What is the order of this subgroup in $GL_2(\mathbb{F}_5)$?

Question

The following subgroup in $GL_2(\mathbb{F}_5)$ is supposed to be of order 20, but I keep calculating that the order is 25?

The subgroup is: $\left\langle\begin{pmatrix}2 & 0\\0 & 1\end{pmatrix}, \begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}\right\rangle$.

I am confused because I am getting all the matrices in the format: \begin{pmatrix}a & b\\0 & 1\end{pmatrix} where $0\le a,b \le 4$. So essentially I am getting 25 elements in this matrix when there are supposed to be only 20. Am I doing any calculations wrong, or must there be a typo somewhere in the problem?

Answer 1

Note that $\begin{pmatrix}a & b\\0 & 1\end{pmatrix} \in \mathsf{GL}_2(\mathbb{F}_5)$ iff $a \neq 0$.

An ad-hoc way to see we obtain all those elements: the two given matrices are in the kernel $K$ of the morphism $B_2(\mathbb F_5) \to \mathbb F_5^\times$ that sends an upper triangular invertible matrix (Borel subgroup) to its right bottom entry. (This is a fancy way of saying that their second row is $0\;1$.)
It is surjective and $|B_2| = 4\cdot4\cdot 5$ and $|F_5^\times|=4$, hence $|K|=20$. (Which one could count directly, as you did.)
Our subgroup has an element of order $4$ and of order $5$, so it must equal the whole of $K$.

Answer 2

The subgroup cannot have $25$ elements, since $$ |GL(2,5)|=(5^2-1)(5^2-5)=480. $$ This follows from Lagrange, because otherwise $25$ would divide $480$, which is not the case.

Answer 3

Notice that the top left entry of a matrix in this group is a power of 2 ; there are 4 distinct powers of 2 modulo 5, namely $2,4,3,1$. We can have at most 5 distinct entries for top right, so there are $4.5 = 20$ elements at most. So if you can show existence of 20 distinct elements you are then done.