (The group $\operatorname{GL}_2(\Bbb{F}_5)$ acts on $\begin{pmatrix}
0\\
1
\end{pmatrix}$ by left multiplication), and determine the order (number of matrices) of the stabilizer.
My attempt:
Let $M=\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}
$ with $a, b, c, d\in\Bbb{F}_5$. Because $M \cdot \begin{pmatrix}
0\\
1
\end{pmatrix} = \begin{pmatrix}
0\\
1
\end{pmatrix}$, we have:
$$ a \times 0 + b \times 1 = 0,$$
and
$$c \times 0 + d \times 1 = 1,$$
and so $b = 0$ and $d = 1$. So the matrix has the form $M=\begin{pmatrix}
a & 0\\
c & 1
\end{pmatrix}$.
Because we have $24$ choices for the 1st column and 1 choice for then 2nd column, then the total number of matrices satisfying is 24.
Thank you.
Best Answer
Your argument is almost correct. Indeed such a matrix must be of the form $$\begin{pmatrix} a & 0\\ c & 1 \end{pmatrix},$$ but you do not have $24$ choices for the first column; you need the matrix to be invertible. Equivalently, the first column must be linearly independent of the second column. How many choices does that leave?