# Prove that $\langle \{(123…n),(12) \} \rangle =S_n$ [duplicate]

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I know that $$\langle \{(123…n),(12) \}\rangle \subset S_n$$. I was thinking that if I could show that $$\langle \{(123…n),(12)\} \rangle$$ contains all the transpositions of $$S_n$$ then it would contain $$S_n$$. How would I go about showing this?

## marked as duplicate by rogerl, Arnaud D., Trevor Gunn, Mees de Vries, Derek Holt 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(); } ); }); }); Oct 3 '18 at 16:11

• All you need is that the subgroup contains all transpositions of the form $(k\,k+1)$. – Lord Shark the Unknown Oct 3 '18 at 15:32
You can first prove that $$S_n$$ is generated by $$(k\;\;\;k+1)$$ for $$k=1,2,\ldots,n-1$$. Then, check that $$(1\;2\;3\;\ldots\;n)^{k-1}(1\;2)(1\;2\;3\;\ldots\;n)^{-(k-1)}=(k\;\;\;k+1).$$
• It should be exponent $k-1$. – Berci Oct 3 '18 at 15:51