Proof for Alternating Group Order |An| = ½(n!)

[Gear300]

Changed to a different question:
Can anyone provide a proof for the order of alternating groups |A_n| = ½(n!)?

 

[matt grime]

It is almost immediate from the definition that it is a group of index 2 inside S_n.

 

[sutupidmath]

Changed to a different question:
Can anyone provide a proof for the order of alternating groups |A_n| = ½(n!)?

The way i like to show this goes something along these lines:

THe whole idea is to show that there are as many odd permutations as there are even permutation in Sn. So, let


$$S_n= ( \alpha_i,\beta_j)$$ which represents the set of even and odd permutations.

where $$\alpha_i, i=1,2,3,...,r$$ and $$\beta_j,j=1,2,3,...,k$$ are even and odd permutations respectively.

Now, let's consider the following:


$$\beta_1\beta_j, j=1,2,3,...,k$$ (there is some extra work here to show that all these elements are indeed unique, but it is not difficult to establish it. a proof by contradiction would work)

So, we know that the multiplication of odd permutations is an even permutation, so we know that in our set Sn, we have the following relation:

$$|k|\leq|r|------(1)$$

Now, consider the following:


$$\beta_1\alpha_i,i=1,2,...,r$$

so all these permutations now are odd. From this we get the following relation:

$$|r|\leq |k|-----(2)$$


From (1) &(2) we get the following:


$$|k|\leq |r| \leq |k|=>|k|=|r|$$

Which means that the number of even and odd permutations in Sn is equal.

Now, since |Sn|=n! => |An|=n!/2

 

[matt grime]

$$\beta_1\beta_j, j=1,2,3,...,k$$ (there is some extra work here to show that all these elements are indeed unique, but it is not difficult to establish it. a proof by contradiction would work)

There is no work in showing this: that multiplication by a group element is a bijection is almost the definition of a group.

Apart from that, your proof is 'the correct one': there exists an injection from the set of odd elements to even elements, and vice versa, hence they have the same cardinality (this has no dependence on their being a finite number of them, which is always nice).

 

[Gear300]

The way i like to show this goes something along these lines:

THe whole idea is to show that there are as many odd permutations as there are even permutation in Sn. So, let$$S_n= ( \alpha_i,\beta_j)$$ which represents the set of even and odd permutations.

where $$\alpha_i, i=1,2,3,...,r$$ and $$\beta_j,j=1,2,3,...,k$$ are even and odd permutations respectively.

Now, let's consider the following:$$\beta_1\beta_j, j=1,2,3,...,k$$ (there is some extra work here to show that all these elements are indeed unique, but it is not difficult to establish it. a proof by contradiction would work) ---------

...multiplication by a group element is a bijection is almost the definition of a group.

So, we know that the multiplication of odd permutations is an even permutation, so we know that in our set Sn, we have the following relation:

$$|k|\leq|r|------(1)$$

Now, consider the following:$$\beta_1\alpha_i,i=1,2,...,r$$

so all these permutations now are odd. From this we get the following relation:

$$|r|\leq |k|-----(2)$$From (1) &(2) we get the following:$$|k|\leq |r| \leq |k|=>|k|=|r|$$

Which means that the number of even and odd permutations in Sn is equal. ---------

...there exists an injection from the set of odd elements to even elements, and vice versa, hence they have the same cardinality...

Now, since |Sn|=n! => |An|=n!/2

I see. The proof was written in good clarity, so I'm thinking I got a good hold from it. Thanks.