Solving Permutation Groups: Odd Permutations Have Even Order

[VeeEight]

I have two questions, they aren't homework questions but I figured this would be the best place to post them (they are for studying for my exam).

Homework Statement

How many elements of S_6 have order 4? Do any elements have order greater than 7?

Homework Equations

S_6 is the permutation group on {1, 2,..,6}. The order, n, of an element here is a permutation f such that f^n = 1

The Attempt at a Solution

I figured that it wouldn't be wise to check all the elements since there are so many of them. I know the order of an element has to divide the order of the group. The order of S_6 is 6!, which seven does not divide so no elements have order 7 (but there must be an element of order 6 since 6 divides 6!). But there must be an element of order 9 since 9|6!. I am not sure if this logic is correct and how to actually find how many elements there are of each order in general.

Homework Statement

Prove that every odd permutation in S_n has even order

Homework Equations

An odd permutation is a permutation that can be written as a product of an odd number of transpositions (2-cycles). If a permutation is odd then it's sgn is -1

The Attempt at a Solution

My instinct is to incorporte the sgn function into the proof but I am unsure of how to use it.

 

[CompuChip]

I have two questions, they aren't homework questions but I figured this would be the best place to post them (they are for studying for my exam).

Homework Statement

How many elements of S_6 have order 4? Do any elements have order greater than 7?

Homework Equations

S_6 is the permutation group on {1, 2,..,6}. The order, n, of an element here is a permutation f such that f^n = 1

The Attempt at a Solution

I figured that it wouldn't be wise to check all the elements since there are so many of them. I know the order of an element has to divide the order of the group. The order of S_6 is 6!, which seven does not divide so no elements have order 7 (but there must be an element of order 6 since 6 divides 6!). But there must be an element of order 9 since 9|6!. I am not sure if this logic is correct and how to actually find how many elements there are of each order in general.

You are right about the first argument: since 7 does not divide 6! there are no elements of order 7. However, it is not true that for every divisor of 6! there must be an element with that order. For example, the group A4 (odd permutations in S4) does not have any element of order 6. In general, it's hard to say something about the number of subgroups that do exist; usually the Sylow theorems are useful in such cases.

Homework Statement

Prove that every odd permutation in S_n has even order

Homework Equations

An odd permutation is a permutation that can be written as a product of an odd number of transpositions (2-cycles). If a permutation is odd then it's sgn is -1

The Attempt at a Solution

My instinct is to incorporte the sgn function into the proof but I am unsure of how to use it.

What order does the identity have? What happens to the sign if you compose n odd permutations?

 

[Dick]

As far as counting elements of order 4, can you relate the order of an element to the cycle structure of the permutation in the case where it has order 4? Actually, that can probably help you with the 7 or greater question as well.

 

[VeeEight]

thanks! I believe I have the answer