# Finding the order of element of symmetric group

• Mr Davis 97
In summary: Well since the RHS consists of compositions of permutations that are disjoint with ##c_1##, the RHS will evaluate to ##a##. So ##c_1^r (a) = a##? And so ##c_1^r =...##Yes, ##c_1^r(a) = a##, so you have ##c_i^r(a)=a## for all ##i##.Yes, ##c_1^r(a) = a##, so you have ##c_i^r(a)=a## for all ##i##.I am not sure how to take it from here. I understand that each element of each cycle is fixed by the power of each cycle, so if an element
Mr Davis 97

## Homework Statement

Let ##n## be a natural number and let ##\sigma## be an element of the symmetric group ##S_n##. Show that if ##\sigma## is a product of disjoint cycles of orders ##m_1 , \dots , m_k##, then ##|\sigma|## is the least common multiple of ##m_1 , \dots , m_k##.

## The Attempt at a Solution

So if I suppose that ##\sigma = c_1 c_2 \dots c_k## where ##|c_i| = m_i##, and ##|\sigma| = n##, then ##\sigma ^n = c_1^n \dots c_k^n = 1##. Now, if I could show that this implies that ##c_i^n = 1## then I would basically be done, but I am not sure how to do this. It's kind of "obvious" to me since after taking powers we still have disjoint cycles, and also since they're disjoint none of the disjoint cycles can be inverses of other possible compositions of other disjoint cycles, so they must all be the identity.

Mr Davis 97 said:

## Homework Statement

Let ##n## be a natural number and let ##\sigma## be an element of the symmetric group ##S_n##. Show that if ##\sigma## is a product of disjoint cycles of orders ##m_1 , \dots , m_k##, then ##|\sigma|## is the least common multiple of ##m_1 , \dots , m_k##.

## The Attempt at a Solution

So if I suppose that ##\sigma = c_1 c_2 \dots c_k## where ##|c_i| = m_i##, and ##|\sigma| = n##, then ##\sigma ^n = c_1^n \dots c_k^n = 1##.
Here you use the fact that disjoint cycles commute, and that for every element ##g## of a finite group ##G## we have ##g^{|G|}=1\,.## But why does ##n## has to be the order of ##\sigma\,?## Conversely if you define ##n:=\operatorname{ord}\sigma ## you shouldn't use the same letter as for ##G=S_n\,.##
Now, if I could show that this implies that ##c_i^n = 1## then I would basically be done, but I am not sure how to do this. It's kind of "obvious" to me since after taking powers we still have disjoint cycles, and also since they're disjoint none of the disjoint cycles can be inverses of other possible compositions of other disjoint cycles, so they must all be the identity.
I would start with ##m := \operatorname{lcm}(m_1,\ldots,m_k)## and show ##\sigma^m=1##, and that for any other ##k##, that is one which isn't a multiple of (w.l.o.g.) ##m_1##, i.e. ##k=qm_1+r_1## has a remainder, ##\sigma^k \neq 1\,.##

fresh_42 said:
Here you use the fact that disjoint cycles commute, and that for every element ##g## of a finite group ##G## we have ##g^{|G|}=1\,.## But why does ##n## has to be the order of ##\sigma\,?## Conversely if you define ##n:=\operatorname{ord}\sigma ## you shouldn't use the same letter as for ##G=S_n\,.##

I would start with ##m := \operatorname{lcm}(m_1,\ldots,m_k)## and show ##\sigma^m=1##, and that for any other ##k##, that is one which isn't a multiple of (w.l.o.g.) ##m_1##, i.e. ##k=qm_1+r_1## has a remainder, ##\sigma^k \neq 1\,.##
Why does being any other ##k## imply that ##k## is not a multiple of ##m_1##?

Mr Davis 97 said:
Why does being any other ##k## imply that ##k## is not a multiple of ##m_1##?
You're right one has to be more careful here. Assume ##k<m## with ##\sigma^k=1.## Then what is ##c_i^k\,?##

fresh_42 said:
You're right one has to be more careful here. Assume ##k<m## with ##\sigma^k=1.## Then what is ##c_i^k\,?##
I don't see what I can say about ##c_i^k## in general. It would seem that it depends on which ##c_i## we're talking about. For example, say we have ##m = \operatorname{lcm} (2,3,5)= 30## where ##m_1=2,m_2=3,m_3=5##. Then, if we take ##k=2<30## ##c_1^2 = 1## but ##c_3^2 \ne 1##

Mr Davis 97 said:
I don't see what I can say about ##c_i^k## in general. It would seem that it depends on which ##c_i## we're talking about. For example, say we have ##m = \operatorname{lcm} (2,3,5)= 30## where ##m_1=2,m_2=3,m_3=5##. Then, if we take ##k=2<30## ##c_1^2 = 1## but ##c_3^2 \ne 1##
I meant, still assume a ##k## with ##\sigma^k=1## and ##k<m##. Compute it and see what it does with the numbers shuffled by ##c_i##. They are still disjoint, which allows to sort the powers by the ##c_i## and all other ##c_j\; , \;(i\neq j)## don't affect the numbers permuted by ##c_i##.

fresh_42 said:
I meant, still assume a ##k## with ##\sigma^k=1## and ##k<m##. Compute it and see what it does with the numbers shuffled by ##c_i##. They are still disjoint, which allows to sort the powers by the ##c_i## and all other ##c_j\; , \;(i\neq j)## don't affect the numbers permuted by ##c_i##.

