Hi

Let $n$ be an odd natural number (sufficiently large), and $ 1\leq t,k < n $. Let $A= \{ a_{1},a_{2},\ldots,a_{t} \} $, $B= \{ b_{1},b_ {2},\ldots,b_{n-t} \} $, $C= \{ c_{1},c_{2} \ldots,c_{k} \} $ and $D= \{ d_{1},d_{2},\ldots,d_{n-k} \} $ be subsets of $ \{ 1,2,\ldots,n \} $ such that $A\cup B=C\cup D= \{ 1,2,\ldots,n \} $ and $A\cap B=C\cap D=\phi$ and $A\neq C$ and $A\neq D$. What can we say about the subgroup generated by $(a_{1},a_{2},\ldots,a_{t})(b_{1},b_{2},\ldots,b_{n-t})$ and $(c_{1},c_{2},\ldots,c_{k})(d_{1},d_{2},\ldots,d_{n-k})$. I think that this subgroup is $S_{n}$. Is it true? Maybe there is a simple counterexample but I've not found it.

Clearly, $n$ should be odd. Otherwise, this group is a subgroup of $A_{n}$.