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This exercise asks to provide elements of order 10, 20, and 30 in $S_{10}$. Thinking that the order of a permutation $\sigma$ is the least common multiple of the length of the disjoint cycles whose product is $\sigma$, I went for the following:

$\sigma_1=(a_1a_2)(b_1b_2b_3b_4b_5)$ of order 10

$\sigma_2=(a_1a_2a_3a_4)(b_1b_2b_3b_4b_5)$ of order 20

$\sigma_2=(a_1a_2a_3)(b_1b_2b_3b_4b_5)(c_1c_2)$ of order 30

The question also asked if there could be a permutation of order 40. This is where I got stuck, mainly because I could not find any combination of cycle lengths which would give 40 as least common multiple. Because the order of $S_{10}=3628800$, and 40 divides that, I figured there would have to be an element of order 40, and much larger orders as well, which it seems to be impossible to obtain considering a permutation written as disjoint cycles. Help in finding the flaw in my reasoning would be appreciated.

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    $\begingroup$ I think you are right to consider disjoint cycles, and that there is no element of order 40. There are many of order 2, many-many of order 3 and so on. $\endgroup$ – Empy2 Dec 27 '14 at 17:51
  • $\begingroup$ You need at least 13 elements to write a permutation of order 40 $\endgroup$ – Andrea Mori Dec 27 '14 at 17:53
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Every permutation (of finitely many elements) can be written as a product of disjoint cycles.

$40$ is divisible by $8=2^3$ and the only way for a least common multiple to be divisible by a prime power is if one of the operands to the lcm is divisible by that same prime power.

So in order to make a permuation of order 40, we need at least one cycle whose length is a multiple of 8. There's not room for that in $S_{10}$ while also getting a factor of 5 from somewhere.

Your assumption that just because 40 divides the order of the group there must be an element of order 40 is not true.

For example, consider the gorup $C_2\times C_2\times C_2$. Its order is 8, but there is no element of order 4 -- all elements except the identity have order 2.

In fact your reasoning would mean that every group $G$ has an element of order $|G|$ (because $|G|$ divides itself!) and so is cyclic. This is obviously not true.

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  • $\begingroup$ I see, so that assumption holds but only for cyclic groups? Because I was quite convinced I'd seen that assumption I made somewhere before but might be mixing up things, I wanted to understand that. $\endgroup$ – Snowflake Dec 27 '14 at 17:58
  • $\begingroup$ @Snowflake: It is the converse of Lagrange's theorem: The order of an element is always a divisor of $|G|$, but not all divisors of $|G|$ are orders of elements). $\endgroup$ – hmakholm left over Monica Dec 27 '14 at 18:02

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