What Are the Right and Left Cosets in Various Groups?

[hsong9]

Homework Statement

In each case find the right and left cosets in G of the subgroups H and K of G.
a) G = A₄; H = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}, K = <(1 2 3)>
b) G= Z₁₂; H = 3Z₁₂, K = 2Z₁₂
c) G = D₄ = D₃ = {1,a,a²
, a³, b, ba, ba², ba³}, |a|=4,|b|=2, and aba = b; H =<a²>, K=<b>
d) G=any group, H is any subgroup of index 2

Homework Equations

A₄ is alternating group degree 4

The Attempt at a Solution

a) H(1 2 3) = {(1 2 3),(2 4 3),(1 4 2),(1 3 4)} = (1 2 3)H
H(1 3 2) = {(1 3 2),(1 4 3),(2 3 4),(1 2 4)} = (1 3 2)H
I know this is correct, but I don't know why
H(1 2 4) = {(1 2 4),(2 3 4),(1 4 3),(1 3 2)}...
H(1 4 2) = {...} are not cosets of G.

b) H + 0 = {3k | k in Z₁₂ }
H + 1 = {3k+1| k in Z₁₂ }
H + 2 = {3k+2 | k in Z₁₂ }
correct??

c) Ha = {1,a,a², a³, b, ba, ba², ba³}
aH = {{1,a,a², a³, b, ab, a²b, a³b}
K1 = {a, ab, a²b, a³b, b, ba²b, ba³b}
1K = {{1,a,a², a³, b, ba, ba², ba³}
Correct?

d) This is just Ha and aH ?? I'm not sure about this quesiton.

 

[mighty2000]

Thank you for your post. I am happy to help you with your question.

a) Your solution for finding the cosets of H and K in A4 is correct. This is because A4 is a subgroup of the symmetric group S4, and the cosets of a subgroup in a symmetric group are equal to the left and right cosets of the subgroup in the group.

b) Your solution for finding the cosets of H and K in Z12 is also correct. In this case, H is the subgroup of all multiples of 3 in Z12, and K is the subgroup of all multiples of 2 in Z12. So, for example, H + 0 is the set of all multiples of 3, which is a subgroup of Z12.

c) Your solution for finding the cosets of H and K in D4 is also correct. In this case, H is the subgroup generated by a2, and K is the subgroup generated by b. So, for example, Ha is the set of all elements in D4 that can be written as a*b^n for some n, and aH is the set of all elements in D4 that can be written as b^n*a for some n.

d) For this question, you are correct that the cosets of H and K in G will just be Ha and aH. This is because the index of a subgroup in a group is equal to the number of distinct left (or right) cosets of the subgroup in the group. So, since the index of H in G is 2, there can only be two distinct left (or right) cosets of H in G, which are Ha and aH.

I hope this helps. Good luck with your studies!