Show that if a transitive subgroup $G\subset S_n$ of the symmetric group $S_n$ contains a n-1 cycle and transposition,then $G=S_n$.

Firstly, without loss of the genralization, we can assume the n-1 cycle is $(2, 3,4,....,n-1)$, and the transposition is (1,2), but how can we prove that (1,2) and (2,3,...,n-1) generates $S_n$


marked as duplicate by user26857 abstract-algebra Mar 26 '17 at 21:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ By conjugating the first generator by the second, the subgroup generated contains $(1,k)$ for all $k$. $\endgroup$ – Derek Holt Mar 26 '17 at 17:03
  • $\begingroup$ you are right! Thanks! $\endgroup$ – user144600 Mar 26 '17 at 17:10