# Basic Question: Order of permutations in Sn

1. Dec 12, 2013

### ZZ Specs

Is there a theorem or any useful application for knowing the order of a permutation belonging to the symmetric group Sn?

For example,

Lets say σ is a permutation belonging to S5; i.e. σ is a permutation of {1,2,3,4,5}. If we are given that σ^7 = I (the identity permutation), then how can we show that necessarily σ = I ?

Thank you all for your time and help.

2. Dec 12, 2013

### jgens

Since 7 is prime this means either σ = id or S5 contains a cyclic group of order 7. The latter is impossible and so the desired conclusion follows.

3. Dec 12, 2013

### ZZ Specs

Thank you for your reply, much appreciated. Could you further explain what property shows that "either σ = id or S5 contains a cyclic group of order 7"?

4. Dec 12, 2013

### jgens

Basic number theory. The order of σ must divide 7.

5. Dec 12, 2013

### ZZ Specs

Is it valid to say that because σ is in S5, the order of sigma is at most 6?

6. Dec 12, 2013

### jgens

That is true, but it takes more work to establish than the argument I sketched above.

7. Dec 12, 2013

### ZZ Specs

In that case, why can we say that a cyclic group of order 7 is impossible? I was thinking that because σ lies in S5 (i.e. |σ| ≤ 6 ), σ^k must equal the identity for some k < 7.

Is there another way to establish this?

8. Dec 12, 2013

### ZZ Specs

Thank you very much for all your help though!

9. Dec 12, 2013

### jgens

Do you know Lagrange's Theorem?

10. Dec 12, 2013

### ZZ Specs

I believe so, that the order of a subgroup must divide the order of its group. I know that <σ> is a group (all cyclic elements form groups, correct?), but do we have any other groups to compare in this situation? I'm not used to working with permutations, especially in view of groups.

11. Dec 12, 2013

### jgens

In more detail: Since σ7 = id the order of <σ> is either 1 or 7. Since |S5| = 120 and 7 does not divide this number that means <σ> = {id}. Now the proof is done.

Last edited: Dec 12, 2013