Exercise :

Find the maximum possible order of an element of the group of permutations $S_7$. Do the same thing for $S_{10}$.

Discussion :

Recalling that any permutation can be written as a product of disjoint cycles :

$$c=c_1 c_2\dots c_r$$

the order of $|σ|=\text{lcm}(|σ_1||σ_2|\dots|c_r|)$ and if $c_i$ has length $k_i$ then it will be $|c_i|=k_i$.

So what I have to do is find all the possible products of disjoint cycles, which will be :

$$(1,2) \space \text{ of order } \space 2$$ $$(1,2,3) \space \text{ of order } \space 3$$ $$(1,2,3,4) \space \text{ of order } \space 4$$ $$(1,2)(3,4) \space \text{ of order } \space 2$$ $$(1,2,3,4,5) \space \text{ of order } \space 5$$ $$(1,2,3)(4,5) \space \text{ of order } \space 6$$ $$(1,2,3,4,5,6) \space \text{ of order } \space 6$$ $$(1,2,3,4)(5,6) \space \text{ of order } \space 4$$ $$(1,2)(3,4)(5,6) \space \text{ of order } \space 2$$ $$(1,2,3)(4,5,6) \space \text{ of order } \space 3$$ $$(1,2,3)(4,5)(6,7) \space \text{ of order } \space 6$$ $$(1,2,3,4,5)(6,7) \space \text{ of order } \space 10$$ $$(1,2,3,4)(5,6,7) \space \text{ of order } \space 12$$ $$(1,2,3,4,5,6,7) \space \text{ of order } \space 7$$

So, the maximum possible order of an element in $S_7$ is $12$.

Question :

What I wanted to ask is $(1)$ am I correct, first of all?

And $(2)$ how am I supposed to find the maximum order of an element in $S_{10}$ ? Judging from all the possible cycle products in $S_7$ it will be a pretty big list for $S_{10}$. Am I missing a trick or some clever way to reach the desired result faster ?

P.S. : I know that Landau's function calculates exactly that thing, but we haven't been taught about it yet.

  • $\begingroup$ Why $(1,2,3)(4,5)(6,7)$ has order $2$? $\endgroup$ – Skills Dec 29 '17 at 11:55
  • $\begingroup$ @Skills Sorry, was typo. $\endgroup$ – Rebellos Dec 29 '17 at 11:57
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    $\begingroup$ Case it by the order of the largest cycle, from high to low, so just 10 cases, and most will be already dominated. Also, no need to write out the actual cycles, just write their orders (e.g. 4-3-3 => 12). $\endgroup$ – quasi Dec 29 '17 at 11:57
  • $\begingroup$ Besides the typo pointed out by Skills, yes, it looks right. You can write a computer program to do it; you are effectively trying to maximize least common multiple of a bunch of cycles such that their lengths add up to 10. It should not be that hard to do it by hand. (For instance, do you think you will have a cycle of length 9?) $\endgroup$ – E-A Dec 29 '17 at 11:58
  • $\begingroup$ @E-A The problem is that this is an exercise that should be hand-written/solved. This is why I'm concerned about larger $n$. $\endgroup$ – Rebellos Dec 29 '17 at 12:03

First of all yes, your answer for $S_7$ is correct. And indeed listing all cycle types quickly becomes unmanageable as $n$ grows.

It is quite helpful and not difficult to prove that the maximum is always assumed for a permutation that is the product of disjoint cycles of prime power lengths. For increasing values of $n$ this eliminates an increasing proportion of permutations; for $n=7$ and $n=10$ it does not eliminate much.

It is also quite helpful and not difficult to prove that the maximum is always assumed for a permutation that has at most one fixed point. This eliminates a large proportion of permutations for any $n$.

With these two facts in mind, for $n=7$ we are left with permutations of cycle types $$(7),\ (5,2),\ (4,3),\ (4,2,1),\ (3,3,1),\ (3,2,2)\quad \text{and}\quad (2,2,2,1).$$ For $n=10$ we are left with permutations of cycle types $$(9,1),\ (8,2),\ (7,3),\ (7,2,1),\ (5,5),\ (5,4,1),\ (5,3,2),\ (4,4,2),\ (4,3,3),\ (4,3,2,1),\ (3,3,3,1),\ (3,3,2,2),\quad\text{and}\quad (2,2,2,2,2).$$ Of course one can already see some recursion in this problem; given that the maxima for $n=3$ and $n=5$ are assumed for cycle types $(3)$ and $(3,2)$, respectively, for $n=10$ we can eliminate the cycle types $$(7,2,1),\ (5,5)\quad\text{and}\quad(5,4,1).$$ There are a lot of simple facts to note that each reduce the number of cycle types to consider by a little bit. For your problem there aren't that many to begin with, and the facts above are enough to make the problem manageable by hand. But it can just as well be done by hand for $n=100$ and $n=1000$, though this requires more careful analysis of the problem which I won't go into here. I'll leave you with this vague intuition though:

Heuristically, for large $n$ the maximum is assumed when writing $n$ as a sum of prime powers close to $\sqrt{n}$, and padding the remainder with small primes. It is not easy to compute the maximum precisely for large values of $n$, though asymptotically it is $\exp((1+o(1))\sqrt{n\ln{n}})$.

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    $\begingroup$ For the observation on prime power lengths: Let $\sigma\in S_n$ be a cycle of length $m=\prod_{p\mid m}p^{m_p}>1$. We want to show that there exists a $\tau\in S_n$ of order $m$ that is a product of disjoint cycles of prime power lengths. If $\sum_{p\mid m}p^{m_p}\leq n$ we can simply take disjoint cycles of lengths $p^{m_p}$ for every prime $p$ dividing $m$. Then the order of the product is $\operatorname{lcm}_{p\mid m}p^{m_p}=\prod_{p\mid m}p^{m_p}=m$. As $n\geq m$ we can do this whenever $$\sum_{p\mid m}p^{m_p}\leq m=\prod_{p\mid m}p^{m_p}.$$ This holds for all $m>1$. (Continued) $\endgroup$ – Servaes Dec 29 '17 at 17:43
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    $\begingroup$ This follows quickly from the fact that $p^{m_p}\geq2$ for all $p$. Now for a general $\sigma\in S_n$ (that is not necessarily a cycle), write $\sigma$ as a product of disjoint cycles and apply the above to each cycle. $\endgroup$ – Servaes Dec 29 '17 at 17:44
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    $\begingroup$ Hello and thanks a lot for your detailed and amazing answer. I checked it fast yesterday (had to attend a Christmas table later on) and I had the time to sit down, write and prove everything. I have to say that your answer was really helpful and written in such a good way with details and mathematical formality that gave me a great insight and understanding ! Also the comments were really helpful regarding the proofs and leading me to a general understanding. Thanks a really lot, I would upvote double times if I could ! $\endgroup$ – Rebellos Dec 30 '17 at 12:20

You see that in $S_7$ the maximum possible order is given by $$(1,2,3,4)(5,6,7)$$ so used element $s_1,s_2$ such that $o(s_1)+o(s_2)=7$ that maximize $l.c.m.$ So you have to choose a similar case in $S_{10}$. The solution is $s_1,s_2,s_3$ such that $o(s_1)=5,o(s_2)=3,o(s_3)=2$ so: $$(1,2,3,4,5)(6,7,8)(9,10).$$ I don't know if it's possible to generalize it for $S_n$.


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