Elements in only 1 left/right coset

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SUMMARY

Every element of a group G belongs to exactly one left coset and one right coset of a subgroup H. This is established through the equivalence relation defined by x~y if xy^-1 is in H, which partitions G into mutually disjoint subsets. If an element X belongs to two right cosets, Ha and Hb, it follows that Ha equals Hb. The proof involves demonstrating that the defined relation is an equivalence relation and that the equivalence classes correspond to the right cosets of H in G.

PREREQUISITES
  • Understanding of group theory concepts, specifically cosets and equivalence relations.
  • Familiarity with subgroups and their properties within a group.
  • Knowledge of the definition and properties of equivalence relations.
  • Basic grasp of isomorphism in the context of groups, particularly with (R/Z) and the unit circle.
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  • Study the properties of equivalence relations in group theory.
  • Learn about the structure and significance of cosets in abstract algebra.
  • Explore the concept of isomorphism and its applications in group theory.
  • Investigate the relationship between (R/Z) and the unit circle in complex analysis.
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Mathematics students, particularly those studying abstract algebra, group theory, and anyone interested in understanding the properties of cosets and equivalence relations in groups.

Wiemster
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My lecture notes state that every element of a group belongs to exactly one right and one left coset of a certain subgroup, but I don't see why this should be the case, so I tries to prove it:

with S and S' elements of the subgroup H of G we then have that if a certain element X of the group G belongs to the coset TH but to T'H as well

i.e. X=TS=T'S' we should have that T=T' right?

This means TS(S'^-1)=T'E with S(S'^-1) again an element of H, but I don't see why T should be T', or did I misintepreted the theorem?

(PS: Somewhat further they use as an example a generelaized torus: Rn / Zn = (R/Z)n which is supposed to b homomorphic to the unit circle. But I don't really see what this 'factor group' represents...:confused: )
 
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The right cosets of H in G are the equivalences classes of the equivalence relation ~ on G given by x~y iff xy^-1 in H. They are equivalence classes, so they partition G into a family of mutually disjoint nonempty subsets. So every element in G belongs to exactly 1 right coset(to exactly 1 equivalence class). So if x belongs to two right cosets say Ha and Hb, then it follows that Ha = Hb.

I just said this in words, didn't prove anything. If you want to prove it, you can,

1) Let R be an equivalence relation on a set S. Show every element of S is in exactly one equivalence class.

2) Show the relation defined, ~, is an equivalence relation on a group G(with H being a subgroup).

3) Show the equivalence classes of ~ are the right cosets of H in G.

Then it follows immediately.
 
Cosets are either equal or disjoint. It is a natural consequence of the definition of a group.

(R/Z) is isomorphic to the unit circle in C. (R/Z)^n is not.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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