Multiplication of matrices of the given form is a group. \begin{equation*}
A_{2,2} = 
\begin{pmatrix}
a & b \\
0 & 1 \\
\end{pmatrix}
\end{equation*} for $a,b \in $ $\mathbf{Q}$. Find $Z(A)$ and all elements and subgroups of finite order in $A$, provided that $A$ is a group under multiplication of matrices.
My attempt so far is that I just multiplied two matrices to check when they commute. Namely, \begin{equation*}
A_1.A_2= 
\begin{pmatrix}
a_1 & b_1 \\
0 & 1 \\
\end{pmatrix} . \begin{pmatrix}
a_2 & b_2 \\
0 & 1 \\
\end{pmatrix} =A_2.A_1 \leftrightarrow a.b_1+b-b_1=a_1.b; 
\end{equation*}
for $b\neq 0$ and then we can form the matrices of $Z(A)$ to be of the form \begin{pmatrix}
a_1 & b_1 \\
0 & 1 \\
\end{pmatrix} for $ a_1 =(a.b_1+b-b_1)/b $ and $b_1 \neq b/(1-a) $ for $ a\neq {0,1} $ and b $\neq 0$.
But what does mean elements of finite order? I mean that the set of rational numbers is only countable and so how I can count all the element of a given subgroup of A?
Any help is welcome!

Note that the identity matrix is the identity element in this group. An element is of finite order if multiplying it finitely many times gives us the identity. Now note that
\begin{equation}
\begin{pmatrix}a&b\\0&1\end{pmatrix}\cdot\begin{pmatrix}a&b\\0&1\end{pmatrix} = \begin{pmatrix}a^2&b(a+1)\\0&1\end{pmatrix}
\end{equation}
By induction you get $n$th power of a matrix to be
\begin{align}
\begin{pmatrix}a&b\\0&1\end{pmatrix}^n = \begin{pmatrix}a^n&b(1+a+a^2+\dots+a^{n-1})\\0&1\end{pmatrix}
\end{align}
For this to be identity for some finite $n$, we have $a^n=1$ and $b(1+a+\dots+a^{n-1})=0$.  We make two cases:
Case 1: If $a=1$ then $b(1+a+\dots+a^{n-1})=nb$, which is zero for some natural $n$ iff $b=0$. So we get the matrix $\begin{pmatrix}a&b\\0&1\end{pmatrix}$ to be identity itself!
Case 2: If $a=-1$, then $b(1+a+a^2+\dots+a^{n-1}) = 0$ for $n=2$. So every element of the form $\begin{pmatrix}-1&b\\0&1\end{pmatrix}$ has order $2$ which is finite.
No other element is of finite order.
