Proving the Subgroup Property of Even Permutations in Permutation Groups

  • Thread starter Thread starter PhysicsUnderg
  • Start date Start date
  • Tags Tags
    Groups Permutation
PhysicsUnderg
Messages
17
Reaction score
0

Homework Statement


Show that if G is any group of permutations, then the set of all even permutations in G forms a subgroup of G.

I am not sure where to start - I know there is a proposition that states this to be true, but I know that is not enough to prove this statement.
 
Physics news on Phys.org
Hi PhysicsUnderg! :smile:

Hint: if H is a subgroup of G, then for any a and b in H, the product ab must also be in H. :wink:
 
Is it really that simple? lol This is what I was thinking, but I wasn't sure how to connect the idea to permutations. Can I just say "a is an even permutation" and "b is an even permutation" thus "a*b is also an even permutation"? Because, if this is true and if I assume that a and b are elements of H, then ab is an element of H and is an even permutation, so G has a subgroup of even permutations. Also, for H to be a subgroup, the identity element must be contained in H, as well as an inverse. How do you connect this to permutations?
 
Hi PhysicsUnderg! :smile:
PhysicsUnderg said:
Is it really that simple?

Yes! :biggrin:

Sometimes, maths really is that simple! :wink:
… Also, for H to be a subgroup, the identity element must be contained in H, as well as an inverse. How do you connect this to permutations?

You ask "Is the identity an even permutation? What is the inverse of an even permutation?" :smile:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top