So am I trying to prove that if the cycles are disjoint, then ##c_1^n c_2^n \dots c_k^n = 1## implies that ##c_i^n = 1##? This is what I'm seeing is necessary, and how I see it in most proofs online. And it seems obvious because the cycles are disjoint, but I don;t see how to make it rigorous...

The formula ##(c_1\ldots c_k)^r = c_1^r\ldots c_k^r## is of general interest. For this you only need to show ##c_1c_2=c_2c_1## and the rest is either by induction or a so forth argument. Given that, you have ##c_1^r\ldots c_k^r=1 \Longrightarrow c_1^r=c_{k}^{-r}\ldots c_2^{-r}##. Now evaluate both sides at an ##a\in c_1##.

fresh_42 said:
The formula ##(c_1\ldots c_k)^r = c_1^r\ldots c_k^r## is of general interest. For this you only need to show ##c_1c_2=c_2c_1## and the rest is either by induction or a so forth argument. Given that, you have ##c_1^r\ldots c_k^r=1 \Longrightarrow c_1^r=c_{k}^{-r}\ldots c_2^{-r}##. Now evaluate both sides at an ##a\in c_1##.
Well since the RHS consists of compositions of permutations that are disjoint with ##c_1##, the RHS will evaluate to ##a##. So ##c_1^r (a) = a##? And so ##c_1^r = 1##?

Yes. That's the idea. And thus ##m_1\,|\,r##. And at last note that you can do this for any index thanks to commutativity.

Mr Davis 97
fresh_42 said:
Yes. That's the idea. And thus ##m_1\,|\,r##. And at last note that you can do this for any index thanks to commutativity.
Alright, I think I see it all coming together. Last thing: Suppose that we have the disjoint cycles ##c_1, \dots, c_k##. Do I need to show that ##\forall r \in \mathbb{Z}## ##c_1^r ,\dots, c_k^r## are still disjoint? Once again, it seems obvious since when you compose a permutation with itself, numbers can leave the cycle but additional ones will never be added, so all the permutation remain disjoint.

Mr Davis 97 said:
Alright, I think I see it all coming together. Last thing: Suppose that we have the disjoint cycles ##c_1, \dots, c_k##. Do I need to show that ##\forall r \in \mathbb{Z}## ##c_1^r ,\dots, c_k^r## are still disjoint? Once again, it seems obvious since when you compose a permutation with itself, numbers can leave the cycle but additional ones will never be added, so all the permutation remain disjoint.
They cannot even leave the cycle, they are only shifted by ##r## places. Again, this depends on how deep you want to get. I'd say a remark will do. At a closer look, you don't need commutativity after they are arranged as ##c_1^r\ldots c_k^r##, which I mistakenly stated. It's only used for this arrangement. And thus I don't think you need that they are still disjoint, only that numbers of one cycle aren't touched by any other.

## 1. What is the symmetric group?

The symmetric group, denoted as Sn, is a mathematical group consisting of all possible permutations of n distinct elements or objects. In simpler terms, it is the group of all possible ways that n objects can be arranged or ordered.

## 2. How many elements are in the symmetric group?

The number of elements in the symmetric group Sn is equal to n!, where n represents the number of distinct elements. For example, the symmetric group S3 has 3! = 6 elements.

## 3. How do you find the order of an element in the symmetric group?

The order of an element in the symmetric group is equal to the number of times the element needs to be applied to itself to return to its original position. For example, in the symmetric group S3, the element (1 2) has an order of 2 because applying it twice results in the original permutation (1 2).

## 4. Can an element in the symmetric group have infinite order?

No, the order of an element in the symmetric group is always finite. This is because there are only a finite number of possible permutations of a finite number of elements.

## 5. How is the order of an element in the symmetric group related to its cycle structure?

The order of an element in the symmetric group is equal to the least common multiple of the lengths of its disjoint cycles. For example, in the symmetric group S4, the element (1 2)(3 4) has an order of 2 because the lengths of its disjoint cycles are 2 and 2, and the least common multiple of 2 and 2 is 2.

### Similar threads

• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
864
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
930
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Linear and Abstract Algebra
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K