Compute Right Coset of (1 2) in S3 X Z2

[FanofAFan]

Compute the right coset of ((1 2), [1])) in S₃ X Z₂.

Do I use the given as H (the subgroup) and find right cosets for just (1 2) and then coset for [1] separately. Basically I'm just confused on what to do.

 

[sephy]

Compute the right coset of ((1 2), [1])) in S₃ X Z₂.

Do I use the given as H (the subgroup) and find right cosets for just (1 2) and then coset for [1] separately. Basically I'm just confused on what to do.

The right coset would be the set of all elements {\alpha} such that \alpha=((1 2),[1])(p,n) where p is an element of S₃ and n is an element of Z₂. I would list the elements of S₃ X Z₂ (which would be like ((1 2 3),[2]),((1 2 3),[2])...- there should be 12 elements I believe?), and operate ((1 2), [1]) with each of them. Then those elements are the elements of the right coset of ((1 2),[1]).

Right? I'm just a group theory student myself... but that's how I would go about that.

 

[vela]

The question doesn't make sense to me. Isn't ((1 2), [1]) just an element of S₃xZ₂, not a subgroup? Is the question referring to the subgroup generated by that element?

 

[FanofAFan]

The question is referring to ((12), [1]) being the subgroup... I don't really know how to explain it further... hence the confusion.

@sephy, I worked it out the way you explain it

 

[vela]

It's no surprise you're confused then. You should ask your instructor for clarification.

 

[sephy]

The question doesn't make sense to me. Isn't ((1 2), [1]) just an element of S₃xZ₂, not a subgroup? Is the question referring to the subgroup generated by that element?

The element is its own inverse, so the subgroup generated by that element would just be itself and the identity element. Maybe you are meant to assume that?

Either that or the word "coset" is being mis-used- vela's right, it's not a coset unless it's made with a subgroup. Without the identity ((1 2), [1]) is just a subset of S₃xZ₂. You could still compute the elements of Ha, but Ha would not technically be considered a coset.