TITLE: Will any two elements of orders 4 and 6 generate SL(2,Z)?
QUESTION [5 upvotes]: Ie, if $A,B$ are matrices in $\text{SL}_2(\mathbb{Z})$ with orders 4 and 6 respectively, is $\text{SL}_2(\mathbb{Z}) = \langle A,B\rangle$?

REPLY [4 votes]: No. If $a$ and $b$ are the standard generators of $G={\rm SL}(2,{\mathbb Z})$ of orders $4$ and $6$, then $a$ and $(ab)^{-1}b(ab)$, for example, do not generate $G$. This is not hard to see from the the fact that, modulo the centre $Z = \langle a^2 \rangle$ of the group, $G/Z \cong C_2 * C_3$ is the free product of groups of order $2$ and $3$ generated by $aZ$ and $bZ$. The subgroup that they generate does not contain $b$.