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Someone has already asked whether an exponent less than $n!$ is possible for a symmetric group $S_n$. It has been answered that it is for $n \ge 4$.

I would like to know if there is a general formula to determine the exponent of $S_n$.

Community
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Omar Shehab
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For each prime $p\le n$ we have to decide what is the maximal $k$ such that $S_n$ contains an element of order $p^k$. We may assume that such an element has only a single cycle because disjoint cycles combine only by taking the lcm, which does not "help" for prime powers. Thus the maximal $k$ is given by the bound $p^k\le n$. Hence the exponent of $S_n$ is $$\prod_{p\le n}p^{\lfloor \ln n/\ln p\rfloor}.$$

Hagen von Eitzen
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  • what is the purpose of $\lfloor \ln n/\ln p\rfloor$ here? – Omar Shehab Dec 15 '15 at 07:54
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    @Omar: this is clearly explained in the previous sentence: it's the largest $k$ such that $S_n$ contains an element of order $p^k$. – Qiaochu Yuan Dec 15 '15 at 08:07
  • @QiaochuYuan, I am afraid I do not understand why $k = \lfloor \ln n/\ln p\rfloor$. I understand that we have to do something related to calculating the lcm of the orders of all $p$'s. – Omar Shehab Dec 15 '15 at 08:13
  • @QiaochuYuan Does $k = \lfloor \ln n/\ln p\rfloor$ have anything to do with the prime number theorem? – Omar Shehab Dec 17 '15 at 13:51
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    @Omar: no. It's just the largest $k$ such that $p^k \le n$. To get there from here, take the natural log of both sides, then remember that $k$ is an integer. – Qiaochu Yuan Dec 17 '15 at 16:39
  • @QiaochuYuan, so the exponent of $S_5$ is $2^{\lfloor \ln 5/\ln 2\rfloor} . 3^{\lfloor \ln 5/\ln 3\rfloor} . 5^{\lfloor \ln 5/\ln 5\rfloor} = 2^2 . 3 . 5 = 60$. Right? – Omar Shehab Dec 17 '15 at 16:49
  • Related question: https://math.stackexchange.com/questions/302278/mathrmlcm1-2-3-ldots-n – Julian Kuelshammer Jan 17 '22 at 13:35