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Show that $A:= \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} $ and $ B:=\begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix} $ generate a finite subgroup of $SL_2(\mathbb{Z})$ (set of all $2 \times2$ matrices with determinant $1$).

Is it enough to show that $AB,BA,A^{-1},B^{-1}$ all have determinant $1$ or am I wrong?

I also need to check whether this subgroup is normal. Is it correct to check if $gAg^{-1}=A,gBg^{-1}=B$ for some $g \in SL_2(\mathbb{Z})$?

Thank you!

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    $\begingroup$ You don't need to check determinant $1$ since that follows directly from multiplicativity. It is the finiteness condition that matters here. You need to show that products of $A$ and $B$ will not generate an infinite subgroup. $\endgroup$
    – EuYu
    Commented May 15 at 8:27
  • $\begingroup$ Thanks! Can you help me with that? $\endgroup$ Commented May 15 at 8:29
  • $\begingroup$ Both your suggested strategies are not what you want. The first item has nothing to do with the property you want (and is in fact automatically fulfilled if $A,B$ are both from $\mathrm{SL}_2{\mathbb{Z}}$). The second one is a bit ambiguous. If you intend to show this for each $g$, then the subgroup is central (and therefore also normal), but this is not necessary for normality. If you just want to find a specific $g$, this will show nothing (since $g=1$ always does this). I suggest that you just work straight with the definitions here. $\endgroup$ Commented May 15 at 8:30
  • $\begingroup$ $A^2=B^3=I_2$ so products of $A$ and $B$ will eventually terminate to the identity matrix? Does this show finiteness of the subgroup? $\endgroup$ Commented May 15 at 8:34
  • $\begingroup$ No, not on its own. There are groups generated by such elements (of orders $2$ and $3$ that is) that are not even torsion (for instance, the closely related $\mathrm{PSL}_2(\mathbb{Z})$ is such a group). To finish the argument, you also need to control how $A$ and $B$ interact. $\endgroup$ Commented May 15 at 8:37

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