# Matrices generate finite subgroup of $SL_2(\mathbb{Z})$

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!

• 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.
– EuYu
Commented May 15 at 8:27
• Thanks! Can you help me with that? Commented May 15 at 8:29
• 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. Commented May 15 at 8:30
• $A^2=B^3=I_2$ so products of $A$ and $B$ will eventually terminate to the identity matrix? Does this show finiteness of the subgroup? Commented May 15 at 8:34
• 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. Commented May 15 at 8:37