Generating set of same size must have common elements Generating set $S_1,  S_2$ generates permutation group $G$ where  the number of elements  in $S_1$ is equal to the number of elements  in $S_2$. Prove, $S_1 \cap S_2 \neq  \emptyset $ (not considering  identity element of $G$).
Or is it a false statement?
If the statement is true, then $\exists \pi_1, \pi_2 \in S_1$ such that $\pi_1 . \pi_2 \in S_2$. 
 A: The statement is false. For a counterexample, let $G=S_3$. If $A, B$ are two-element subsets of $G\setminus\{e\}$ with both $A$ and $B$ containing at most one element of order $3$, then $\langle A\rangle=\langle B\rangle=S_3$. For instance, $A=\{(1,2),(1,2,3)\}$ and $B=\{(1,3),(1,3,2)\}$ satisfy $\langle A\rangle=\langle B\rangle=S_3$ and $A\cap B=\varnothing$.
A: The statement is false in general. Consider the permutation group $S_4$.  One generator set for the group is the set $S_1=\{(12),(23),(34)\}$ of three transpositions.  Another generating set of size 3 is $S_2=\{(14),(24),(13)\}$.  These two sets have the same cardinality and yet are disjoint.  
An example where $S_1$ and $S_2$ overlap is: take the permutation group to again be $S_4$, let $S_1 = \{(12),(23),(34)\}$ and $S_2=\{(12),(13),(14)\}$.  Then $S_1$ and $S_2$ overlap.  However, the product of any two elements in $S_1$ does not give an element in $S_2$ since the product is an even permutation while $S_2$ is a set of odd permutations.
