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I want to find the maximum possible order of an element in $S_{13}$. I have found that disjoint cycles of length $(2)(3)(7)$ give order 42,

However ,answer in my book is given to be $60$

Can someone tell me how can I achieve given order ?

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    $60=3\times4\times5$. – Angina Seng Aug 29 '19 at 16:15
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    @Lord Shark the Unknown : but, 3+4+5 = 12 –  Aug 29 '19 at 16:18
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    Yes, and $3\times 4\times 5=60$ which is the order you're after! – Angina Seng Aug 29 '19 at 16:19
  • @Lord Shark the Unknown: oh,I see there is a cycle of length 1 as well. Thank you. –  Aug 29 '19 at 16:20
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    @Lord Shark the Unknown: I have just one more question to ask, Does there exist any algorithm to quickly calculate the maximum order in $S_n$ .For smaller n like 3,4 and 5 ,I can easily calculate by writing their partitions but for higher order does there exist any algorithm/trick to calculate this efficiently ? –  Aug 29 '19 at 16:25
  • Python algorithm: https://tio.run/##pZLBbsIwDIbvPMUvLiSjG0PcJrGH2I4IoSpNaUSTVEk6xl6eOWkphU27jAMU15/927@bU6isWZ3PpbMapctFUNZ4KN1YF7AXxWRSyBK10ExnMPxlAvo4GVpnoPEAZrBYxMTuPe/ydf5pXSHdrpJ1Ix2zV3Y6nabfNyla59WHhA@ygS0RKjmAEHkt2jqPcp4mPRCbejA7y1BzHCvpJGoojxy18iHWECdRU1Cafah8wnyrtTJ7BAuTITcF7CwysRuNFSGLowpViqTmiesFkXrdhqQjo1oipuWhr0FqlVZfshgmSw@qhMF6jedu4tHG4iI2W57CBDvpQWmjvzsaIMY22xQsrcMBysDlZi/ZktY4X/JrWUZIhgvHCbzffHSOmh44oY8HPpCkMXZ/7fteS95Io@9fXgwiD1vMh@7jy2BdYjYG7k6D/TyIi7//tPfibndIvY/COhLSWFNQdiJHzsYNv@9uDf7D38v53606ukNDNk6ZwIYxlyvOz@dv – Daniel Schepler Aug 29 '19 at 16:32
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    See also: https://en.wikipedia.org/wiki/Landau%27s_function – lisyarus Aug 29 '19 at 16:35
  • Possible duplicate of Maximal order of elements in $S_n$. – Dietrich Burde Aug 30 '19 at 18:54

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Every element in $S_{13}$ can be written as a product of disjoint (meaning every number appears only once) cycles. Clearly, the order of a product of disjoint cycles (which, being disjoint, commute) is the least common multiple of the order of each cycle.

(clarifying example: $(12)(345)$ has order $6 = 2 \cdot 3$, because $(12)^2 =1$ and $(345)^3 = 1$)

So to find the maximal order of an element in $S_{13}$ you need to find the partition of $13$ with the highest l.c.m. of the summands. Certainly $3+4+5(+1)$ gives you l.c.m. $60$. The question is: are there bigger ones? I am afraid that "smart" trial and error is the way to go: ideally you want as many different primes as you can get, with the highest possible powers. Possible primes are $2,3,5,7,11,13$: easily, $11$ and $13$ are too big. $7$ would be a possibility, but the biggest l.c.m.s you can get with it are $2\cdot 3 \cdot 7 = 42$, or $5\cdot 7 = 12$, or $2 \cdot 4 \cdot 7 = 56$, all lower than $60$. So, after observing that $2^3 \cdot 5 = 40$, it has to be $3+4+5+1$.

(an interesting fact: in a somewhat counterintuitive way, the highest order in $S_{13}$ is achieved by an element of $S_{12}$!)

AnalysisStudent0414
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