Homework: find right coset of a group

  • Thread starter Thread starter annoymage
  • Start date Start date
  • Tags Tags
    Group Homework
annoymage
Messages
360
Reaction score
0

Homework Statement



Let G be a group, H is subgroup of G, and [G:H]=2
find all the right and left coset of H in G

Homework Equations



n/a

The Attempt at a Solution



(finding right coset)

so there exist 2 distinct right coset, but how to find the 2 right coset?

let a,b in G

so Ha and Hb are the right coset if Ha\capHb={}

then?
 
Physics news on Phys.org


Pick an a in G that is not in H. Then the right cosets are H and Ha, right?
 


yes, and is that the answer?

let a in G not in H

H and Ha are the right coset of H in G
H and aH are the left coset of H in G
is that correct and sufficient?

what if i do like this

let ab-1 in G not in H

Ha and Hb are the right coset of H in G
aH and bH are the left coset of H in G
it's the same thing right?
 


Same thing, yes. But I don't know that it really helps you in any way. I think the point is that there is only one right coset that is not equal to H and there is only one left coset that is is not equal to H. So they must be equal. Isn't that the point?
 


yea, haha, like you said, i wanted to show Hx=xH for all x in G, thank you very much,
 
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