Permutation \psi in S_{10} of Order 20: Odd or Even?

[ElDavidas]

Here's another question I'm lost on:

"Show that the permutation [itex]\psi [/itex] contained in [itex]S_{10}[/itex] of order 20 is odd"

I know that the order of the permutation is the least positive integer n such that [itex]\psi^n [/itex] is the identity permutation.

I don't know how to check whether the permutation is odd. Even if I did, I still doubt I'd know what to do from there on using the information I've got.

Thanks again.

 

[fourier jr]

I don't know how to check whether the permutation is odd. Even if I did, I still doubt I'd know what to do from there on using the information I've got.

Thanks again.

every cycle (& therefore every permutation) can be expressed as a product of transpositions (2-cycles). if a permutation can be expressed as an even number of transpositions it's called even; otherwise it's called an odd permutation. in this problem, since the order of the permutation is 20 and every permutation can be expressed as a product of disjoint cycles, and the order of a cycle is its length, I'm guessing that 20 is the lcm of the orders of the cycles. if that's the case i think it's possible to figure out how many cycles & their possible lengths (won't be longer than 10) and then figure out how many transpositions are necessary to express the cycles as products of transpositions. i think that will work, but it seems long. it' smore typing than i usually do anyway.

 

[ElDavidas]

every cycle (& therefore every permutation) can be expressed as a product of transpositions (2-cycles).

I'm not quite sure how to do that for a permutation. It could just be my general ignorance. Suppose for example that a permutation is:

1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 1 3 5 7 9

How do you write this as a product of transpositions?

(Sorry about the bad notation, the permutation is meant to be in Cauchy form)

 

[fourier jr]

I'm not quite sure how to do that for a permutation. It could just be my general ignorance. Suppose for example that a permutation is:

1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 1 3 5 7 9

How do you write this as a product of transpositions?

(Sorry about the bad notation, the permutation is meant to be in Cauchy form)

written as one cycle (rather than the 2-line notation) it looks like
(1 2 4 8 5 10 9 7 3 6)

& as transpositions
(1 2 4 8 5 10 9 7 3 6) = (1 6)(1 3)(1 7)(1 9)(1 10)(1 5)(1 8)(1 4)(1 2)

can you see the pattern? that isn't a unique way of doing it but it doesn't really matter. the important thing is that the parity (even/odd) of a permutation won't change no matter how you write it down.
edit: count the transpositions & see if you get an even or odd number

 

[AKG]

A better thing to note is that every permutation can be written as a product of disjoint cycles. Since the cycles are disjoint, they commute, and so if you express a permutation as a product of disjoint cycles, the order of that permutation is the least common multiple of the orders of the cycles. So you can see that a permutation of order 20 from S₁₀ has only one possible cycle structure. Moreover, the parity of a cycle is the opposite of the parity of its length. In fact, so far this thread has been about decomposing a 10-cycle into a product of transpositions. If you can figure out the pattern, you can see that in general an n-cycle has a decomposition into a product of n-1 transpositions, and thus has the parity opposite to its length.

EDIT: I see Fourier jr already said this.

 

[ElDavidas]

A better thing to note is that every permutation can be written as a product of disjoint cycles.

I understand this. So the above example would be written as
(1 2 4 8 5 10 9 7 3 6) and this commutes.

But (1 6)(1 3)(1 7)(1 9)(1 10)(1 5)(1 8)(1 4)(1 2) doesn't commute then.

Since the cycles are disjoint, they commute, and so if you express a permutation as a product of disjoint cycles, the order of that permutation is the least common multiple of the orders of the cycles.

I can follow this, although I'm not entirely sure why this is true.

So you can see that a permutation of order 20 from S10 has only one possible cycle structure

I can't follow this.

Moreover, the parity of a cycle is the opposite of the parity of its length.

Again, I think I can accept this.

Please bear with me on this one!

 

[matt grime]

I understand this. So the above example would be written as
(1 2 4 8 5 10 9 7 3 6) and this commutes.

this makes no sense. something can only commute with *something else*, or not. it doesn't make sense to ask if x commutes at all.

[quoteBut (1 6)(1 3)(1 7)(1 9)(1 10)(1 5)(1 8)(1 4)(1 2) doesn't commute then.[/quote[

ditto.

 

[AKG]

I understand this. So the above example would be written as
(1 2 4 8 5 10 9 7 3 6) and this commutes.

But (1 6)(1 3)(1 7)(1 9)(1 10)(1 5)(1 8)(1 4)(1 2) doesn't commute then.

The point is that these transpositions don't commute with one another. The permutation of S₅ (1 2)(3 4 5) is the exact same as (3 4 5)(1 2). However, you can decompose this into transpositions and get (1 2)(3 5)(3 4), but it's not the same as (1 2)(3 4)(3 5).

I can follow this, although I'm not entirely sure why this is true.

Figure it out.

I can't follow this.

It's very easy to prove, but it's not obvious. You'll have to do a little thinking on your own, but really not that much. Just figure it out yourself, it's easy.

Again, I think I can accept this.

Don't just accept it, prove it. It's easy since this thread has already explained to you how to decompose a cycle into a product of transpositions.