Prove that the order of An is (1/2)n!

For each even element there is a unique odd element.
That is, there is a 1-1 relation between the two (a "bijection").
So the number of even elements must be equal to the number of odd elements.

 

one way to see this is using the "sign" function:

sgn(p) = 1, if p is even
sgn(p) = -1, if p is odd.

this is, in fact a homomorphism from Sn to {-1,1}, which is cyclic of order 2.

even more nicely, this homomorphism is onto, which means that {-1,1} ≅ Sn/ker(sgn).

and, ker(sgn) = {even permutations} = An.

so we know that |Sn/An| = 2.

can you do the rest?

 

Ah, forget 1-1. It's true, but it makes it more complex.


An element is either even or odd.

Take the mapping from the even permutations to odd permutations, given by a → (1 2)a.
Each even element is mapped onto a (different) odd element.
So there are at least as many odd elements as even elements.

Similarly take the mapping from the odd permutations to the even permutations, given by b → (1 2)b.
Each odd element is mapped onto a (different) even element.
So there are at least as many even elements as odd elements.

Therefore there must be the same number of even as odd elements.

 

then you'll have to use I like Serena's method.

the "hard part" is showing that p ---> (1 2)p is a bijection between the odd permutations and even permutations.

 

(1 2) is an odd permutation.

An even permution is one that can be written as the product of an even number of transpositions.
If we add another transposition (an odd permutation), we get the product of an odd number of transpositions.

 

suppose q is any even permutation. we want to find some odd permutation p, such that:

(1 2)p = q.

this will show that p ---> (1 2)p is onto, and since Sn is finite, a surjection on a finite set is bijective.

well, if q is even, then (1 2)q is odd (adding another transpositon changes the parity, or sign).

so we can choose as our odd permutation p = (1 2)q:

(1 2)p = (1 2)(1 2)q = q.

 

You have it right. Even length means odd permution.
Think of (1 2) which has an even length, but is only 1 transposition (odd!).

Perhaps I should remark that the number of transpositions is not fixed.
An odd permutation can be written as any odd number of permutations (with a certain minimum).
For instance (1 2)=(1 2)(1 2)(1 2).

Very sharp!

Suppose a and b are different odd permutations that map to the same even permutation.
So (1 2)a = (1 2)b.
Then (1 2)(1 2)a = (1 2)(1 2)b, that is, a=b, which is a contradiction.
So each odd permutation maps onto a different even permutation.

 

Thanks!

Actually, I'm a software engineer who doesn't do math or physics (that's why I'm so active on PF! ).

But in my spare time I'm a private tutor.
Currently I have a student who is already a math teacher, but who needs to learn algebra to upgrade her level.
I also have a number of psychology students who have difficulty learning statistics.

 

Not really a coincidence.
I also do mechanics, electronics, linear algebra, differential equations, etcetera.
I'm currently (re)practicing/learning thermodynamics on PF.
And every now and then I try my hand at chemistry, but I'm a bit scared someone (Borek?) will find out I'm not so good at it.

I saw you were helping people on PF as well!

I'll look forward to seeing you with diffyQ.