Math Is Fun Forum

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Quick question: alternating group

I was reading some source on the internet and they mentioned that all the elements of A5 (even permutations of S5) have one of the following forms:

(i) 3-cycles

(ii) products of two mutually disjoint transpositions

(iii) 5-cycles

However, they do not explain why this is so. What is the reason behind this?

Thanks.

Last edited by LuisRodg (2009-03-28 16:50:15)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Quick question: alternating group

This is most likely case division.  Perhaps a clever argument exists, but dividing it up into cases will do.  For permutations in S_5, you have fairly limited choices.  Write your permutation in a disjoint fashion and so you have:

2, 3, 4, 5 cycles

And combinations of them:

2 and 2 (i.e. (12)(34) )
2 and 3 (i.e. (12)(345) )

Remember that it is fairly easy to go from disjoint cycles to a product of transpositions.  For example:

(a b c d) = (a d) (a c) (a b)

Now just eliminate all the ones except those given in the decomposition.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Quick question: alternating group

Execellent!

Thanks.

Edit:

Can A5 contain a 4-cycle or a 2-cycle? Given alpha a 4-cycle or a 2-cycle then its signum will be -1 and hence it is an odd permutation, not in A5?
Also, it cannot contain a combination of 2 and 3 such as (12)(345) because of the above argument as well?

Last edited by LuisRodg (2009-03-29 06:25:41)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Quick question: alternating group

Correct.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Quick question: alternating group

So then by that reasoning it reduces to the first 3 forms I gave. Nice.

Thanks.

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Quick question: alternating group

I would make the small complaint that the identity does not have one of the given forms.

In general, the problem is to find the partitions of n with an even number of even parts.