I read this link . In Theorem $2.7 $, it is mentioned that for $n\geq 3$ except for $n = 5, 6, 8$, symmetric group $S_n$ is generated by an element of order $2$ and an element of order $3$. However, we also know that for $n\geq 2$, $S_n$ is generated by the transposition $(1 2)$ and the $n$-cycle $(12\ldots n)$. If we use later result then $S_4$ is generated by transposition $(1 2)$ and $4$-cycle $(1234)$ which contradicts result of Theorem $2. 7$ since order of four cycle is four.

Could anybody explain me where I am going wrong? I would be very much grateful.

Thanks for your time.

  • $\begingroup$ I can not see the contradiction, why do you think that $|(1,2,3,4)|=4$ cause a contradiction. $\endgroup$ – mesel May 24 '14 at 11:51
  • 2
    $\begingroup$ You get $(1234)$ as the product $(12)(234)$, so $(12)$ and $(234)$ generate all of $S_4$ confirming that Theorem in the case $n=4$. You have to be extra careful in selecting the elements of order two and three that do generate the group. Also - what amWhy says. $\endgroup$ – Jyrki Lahtonen May 24 '14 at 11:51
  • $\begingroup$ @JyrkiLahtonen Thank you for clearing my doubt. $\endgroup$ – srijan May 24 '14 at 11:57
  • $\begingroup$ @mesel Thanks for the response. $\endgroup$ – srijan May 24 '14 at 11:57
  • $\begingroup$ you are welcome. $\endgroup$ – mesel May 24 '14 at 12:01

The theorem doesn't claim that $S_n$ is only generated by a 2-cycle and a 3-cycle. It claims that it can be generated by these.

The fact that $S_4$ can also be generated by a 2-cycle and a 4-cycle is not a contradiction. Note that $S_4$ can very well be generated by a $2$-cycle and a $3$-cycle: Take, e.g., $\langle (1, 2), (2, 3, 4)\rangle$.

  • $\begingroup$ Thanks for the answer. My doubt is cleared. $\endgroup$ – srijan May 24 '14 at 11:57
  • $\begingroup$ You're welcome, srijan! $\endgroup$ – Namaste May 24 '14 at 11:58

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