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

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SUMMARY

The discussion focuses on computing the right coset of the element ((1 2), [1]) in the group S3 X Z2. Participants clarify that ((1 2), [1]) is not a subgroup but rather an element of S3 X Z2, and the right coset can be computed by applying the element to all members of the group. The correct approach involves listing all 12 elements of S3 X Z2 and performing the operation with ((1 2), [1]) to find the resulting set. The confusion arises from the terminology, as the term "coset" is misapplied without a proper subgroup context.

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  • Familiarity with the symmetric group S3 and the group Z2.
  • Knowledge of the operation in the direct product of groups.
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  • Study the properties of the symmetric group S3, including its elements and structure.
  • Learn about the direct product of groups, specifically S3 X Z2.
  • Explore the concept of cosets and their relationship to subgroups in group theory.
  • Investigate the subgroup generated by an element and its implications in group operations.
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FanofAFan
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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.
 
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FanofAFan said:
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.
 
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 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
 
Last edited:
It's no surprise you're confused then. You should ask your instructor for clarification.
 
vela said:
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.
 

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