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Given the following groups, what is the maximum possible order for an element for

(a) $S_5$ (b) $S_6$ (c) $S_7$ (d) $S_{10}$ (e) $S_{15}$

My book justifies the answer as

(a) The greatest order is $6$ and comes from a product of disjoint cycles of length 2 and 3

<p>(b) The greatest order is $6$ and comes from a cycle of length $6$</p>

The other answers were justified exactly the same way, that is (c) 12, (d) 30, (e) 105

I do not understand how in (a) we even got the number "6" from $S_5$ and what disjoint cycles they are referring to. Could someone at least justify one for me?

Lemon
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    The order of a product of disjoint cycles is the lcm of the order of the individual cycles which is where the 6 comes from. $\operatorname{lcm}(2,3)=6$ To get the max element, you want to max $\operatorname{lcm}(|\sigma_1||\sigma_2|\ldots|\sigma_n|)$ such that $\sum{|\sigma_i|}=n$ in $S_n$. – Jemmy Nov 07 '12 at 03:38
  • How we can finde individual cycles in $S_n$? – erfan soheil Jan 29 '15 at 16:31
  • See https://math.stackexchange.com/questions/221211/maximal-order-of-an-element-in-a-symmetric-group – Gerry Myerson Oct 31 '17 at 04:16

2 Answers2

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You will have to write out the possible forms a given permutation (expressed as the product of disjoint cycles) can take, and then use the convenient fact that for disjoint cycles $\sigma_{1}, \dots, \sigma_{k} \in S_{n}$,

$$|\sigma_{1} \dots \sigma_{k}| = \textrm{lcm}(|\sigma_{1}|, \dots, |\sigma_{k}|).$$

For example, in $S_{5}$, you have (up to isomorphism) the following forms that a given permutation (written as the product of disjoint cycles) can take:

  1. $(1 2 3 4 5)$
  2. $(1 2 3)(4 5)$
  3. $(1 2 3 4)$
  4. $(1 2)(34)$
  5. $(1 2 3)$
  6. $(1 2)$

Then figure out which of the above forms will have the greatest order.

There is a sequence of values (of Landau's function, $g(n)$) that you can refer to for many values of $n$.

There is a known upper bound on the function:

$$g(n) < e^{n/e}.$$

A0000793: Landau's function g(n): largest order of permutation of n elements, Equivalently, largest lcm of partitions of n.

Charles Boyd
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Consider the permutation $p = (1 2)(3 4 5)$. It is an element of $S_5$, but it has order 6. The "disjoint cycles" are $(1 2)$ and $(3 4 5)$, which have lengths of 2 and 3, respectively.

If you don't understand the "cycle notation" $(1 2)(3 4 5)$ leave a comment and I will explain further. The short version is that $(1 2)(3 4 5)$ is the permutation which sends $1\mapsto 2$, $2\mapsto 1$, $3\mapsto 4$, $4\mapsto 5$, and $5\mapsto 3$.

MJD
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  • No my question is do they expect you to come up with your own cycles? – Lemon Nov 07 '12 at 03:35
  • They do, but it's easy to do that. – MJD Nov 07 '12 at 03:36
  • OKay maybe a better question would be, "what is the correct approach or thinking to solving to this problem?" instead of asking you to explain the solution to me. – Lemon Nov 07 '12 at 03:37
  • It's easy to figure out the order of a permutation once you know its cycle structure. There aren't that many different cycle structures. It's not hard to check all the possible cycle structures for elements in $S_6$ and calculate the order that elements with that structure will have. For example, any element of $S_3$ must either look like $(1)(2)(3)$, with order 1, or like $(1 2)(3)$, with order 2, or like $(1 2 3)$, with order 3. So elements of $S_3$ have order at most 3. – MJD Nov 07 '12 at 03:40
  • So you basically just lised all the possible products out (at least the possible orders)? They aren't using some shortcuts I'll learn in a few months right? – Lemon Nov 07 '12 at 03:46
  • They know a trick, which is the trick Jeremy mentioned in his comment. But it doesn't save much much work for these small cases, and by working out the small cases the long way you're likely to figure out the trick by yourself anyway. – MJD Nov 07 '12 at 03:47
  • No it wasn't Jeremy's trick I mentioned. I am referring to the big case like $S_{15}$, surely they aren't writing out all the possible maximum orders of disjoint sets now are they? – Lemon Nov 07 '12 at 03:53
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    Work out some small cases and you'll start to see how to shortcut the big cases. Try $S_4$, then $S_5$. – MJD Nov 07 '12 at 03:58
  • So the moral of this question is "there is just no shortcut"? – Lemon Nov 07 '12 at 04:03
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    I don't know where you got that from. I said if you work out some small cases you'll start to see how to shortcut the big cases. How could that be true if there was no shortcut? – MJD Nov 07 '12 at 04:04