# Maximum element order in $S_n$ [closed]

Denote by $$S_n$$ the group of permutations of the set $$\{1,\ldots,n\}$$ with composition as binary operation. Let $$m_n$$ denote the maximum order that an element of $$S_n$$ can have. What is the smallest positive integer $$k$$ such that $$\lim_{n\to\infty} \frac{m_n}{n^k} < \infty$$?

## closed as off-topic by Wojowu, Chris Godsil, Dan Petersen, Ben Barber, Pace NielsenJan 25 at 17:38

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• A simple OEIS check yields oeis.org/A000793, where also your question is answered as $lim_{n->\infty} (\log a(n)) / \sqrt{n \log n} = 1$. – Christian Stump Jan 25 at 12:05
• From this post on Mathematics Maximal order of an element in a symmetric group we can learn that this is Landau's function. – Martin Sleziak Jan 25 at 12:06
• This is Landau's function. There is no such $k$. – Derek Holt Jan 25 at 12:09
• How strange! The question gets three downvotes while the (brief and correct) answer gets 10 upvotes. – Derek Holt Jan 26 at 15:57

Landau proved in 1902 that the maximal order of an element in $$S_n$$ is $$e^{(1+o(1))\sqrt{n\log n}}$$. In particular, there is no integer $$k$$ with the property you ask for.