Let $G=\langle U,J\rangle$ be the subgroup of $\mathrm{GL}_2(\mathbb{Q})$ generated by $$U=\begin{pmatrix}-1 & 1 \\ -1 & 0\end{pmatrix} \text{ and } J=\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}.$$ I have to determine $\mathrm{ord}(U)$, $\mathrm{ord}(J)$, $\mathrm{ord}(UJ)$ and show that $G\ne \{U^kJ^l\colon k,l\in\mathbb{Z}\}$.
Although, I think I solved it, I am wondering, since I am a beginner in this subject, if there would have been any more systematic or elegant way to do this.
Here is what I did:
Firstly, I computed the different powers of the generators $U$ and $J$.
$U^2=\begin{pmatrix} 0 & -1\\ 1 & -1\end{pmatrix}; U^3=\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix};U^{-1}=U^2$
I can conclude from this that $\mathrm{ord}(U)=3$. I then did the same for $J$:
$J^2=\begin{pmatrix} -1 & 0\\ 0 & -1\end{pmatrix}; J^3=\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix};J^4=\mathrm{id};J^{-1}=J^3$
So the order must be 4. The same goes for $UJ$ where the order is non finite.
Would there be any other way to do this?
Then for the last issue: To show the inequality I sought to get an element of $G$ which is not of the described form. I tried $UJU=\begin{pmatrix}2 & -1\\-1 & 0\end{pmatrix}$. Since $J^{-1}=J^3$ and similar for $U$, I can restrict to looking if $UJU\in \{U^kJ^l\colon k,l\in\{1,2,3\}\}$, also because of the order of $J$ and $U$.
Is my strategy reasonable?
Clearly the powers $(UJ)^n$ or $(JU)^n$ are not all of the form $U^kJ^l$, because $U^3=J^4=I_2$, but $$ (JU)^n=\begin{pmatrix} -1 & (-1)^n n\cr 0 & -1 \end{pmatrix}. $$ So there are only finitely many matrices of the form $U^kJ^l$, but infinitely many matrices of the form $(UJ)^n$. So you are done.