Proof: Cosets Partition the Group | Abstract Algebra

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We prove, for a group G with subgroup H, the family of cosets Ha as a ranges over G, forms a partition of G. This means any two cosets Ha and Hb will be disjoint or equal, and also that every element of G is in a coset Ha. This is a significant but easily proven result. #abstractalgebra #grouptheory

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And the drawings are great.

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m.caeben

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sumittete

Hi !! Love your content, please keep it up. Especially your series on Graph Theory helped me a lot! Speaking of that, I would like to request a proof in Graph Theory as well : “every planar graph can be expressed as a union of five edge-disjoint forests”. Thanks!

SaiyaraLBS

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ssh

Hi, I have a question. In the proof that every element of G is in some Ha, couldn't the statement ec=c∈Hc imply that e∈Ha for all a∈G ?
I mean, I'm convinced by the other video that e∈Ha for one single coset, but this last proof gave me that idea and now I'm a bit confused 🙁

golden_smaug

02:43 Why can't one just say that if x ∈ (Ha ∩ Hb) then x ∈ Ha and x ∈ Hb, and therefore Hx = Ha = Hb?

SimchaWaldman

Wouldn't this be also proving why the Quotient Group of H/G partitions G?

themariobros

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