I am given that $G= \langle (1,2,3,4,5),(1,2,3,5,4) \rangle$, and am asked to show that it's order is 60. I started by showing that $G$ was transive on the set $X = \{ 1,2,3,4,5 \}$ as $a$ takes $1$ to $2$, $a^{2}$ takes $1$ to $3$, ect. So we can apply the orbit stabilizer theorem:
$$|G| = |X| |\text{Stab}_{G}(5)| = 5|\underbrace{\text{Stab}_{G}(1)}_{=H}|.$$
This is a technique I saw in my lecture notes but I'm not really sure how it works.. It then seems that we would consider the actions inside of $H$ to determine another relation, eventually showing that its order is $\geq 60$. And then, since it is obvious that $G < S_{5}$ but not the whole thing, its order must be 60. But I'm unsure of how to get from what I've written so far to that conclusion.