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

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 26 '17 at 21:47

• By conjugating the first generator by the second, the subgroup generated contains $(1,k)$ for all $k$. – Derek Holt Mar 26 '17 at 17:03