1. ## Cycle Problem

Find the number of cycles of each possible length in $S_5$. Then find all possible orders of elements of in $S_5$. (Do this without writing out all 120 possible permutations.

I'm a little confused as to what it's asking. I know the order of a cycle is the length of that cycle. I'm guessing their asking how many 1 cycles are there, how many 2 cycles are there, how many 3 cycles, etc.

For the second part when you have to find all possible orders of the elements, I'd say it was the LCM of all the cycles. So my answer is: 60, 20, 15, 12, 10, 6, 5, 4, 3, 2, 1. Not sure if that's right though. Any help is appreciated.

2. Originally Posted by eXist
Find the number of cycles of each possible length in $S_5$. Then find all possible orders of elements of in $S_5$. (Do this without writing out all 120 possible permutations.

I'm a little confused as to what it's asking. I know the order of a cycle is the length of that cycle. I'm guessing their asking how many 1 cycles are there, how many 2 cycles are there, how many 3 cycles, etc.

For the second part when you have to find all possible orders of the elements, I'd say it was the LCM of all the cycles. So my answer is: 60, 20, 15, 12, 10, 6, 5, 4, 3, 2, 1. Not sure if that's right though. Any help is appreciated.

Indeed, not right because you're forgetting the disjoint thingy: the order of a permutation written as a product of DISJOINT CYCLES is the lcm of their lenghts.
Now try again to find out the possible orders of elements in S_5.

Tonio

3. Originally Posted by eXist
Find the number of cycles of each possible length in $S_5$. Then find all possible orders of elements of $S_5$. (Do this without writing out all 120 possible permutations.

I'm a little confused as to what it's asking. I know the order of a cycle is the length of that cycle. I'm guessing they're asking how many 1 cycles are there, how many 2 cycles are there, how many 3 cycles, etc.

For the second part when you have to find all possible orders of the elements, I'd say it was the LCM of all the cycles. So my answer is: 60, 20, 15, 12, 10, 6, 5, 4, 3, 2, 1. Not sure if that's right though. Any help is appreciated.
I can't quite remember if $(1 2 3)(4 5)$ would count as a cycle or not, but I will assume it will.

So, if we take any element of $S_5$ it will have one of 7 different disjoint cycle shapes (what are these?). Now, the order of an element is intricately linked to its cycle shape. So this question has two parts:

1) How many element have each disjoint cycle shape? Note that the sum of these should sum to 120.

2) What is the order of a cycle? For example, what is the order of $(1 2 3)(4 5)$?

Part two isn't too hard, but is a worthwhile exercise to sit down and think about. If you don't get it after a while, come back and ask.

Part one is slightly harder. Basically, you have to fit in the 5 numbers into each cycle such that you won't get the same cycle appearing twice. So, clearly we have no more than $5*4=20$ 2-cycles, as we have 5 choices for the first number and then 4 for the second number (the numbers cannot be the same). However, the order we enter elements does have some bearing: for instance, $(1 2)=(2 1)$. So, to get rid of such symmetry we must divide by 2. Similarly, in a 3-cycle you must divide by 3 as $(1 2 3)=(2 3 1)=(3 1 2)$. Now, note that in the cycles of the form $(12)(34)$ we have that $(12)(34)=(34)(12)$ so we have an extra degree of symmetry do deal with. Thus, we have to divide by two again.

That is, the number of cycles of the form $(12)(34)$ is $(5*4*3*2)*1/2*1/2*1/2=(5*4*3*2)/2^3=15$. One $1/2$ for the symmetry in the first part, one for the symmetry in the second, and one for the symmetry between the two.

Do you think you could finish the problem off now?

4. I understand part 2 now. If they are disjoint cycles then the only possible orders are: 1, 2, 3, 4, 5, and 6. And yes your right $(123)(45)$ counts and therefor the order of that cycle is 6.

For the first part, I think I have it:

1 Cycles = 1 (The identity cycle)
2 Cycles = 10 $= (\frac{5 * 4}{2})$
3 Cycles = 20 $= (\frac{5 * 4 * 3}{3})$
4 Cycles = 30 $= (\frac{5 * 4 * 3 * 2}{4})$
5 Cycles = 24 $= (\frac{5 * 4 * 3 * 2 * 1}{5})$

Swal I see what your saying, but we are only looking at single cycles I think, so something like $(12)(34)$ doesn't count, but that's just my thoughts.

5. Originally Posted by eXist
I understand part 2 now. If they are disjoint cycles then the only possible orders are: 1, 2, 3, 4, 5, and 6. And yes your right $(123)(45)$ counts and therefor the order of that cycle is 6.

For the first part, I think I have it:

1 Cycles = 1 (The identity cycle)
2 Cycles = 10 $= (\frac{5 * 4}{2})$
3 Cycles = 20 $= (\frac{5 * 4 * 3}{3})$
4 Cycles = 30 $= (\frac{5 * 4 * 3 * 2}{4})$
5 Cycles = 24 $= (\frac{5 * 4 * 3 * 2 * 1}{5})$
Yes, they are all correct. However, I suspect the question is also asking you the number of the other types of cycles as well, such as those of the form $(12)(345)$ and $(12)(34)$

6. Thanks for your help Swal. The book says that what I have is correct. Although I do see what you're saying.

7. Originally Posted by eXist
Thanks for your help Swal. The book says that what I have is correct. Although I do see what you're saying.
Okay. It's just a really bizarre, incomplete question without that bit!

(EDIT: You're welcome.)