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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 Oct 3 '18 at 16:11

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  • $\begingroup$ All you need is that the subgroup contains all transpositions of the form $(k\,k+1)$. $\endgroup$ – 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 $$(1\;2\;3\;\ldots\;n)^{k-1}(1\;2)(1\;2\;3\;\ldots\;n)^{-(k-1)}=(k\;\;\;k+1).$$

  • 1
    $\begingroup$ It should be exponent $k-1$. $\endgroup$ – Berci Oct 3 '18 at 15:51
  • $\begingroup$ @Berci Edited, thanks! $\endgroup$ – user593746 Oct 3 '18 at 15:53

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