Why group is primitive

Question

Transitive substitution group $G$ of degree $n$ contain the substitution $g$, represented as the product of two independent cycles of lengths $n_1$, $n_2$, where $n_1+n_2=n$ and $(n_1 , n_2)=1$. Why $G$ should be primitive.
I think the criterion of primitiveness will do. So it must be shown that there is no subgroup between the stabilizer of any element and the group ($G$ is primitive $\Leftrightarrow \nexists H\lt G:G_\alpha \lt H \lt G$ and $G\neq H \neq G_\alpha$ )

Answer 1

This is not true.

There is a transitive imprimitive subgroup of $S_6$ that contains the permutation $(1,3,2,4)(5,6)$ and preserves the blocks system $\{\{1,2\}, \{3,4\}, \{5,6\}\}$.