I am trying to show that $PSL(2, \Bbb{Z})\cong \Bbb{Z}_2 \ast \Bbb{Z}_3$. What I've shown is that, for $$ A= \left(\begin{matrix} 0 & 1 \\ -1 & 0 \\ \end{matrix}\right) $$ and $$ B= \left(\begin{matrix} 1 & 1 \\ -1 & 0 \\ \end{matrix}\right), $$ $\{A, B \}$ generates $PSL(2, \Bbb{Z})$ with $A^2=B^3=I$.
Here I can't proceed further as I don't see how to ensure that there exists no more restriction on the elements so that I can conclude that $A^2, B^3$ are the only relators and get it done.
Any help will be appreciated.
Edit:
Thanks to the comment I've become aware of the ping-pong lemma, and as it is my first application of this theorem I want my solution below to be examined if it is not flawed.
Claim: $PSL(2, \Bbb{Z})$ is the free product of $\langle{A}\rangle$=$\{I, A \}$ and $\langle{B}\rangle$=$\{I, B, B^2\}$.
proof)
Let $X=\Bbb{R}^2/\sim$ be the real plane with antipodals identified on which $PSL(2, \Bbb{Z})$ acts by the left multiplication. (Regarding each point of $X$ as a column vector.) And let $X_1$ be the 1st (and 3rd) quadrant and $X_2$ be the 2nd (and 4th) quadrant of $X$. Then $A$ maps $X_2$ into $X_1$, and both $B$ and $B^2$ map $X_1$ into $X_2$ as the image below. Hence it follows that $PSL(2, \Bbb{Z})$=$\langle{A}\rangle \ast \langle{B}\rangle$ by the ping-pong lemma.
