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 $A\_{n}$.



Generators of symmetric group
Counterexample: Everything in $\langle(123)(456789),(123456)(789)\rangle$ acts cyclically on the set partition $\lbrace1,4,7\rbrace \lbrace2,5,8\rbrace \lbrace 3, 6, 9 \rbrace$. This type of counterexample works for any odd composite $n$. 


Thanks to Michael Zieve for pointing out that the following is incorrect:


~~For prime $p$ the result is true by the [O'Nan-Scott Theorem](http://www.maths.qmul.ac.uk/~raw/talks_files/ONStalk.pdf), the classification of maximal subgroups of the symmetric (and alternating) groups.~~ 


One of the cases of the O'Nan-Scott classification of maximal subgroups of the symmetric group is allowed: $AGL(1,p)$ acting on the affine line.


