Vtyjkyuk

Let $G=langle U,Jrangle$ be the subgroup of $mathrmGL_2(mathbbQ)$ generated by $$U=beginpmatrix-1 & 1 \ -1 & 0endpmatrix text and J=beginpmatrix 0 & 1 \ -1 & 0endpmatrix.$$
I have to determine $mathrmord(U)$, $mathrmord(J)$, $mathrmord(UJ)$ and show that $Gne U^kJ^lcolon k,linmathbbZ$.

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 $J$.

$U^2=beginpmatrix 0 & -1\ 1 & -1endpmatrix; U^3=beginpmatrix 1 & 0\ 0 & 1endpmatrix;U^-1=U^2$

I can conclude from this that $mathrmord(U)=3$. I then did the same for $J$:

$J^2=beginpmatrix -1 & 0\ 0 & -1endpmatrix; J^3=beginpmatrix 0 & -1\ 1 & 0endpmatrix;J^4=mathrmid;J^-1=J^3$

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 $UJU=beginpmatrix2 & -1\-1 & 0endpmatrix$. Since $J^-1=J^3$ and similar for $U$, I can restrict to looking if $UJUin U^kJ^lcolon k,lin1,2,3$, 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
$$
(JU)^n=beginpmatrix -1 & (-1)^n ncr 0 & -1 endpmatrix.
$$
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.