I am asked to prove that the subgroup $G$ of $S_{12}$ generated by elements $\pi^{2}$ and $\tau=(3\ 10)$ is a cyclic group of order $10$. Here $\pi=(1\ 8\ 12\ 11\ 6\ 10)(2\ 9)$.
My problem is that I can't find the generator of the group, nor can I find the element with order $10$. For example, if my group contains $\pi^{2}$ and $\tau$, it must contain $\pi^{2} \tau$ which is of order $12$. Hence there are more than $10$ elements in my group.