Subgroup generated by two permutations

Question

Given the permutation: $$\sigma:= \left( \begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14\\ 10 & 13 & 6 & 1 & 14 & 2 & 11 & 12 & 4 & 7 & 9 & 5 & 3 & 8 \end{matrix} \right)\in S_{14} $$

1) Determine the period of $\alpha:=\sigma^{27797848}$
2) Determine and motivate if $\beta:=(273)$ is or not in the subgroup of $S_{14}$ generated by $\alpha$ and the 3-cycle $(123)$

I can do 1:
$\sigma=(1\;10\;7\;11\;9\;4)(2\;13\;3\;6)(5\;14\;8\;12)$
period of $\sigma$ is $lcm(6,4)=12$
so $27797848=4\;(mod\;12)$
so $\alpha=\sigma^4=(1\;9\;7)(10\;4\;11)$
and period of $\alpha$ is $lcm(3,3)=3$

Now for 2:
My first question is: if the order of a subgroup generated by one permutation is equal to the period of the permutation, what about a subgroup generated by two permutations? Is it the $lcm$ of the periods?
Second: the two permutations haven't disjoint orbits so the product is not commutative. I guess I'll quite some elements in the subgroup so I can't calculate it to check if $(2\;7\;3)$ is in it.
Third: Can I deduce if $(2\;7\;3)$ is in it?

I'll appreciate any hint or suggestion or answer. Thank you.

Answer 1

Let $\gamma = (1 2 3)$. Then $\gamma^2 = (2 1 3)$. We have $\alpha^{-1}(1) = 7, \alpha^{-1}(2) = 2, \alpha^{-1}(3) = 3$, so $\alpha^{-1}\gamma^2\alpha = (2 7 3) = \beta$.